Acid-Base Equilibria
Introduction to Acid-Base Equilibria
Acid-base equilibria involve the balance between acids and bases in a solution and the reversible dissociation of acids and bases in water. This concept is fundamental to understanding solution chemistry, pH, buffers, and the behavior of strong and weak acids and bases.
In acid-base equilibria, an acid donates a proton (H\(^+\)) to a base, establishing an equilibrium state where the concentrations of the acid, base, and their conjugates remain constant. The position of equilibrium depends on the strength of the acid or base and can be quantitatively described using dissociation constants.
Theories of Acids and Bases
1. Arrhenius Theory:
- Acid: A substance that releases H\(^+\) ions in aqueous solution.
- Base: A substance that releases OH\(^-\) ions in aqueous solution.
2. Bronsted-Lowry Theory:
- Acid: A proton (H\(^+\)) donor.
- Base: A proton (H\(^+\)) acceptor.
- This theory applies to both aqueous and non-aqueous solutions and provides a broader definition than the Arrhenius theory.
3. Lewis Theory:
- Acid: An electron pair acceptor.
- Base: An electron pair donor.
- This theory explains acid-base behavior in reactions that do not involve protons, such as the interaction between BF\(_3\) and NH\(_3\).
Strong vs. Weak Acids and Bases
1. Strong Acids and Bases:
- Strong acids (e.g., HCl, H\(_2\)SO\(_4\)) and strong bases (e.g., NaOH, KOH) completely dissociate in water, producing a high concentration of ions. The equilibrium for strong acids and bases lies far to the right, as they dissociate fully.
2. Weak Acids and Bases:
- Weak acids (e.g., acetic acid, CH\(_3\)COOH) and weak bases (e.g., NH\(_3\)) only partially dissociate in water, establishing an equilibrium between the undissociated molecules and ions. This equilibrium position is closer to the reactants, indicating only partial ionization.
Dissociation Constants and pH
1. Acid Dissociation Constant (K\(_a\)):
- The acid dissociation constant \(K_a\) measures the strength of a weak acid:
\[
K_a = \frac{[H^+][A^-]}{[HA]}
\]
- Higher \(K_a\) values indicate stronger acids (greater dissociation), while lower values indicate weaker acids.
2. Base Dissociation Constant (K\(_b\)):
- The base dissociation constant \(K_b\) measures the strength of a weak base:
\[
K_b = \frac{[B^+][OH^-]}{[BOH]}
\]
- Higher \(K_b\) values indicate stronger bases.
3. pH and pOH:
- pH is a logarithmic measure of the H\(^+\) ion concentration:
\[
pH = -\log[H^+]
\]
- pOH is a logarithmic measure of the OH\(^-\) ion concentration:
\[
pOH = -\log[OH^-]
\]
- For aqueous solutions at 25°C, \( pH + pOH = 14 \).
4. Relation between K\(_a\) and K\(_b\):
- For a conjugate acid-base pair:
\[
K_a \cdot K_b = K_w
\]
- Where \(K_w\) is the ion-product constant for water, equal to \(1.0 \times 10^{-14}\) at 25°C.
Buffer Solutions
Buffer solutions are mixtures that resist changes in pH upon the addition of small amounts of acids or bases. Buffers consist of a weak acid and its conjugate base (e.g., acetic acid and sodium acetate) or a weak base and its conjugate acid (e.g., ammonia and ammonium chloride).
1. Buffer Action:
- When H\(^+\) is added, the conjugate base neutralizes it, preventing a significant pH change.
- When OH\(^-\) is added, the weak acid neutralizes it, maintaining pH stability.
2. Henderson-Hasselbalch Equation:
- The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:
\[
pH = pK_a + \log \frac{[A^-]}{[HA]}
\]
- This equation is essential for designing buffers with a specific pH.
Le Chatelier's Principle in Acid-Base Equilibria
According to Le Chatelier's Principle, if an equilibrium system is disturbed, it will shift to counteract the disturbance:
- Adding more H\(^+\) ions (e.g., from an acid) shifts the equilibrium toward the reactants, reducing H\(^+\) concentration.
- Adding OH\(^-\) ions (e.g., from a base) shifts the equilibrium toward the products, reducing OH\(^-\) concentration and potentially producing more H\(^+\) ions.
This principle explains how buffers maintain pH stability and how acid-base equilibria respond to changes in concentration.
Applications of Acid-Base Equilibria
1. Biological Systems:
- Acid-base equilibria are crucial in regulating the pH of blood and other bodily fluids. Buffers like the bicarbonate buffer system help maintain a stable pH necessary for physiological functions.
2. Industrial Processes:
- Precise pH control is essential in industries like pharmaceuticals, where pH affects drug stability and activity.
3. Environmental Chemistry:
- Understanding acid-base equilibria is essential in addressing environmental issues like acid rain and the impact of CO\(_2\) on ocean acidity.
4. Analytical Chemistry:
- Acid-base equilibria play a vital role in titrations, where acids or bases of known concentration are used to determine the concentration of an unknown solution.
Sample Calculation in Acid-Base Equilibria
Example: Calculate the pH of a 0.1 M acetic acid solution with \(K_a = 1.8 \times 10^{-5}\).
1. Set up the dissociation:
\[
CH_3COOH \rightleftharpoons H^+ + CH_3COO^-
\]
2. Use \(K_a\) to find \([H^+]\) and then calculate pH:
\[
K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}
\]
Solve for \([H^+]\) and use \( pH = -\log[H^+]\).
Conclusion
Acid-base equilibria are fundamental in understanding the behavior of acids, bases, and buffers in solution. By studying dissociation constants, pH, and buffer systems, chemists can predict and control the pH of solutions, which is crucial in fields like biology, industry, and environmental science. Acid-base equilibria provide valuable insights into the dynamic balance of reactions involving proton transfer, enabling precise pH control in various applications.
|
Q 1. Boric acid is an acid because its molecule; |
|
(D) combines with proton from water molecule; |
|
(C) accepts \(\mathrm{OH}^{-}\)from water releasing proton.; |
|
(B) gives up a proton.; |
|
(A) contains replaceable \(\mathrm{H}^{+}\)ion.; |
|
Q 2. When a buffer solution, sodium acetate and acetic acid is diluted with water; |
|
(D) \(\mathrm{H}^{+}\)ion concentration remains unaltered; |
|
(C) \(\mathrm{OH}^{-}\)ion conc. Increases; |
|
(B) \(\mathrm{H}^{+}\)ion concentration increases; |
|
(A) Acetate ion concentration increase; |
|
Q 3. Boric acid is an acid because its molecule; |
|
(D) combines with proton from water molecule; |
|
(C) accepts \(\mathrm{OH}^{-}\)from water releasing proton.; |
|
(B) gives up a proton.; |
|
(A) contains replaceable \(\mathrm{H}^{+}\)ion.; |
|
Q 4. When a buffer solution, sodium acetate and acetic acid is diluted with water; |
|
(D) \(\mathrm{H}^{+}\)ion concentration remains unaltered; |
|
(C) \(\mathrm{OH}^{-}\)ion conc. Increases; |
|
(B) \(\mathrm{H}^{+}\)ion concentration increases; |
|
(A) Acetate ion concentration increase; |
|
Q 5. Which of the following is/are electrolytes? (i) Sugar solution (ii) Sodium chloride (iii) Acetic acid (iv) Starch solution; |
|
(d) (i) and (iii); |
|
(c) (ii) and (iii); |
|
(b) (ii) and (iv); |
|
(a) (i) and (iv); |
|
Q 6. Which of the following statements are correct regarding Arrhenius theory of acid and base?; |
|
(d) Both (a) and (c); |
|
(c) This theory could not explain the basicity of substances like ammonia which do not possess a hydroxyl group; |
|
(b) This theory was applicable to all solutions; |
|
(a) This theory was applicable to only aqueous solutions; |
|
Q 7. Would gaseous \(\mathrm{HCl}\) be considered as an Arrhenius acid ?; |
|
(a) Yes; |
|
(b) No; |
|
(c) Not known; |
|
(d) Gaseous \(\mathrm{HCl}\) does not exist; |
|
Q 8. \(\mathrm{BF}_{3}\) is an acid according to; |
|
(a) Arrhenius concept; |
|
(b) Bronsted-Lowry concept; |
|
(c) Lewis Concept; |
|
(d) Both (b) and (c); |
|
Q 9. Which of the following can act as both Bronsted acid and Bronsted base?; |
|
(a) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\); |
|
(b) \(\mathrm{OH}^{-}\); |
|
(c) \(\mathrm{HCO}_{3}^{-}\); |
|
(d) \(\mathrm{NH}_{3}\); |
|
Q 10. Conjugate acid of \(\mathrm{NH}_{2}^{-}\)is :; |
|
(a) \(\mathrm{NH}_{4}^{+}\); |
|
(b) \(\mathrm{NH}_{3}\); |
|
(c) \(\mathrm{NH}_{2}\); |
|
(d) \(\mathrm{NH}\); |
Law of Chemical Equilibrium
Definition of the Law of Chemical Equilibrium
The Law of Chemical Equilibrium states that at a given temperature, a chemical reaction reaches a state of balance in which the concentrations of reactants and products remain constant over time. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, leading to a dynamic but stable balance. The concentrations of reactants and products at equilibrium are related by a constant known as the equilibrium constant (K), which is specific to each reaction at a given temperature.
For a general reversible reaction:
\[
aA + bB \rightleftharpoons cC + dD
\]
The equilibrium constant \(K_c\) for this reaction in terms of concentration is given by:
\[
K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}
\]
Where:
- \([A]\), \([B]\), \([C]\), and \([D]\) are the molar concentrations of the reactants and products at equilibrium,
- \(a\), \(b\), \(c\), and \(d\) are the stoichiometric coefficients.
Types of Equilibrium Constants
1. Equilibrium Constant in Terms of Concentration (K\(_c\)):
- When the equilibrium concentrations of reactants and products are expressed in molarity (mol/L), the equilibrium constant is denoted as \(K_c\).
2. Equilibrium Constant in Terms of Partial Pressure (K\(_p\)):
- For gaseous reactions, the equilibrium constant can also be expressed in terms of the partial pressures of gases, denoted as \(K_p\):
\[
K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}
\]
- \(K_p\) and \(K_c\) are related by the equation:
\[
K_p = K_c (RT)^{\Delta n}
\]
where \(R\) is the gas constant, \(T\) is the temperature in Kelvin, and \(\Delta n\) is the difference in moles of gaseous products and reactants (\( \Delta n = \text{moles of products} - \text{moles of reactants} \)).
Characteristics of Chemical Equilibrium
1. Dynamic Nature:
- At equilibrium, the forward and reverse reactions continue to occur, but they proceed at equal rates, making the concentrations of reactants and products remain constant.
2. Dependence on Temperature:
- The value of the equilibrium constant (\(K\)) depends only on temperature. Changes in concentration or pressure affect the position of equilibrium but not the value of \(K\).
3. Independence from Initial Concentrations:
- The equilibrium constant \(K\) is independent of the initial concentrations of reactants and products; it only depends on the nature of the reaction and the temperature.
Factors Affecting Equilibrium Position
While the equilibrium constant remains constant at a given temperature, the position of equilibrium can shift in response to changes in concentration, pressure, or temperature. These changes are explained by Le Chatelier's Principle, which states that if a dynamic equilibrium is disturbed by changing the conditions, the system shifts to counteract the disturbance and re-establish equilibrium.
1. Concentration Changes:
- Increasing the concentration of a reactant will shift the equilibrium towards the products to reduce the concentration of the added reactant.
- Increasing the concentration of a product will shift the equilibrium towards the reactants.
2. Pressure Changes (for Gaseous Reactions):
- If the pressure is increased by decreasing the volume, the equilibrium shifts towards the side with fewer moles of gas.
- If the pressure is decreased by increasing the volume, the equilibrium shifts towards the side with more moles of gas.
3. Temperature Changes:
- For exothermic reactions (\(\Delta H < 0\)), increasing temperature shifts the equilibrium towards the reactants (left).
- For endothermic reactions (\(\Delta H > 0\)), increasing temperature shifts the equilibrium towards the products (right).
Significance of the Equilibrium Constant (K)
1. Predicting Reaction Extent:
- A large \(K\) value (\(K \gg 1\)) indicates that the equilibrium lies far to the right, favoring the formation of products.
- A small \(K\) value (\(K \ll 1\)) indicates that the equilibrium lies far to the left, favoring the reactants.
2. Calculating Equilibrium Concentrations:
- By knowing \(K\) and the initial concentrations or partial pressures of reactants and products, one can use the ICE (Initial, Change, Equilibrium) method to calculate the equilibrium concentrations.
3. Predicting Reaction Direction:
- The reaction quotient (Q) is used to predict the direction of a reaction:
\[
Q = \frac{[C]^c [D]^d}{[A]^a [B]^b}
\]
- If \(Q < K\), the reaction will proceed forward to reach equilibrium.
- If \(Q > K\), the reaction will proceed in the reverse direction to reach equilibrium.
- If \(Q = K\), the reaction is at equilibrium.
Applications of the Law of Chemical Equilibrium
1. Industrial Synthesis:
- The Haber process for ammonia synthesis uses the principles of chemical equilibrium to optimize conditions (pressure, temperature, and concentration) for maximum yield.
2. Environmental Chemistry:
- Understanding equilibrium is essential in atmospheric and environmental chemistry, such as the balance between CO\(_2\) and dissolved carbonates in oceans.
3. Biochemical Reactions:
- Equilibrium concepts are applied in cellular reactions, where enzyme-catalyzed processes often reach equilibrium, balancing substrates and products for efficient functioning.
4. Analytical Chemistry:
- Acid-base titrations and complexometric titrations depend on equilibrium principles to calculate unknown concentrations.
Example Calculation Using the Law of Chemical Equilibrium
For a reaction:
\[
N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)
\]
If the equilibrium constant \(K_c\) is known, and the initial concentrations of \(N_2\), \(H_2\), and \(NH_3\) are given, we can set up an ICE table to calculate the equilibrium concentrations of each component based on the equilibrium expression:
\[
K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}
\]
Using this approach, we can determine the concentrations at equilibrium for a given set of initial conditions.
Conclusion
The Law of Chemical Equilibrium is fundamental in understanding reversible reactions and the conditions under which they reach equilibrium. The equilibrium constant (\(K\)) allows chemists to predict the extent of a reaction, calculate equilibrium concentrations, and determine the effect of changing conditions on the equilibrium position. By applying Le Chatelier’s Principle, the Law of Chemical Equilibrium provides a framework for optimizing reaction conditions in industrial processes, understanding natural chemical systems, and conducting analytical determinations in chemistry.
|
Q 1. At \(277 \mathrm{~K}\), degree of dissociation water is \(1 \times 10^{-7} \%\). The value of ionic product of water is; |
|
(A) \(3.0 \times 10^{-14}\); |
|
(B) \(3.085 \times 10^{-15}\); |
|
(C) \(1 \times 10^{-16}\); |
|
(D) \(1 \times 10^{-14}\); |
|
Q 2. If the concentration of \(\left[\mathrm{NH}_{4}^{+}\right]\)in a solution having \(0.02 \mathrm{MNH}_{3}\) and \(0.005 \mathrm{MCa}(\mathrm{OH})_{2}\) is \(\mathrm{a} \times 10^{-6} \mathrm{M}\), determine a \[\n\left[\mathrm{k}_{\mathrm{b}}\left(\mathrm{NH}_{3}\right)=1.8 \times 10^{-5}\right]\]; |
|
(A)24; |
|
(B)30; |
|
(C)36; |
|
(D)42; |
|
Q 3. \(1 \mathrm{MNaCl}\) and \(1 \mathrm{MHCl}\) are present in an aqueous solution. The solution is; |
|
(A) An acidic solution with \(\mathrm{pH}<7\); |
|
(B) An acidic solution with \(\mathrm{pH}>7\); |
|
(C) Buffer solution with \(\mathrm{pH}<7\); |
|
(D) Buffer solution with \(\mathrm{pH}>7\); |
|
Q 4. At \(277 \mathrm{~K}\), degree of dissociation water is \(1 \times 10^{-7} \%\). The value of ionic product of water is; |
|
(A) \(3.0 \times 10^{-14}\); |
|
(B) \(3.085 \times 10^{-15}\); |
|
(C) \(1 \times 10^{-16}\); |
|
(D) \(1 \times 10^{-14}\); |
|
Q 5. If the concentration of \(\left[\mathrm{NH}_{4}^{+}\right]\)in a solution having \(0.02 \mathrm{MNH}_{3}\) and \(0.005 \mathrm{MCa}(\mathrm{OH})_{2}\) is \(\mathrm{a} \times 10^{-6} \mathrm{M}\), determine a \[\n\left[\mathrm{k}_{\mathrm{b}}\left(\mathrm{NH}_{3}\right)=1.8 \times 10^{-5}\right]\]; |
|
(A)24; |
|
(B)30; |
|
(C)36; |
|
(D)42; |
|
Q 6. \(1 \mathrm{MNaCl}\) and \(1 \mathrm{MHCl}\) are present in an aqueous solution. The solution is; |
|
(A) An acidic solution with \(\mathrm{pH}<7\); |
|
(B) An acidic solution with \(\mathrm{pH}>7\); |
|
(C) Buffer solution with \(\mathrm{pH}<7\); |
|
(D) Buffer solution with \(\mathrm{pH}>7\); |
|
Q 7. Which of the following is not a general characteristic of equilibria involving physical processes?; |
|
(a) Equilibrium is possible only in a closed system at a given temperature.; |
|
(b) All measurable properties of the system remain constant.; |
|
(c) All the physical processes stop at equilibrium.; |
|
(d) The opposing processes occur at the same rate and there is dynamic but stable condition.; |
|
Q 8. The liquid which has a _ _ _ _ vapour pressure is more volatile and has a _ _ _ _ boiling point; |
|
(a) Higher, higher; |
|
(b) Lower, lower; |
|
(c) Higher, lower; |
|
(d) Lower, higher; |
|
Q 9. Boiling point of the liquid depends on the atmospheric pressure It depends on the altitude of the place; at high altitude the boiling point \(\qquad\); |
|
(a) increases; |
|
(b) decreases; |
|
(c) either decreases or increases; |
|
(d) remains same; |
|
Q 10. In an experiment three watch glasses containing separately \(1 \mathrm{~mL}\) each of acetone, ethyl alcohol, and water are exposed to atmosphere and the experiment with different volumes of the liquids in a warmer room is repeated, it is observed that in all such cases the liquid eventually disappears and the time taken for complete evaporation in each case was different The possible reason is/are; |
|
(a) the nature of the liquids is different; |
|
(b) the amount of the liquids is different; |
|
(c) the temperature is different; |
|
(d) All of the above; |
Buffer Solutions and Their Role
Definition of Buffer Solutions
A buffer solution is a solution that resists changes in its pH when small amounts of acid or base are added. Buffer solutions are essential in maintaining a stable pH in various chemical and biological systems. They consist of a weak acid and its conjugate base or a weak base and its conjugate acid. This unique composition allows buffers to neutralize small amounts of added H\(^+\) or OH\(^-\), stabilizing the pH of the solution.
Common examples of buffer solutions include:
- Acetic acid and sodium acetate (an acid buffer),
- Ammonia and ammonium chloride (a basic buffer).
How Buffer Solutions Work
The ability of a buffer solution to resist pH changes is based on the reversible reactions of the weak acid/base and its conjugate. When an acid or base is added to a buffer, the following processes occur:
1. Addition of Acid (H\(^+\)):
- The conjugate base in the buffer (e.g., \( \text{CH}_3\text{COO}^- \) in an acetic acid buffer) reacts with the added H\(^+\) ions, neutralizing them:
\[
\text{CH}_3\text{COO}^- + H^+ \rightarrow \text{CH}_3\text{COOH}
\]
2. Addition of Base (OH\(^-\)):
- The weak acid in the buffer (e.g., \(\text{CH}_3\text{COOH}\)) reacts with the added OH\(^-\) ions, preventing a significant pH increase:
\[
\text{CH}_3\text{COOH} + OH^- \rightarrow \text{CH}_3\text{COO}^- + H_2O
\]
Through these reactions, the buffer can neutralize added acids or bases, maintaining a nearly constant pH.
Types of Buffers
1. Acidic Buffers:
- Made from a weak acid and its conjugate base, acidic buffers maintain a pH below 7.
- Example: Acetic acid (\(\text{CH}_3\text{COOH}\)) and sodium acetate (\(\text{CH}_3\text{COONa}\)).
2. Basic Buffers:
- Made from a weak base and its conjugate acid, basic buffers maintain a pH above 7.
- Example: Ammonium hydroxide (\(\text{NH}_4\text{OH}\)) and ammonium chloride (\(\text{NH}_4\text{Cl}\)).
Buffer Capacity
Buffer capacity is the measure of a buffer’s ability to resist pH changes upon the addition of an acid or base. It depends on two factors:
- The concentration of the weak acid and its conjugate base (or vice versa): Higher concentrations mean a greater capacity to neutralize added H\(^+\) or OH\(^-\).
- The ratio of acid to base: A 1:1 ratio provides maximum buffering capacity at the pH equal to the pKa of the acid.
Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is used to calculate the pH of a buffer solution. It relates the pH of the buffer to the pKa (or pKb) of the acid (or base) and the ratio of concentrations of the conjugate base to the acid:
\[
pH = pK_a + \log \frac{[\text{A}^-]}{[\text{HA}]}
\]
Where:
- \(pK_a\) is the acid dissociation constant,
- \([\text{A}^-]\) is the concentration of the conjugate base,
- \([\text{HA}]\) is the concentration of the weak acid.
This equation helps design buffer solutions with a desired pH by adjusting the concentrations of the acid and its conjugate base.
Applications of Buffer Solutions
1. Biological Systems:
- Buffer solutions are crucial in maintaining the pH of biological fluids, like blood, which needs a stable pH around 7.4. The bicarbonate buffer system (HCO\(_3^-\)/H\(_2\)CO\(_3\)) is essential for blood pH regulation.
- Enzymes in living organisms function within narrow pH ranges; buffers help maintain optimal pH for enzymatic activity.
2. Industrial Applications:
- Buffers are used in chemical manufacturing, pharmaceuticals, and food production to control the pH of solutions and ensure product stability.
3. Analytical Chemistry:
- Buffers are used in titrations and electrophoresis to maintain a stable pH, allowing accurate and reproducible measurements.
4. Environmental Science:
- Natural water bodies contain buffers that help resist pH changes, which is crucial for aquatic life. Acid rain can overwhelm these natural buffers, causing harmful pH shifts in lakes and rivers.
Example Calculation Using the Henderson-Hasselbalch Equation
For a buffer solution with 0.1 M acetic acid (\(\text{CH}_3\text{COOH}\)) and 0.1 M sodium acetate (\(\text{CH}_3\text{COONa}\)), and knowing that the \(pK_a\) of acetic acid is 4.76, we can calculate the pH as follows:
\[
pH = pK_a + \log \frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}
\]
\[
pH = 4.76 + \log \frac{0.1}{0.1} = 4.76
\]
The pH of this buffer solution is 4.76, showing how the buffer maintains a stable pH close to the \(pK_a\) of the weak acid.
Limitations of Buffer Solutions
1. Limited Buffering Range:
- Buffers are effective only within a pH range of approximately \( pK_a \pm 1 \).
2. Limited Buffer Capacity:
- The buffer’s ability to neutralize added acid or base is finite. Adding too much acid or base can exceed the buffer capacity, leading to significant pH changes.
3. Dependence on Temperature:
- The \(pK_a\) of weak acids and bases can vary with temperature, affecting the pH of the buffer solution.
Conclusion
Buffer solutions play a crucial role in stabilizing pH in various chemical and biological systems. They work by neutralizing small amounts of added acid or base, thus maintaining a nearly constant pH. Understanding buffer capacity and using the Henderson-Hasselbalch equation enables chemists to design buffers with precise pH values. Buffers are essential in fields ranging from biological systems to industrial applications and environmental science, highlighting their importance in both natural and artificial processes.
|
Q 1. Which of the following pairs constitutes a buffer?; |
|
(A) \(\mathrm{HCl}\) and \(\mathrm{KCl}\); |
|
(B) \(\mathrm{HNO}_{2}\) and \(\mathrm{NaNO}_{2}\); |
|
(C) \(\mathrm{NaOH}\) and \(\mathrm{NaCl}\); |
|
(D) \(\mathrm{HNO}_{3}\) and \(\mathrm{NH}_{4} \mathrm{NO}_{3}\); |
|
Q 2. Which of the following pairs constitutes a buffer?; |
|
(A) \(\mathrm{HCl}\) and \(\mathrm{KCl}\); |
|
(B) \(\mathrm{HNO}_{2}\) and \(\mathrm{NaNO}_{2}\); |
|
(C) \(\mathrm{NaOH}\) and \(\mathrm{NaCl}\); |
|
(D) \(\mathrm{HNO}_{3}\) and \(\mathrm{NH}_{4} \mathrm{NO}_{3}\); |
|
Q 3. A buffer solution is prepared in which the concentration of \(\mathrm{NH}_{3}\) is \(0 30 \mathrm{M}\) and the concentration of \(\mathrm{NH}_{4}{ }^{+}\)is \(0 20 \mathrm{M}\) If the equilibrium constant, \(\mathrm{K}_{\mathrm{b}}\) for \(\mathrm{NH}_{3}\) equals \(1 8 \times 10^{-5}\), what is the \(\mathrm{pH}\) of this solution \(?(\log 2 7=0 433)\); |
|
(a) 9.08; |
|
(b) 9.43; |
|
(c) 11.72; |
|
(d) 8.73; |
|
Q 4. Which of the following pairs constitutes a buffer?; |
|
(a) \(\mathrm{NaOH}\) and \(\mathrm{NaCl}\); |
|
(b) \(\mathrm{HNO}_{3}\) and \(\mathrm{NH}_{4} \mathrm{NO}_{3}\); |
|
(c) \(\mathrm{HCl}\) and \(\mathrm{KCl}\); |
|
(d) \(\mathrm{HNO}_{2}\) and \(\mathrm{NaNO}_{2}\); |
|
Q 5. A buffer solution is prepared in which the concentration of \(\mathrm{NH}_{3}\) is \(0 30 \mathrm{M}\) and the concentration of \(\mathrm{NH}_{4}^{+}\)is \(0 20 \mathrm{M}\) If the equilibrium constant, \(\mathrm{K}_{\mathrm{b}}\) for \(\mathrm{NH}_{3}\) equals \(1 8 \times 10^{-5}\), what is the \(\mathrm{pH}\) of this solution \(?(\log 2 7=0 433)\) [2011]; |
|
(a) 9.08; |
|
(b) 9.43; |
|
(c) 11.72; |
|
(d) 8.73; |
|
Q 6. In a buffer solution containing equal concentration of \(\mathrm{B}^{-}\)and \(\mathrm{HB}\), the \(K_{\mathrm{b}}\) for \(\mathrm{B}^{-}\) is \(10^{-10}\) The \(\mathrm{pH}\) of buffer solution is: [2010]; |
|
(a) 10; |
|
(b) 7; |
|
(c) 6; |
|
(d) 4; |
|
Q 7. Which of the following pairs constitutes a buffer? [2006]; |
|
(a) \(\mathrm{NaOH}\) and \(\mathrm{NaCl}\); |
|
(b) \(\mathrm{HNO}_{3}\) and \(\mathrm{NH}_{4} \mathrm{NO}_{3}\); |
|
(c) \(\mathrm{HCl}\) and \(\mathrm{KCl}\); |
|
(d) \(\mathrm{HNO}_{2}\) and \(\mathrm{NaNO}_{2}\); |
Solubility Product Constant (\(K_{sp}\))
Definition of the Solubility Product Constant
The solubility product constant (\(K_{sp}\)) is an equilibrium constant that applies to the dissolution of sparingly soluble ionic compounds in water. It quantifies the extent to which a compound dissolves, representing the product of the molar concentrations of its ions at saturation (when the solution is at equilibrium with the solid substance). \(K_{sp}\) values help predict whether a precipitate will form in a solution and determine the solubility of an ionic compound.
For a salt \(AB\) that dissociates as follows:
\[
AB(s) \rightleftharpoons A^+(aq) + B^-(aq)
\]
The solubility product constant (\(K_{sp}\)) is given by:
\[
K_{sp} = [A^+][B^-]
\]
where \([A^+]\) and \([B^-]\) are the molar concentrations of the ions at equilibrium.
Expression of \(K_{sp}\) for Different Types of Salts
1. For Binary Salts (\(AB\)):
- Dissociation: \(AB \rightarrow A^+ + B^-\)
- \(K_{sp} = [A^+][B^-]\)
2. For Salts like \(A_2B\) (e.g., \(CaF_2\)):
- Dissociation: \(A_2B \rightarrow 2A^+ + B^{2-}\)
- \(K_{sp} = [A^+]^2[B^{2-}]\)
3. For Salts like \(AB_2\) (e.g., \(PbCl_2\)):
- Dissociation: \(AB_2 \rightarrow A^{2+} + 2B^-\)
- \(K_{sp} = [A^{2+}][B^-]^2\)
Each compound has a unique \(K_{sp}\) value, which varies with temperature and is typically provided for standard conditions (e.g., 25°C).
Relationship Between \(K_{sp}\) and Solubility
The molar solubility of a compound refers to the number of moles of the compound that can dissolve per liter of solution to reach saturation. \(K_{sp}\) is directly related to the molar solubility of a compound.
For example, for a simple salt like \(AB\), if the solubility is \(s\) mol/L:
\[
K_{sp} = s \times s = s^2
\]
Thus, \(s = \sqrt{K_{sp}}\).
For more complex salts, such as \(A_2B\), the relationship between \(K_{sp}\) and \(s\) will involve powers corresponding to the dissociation stoichiometry.
Factors Affecting \(K_{sp}\) and Solubility
1. Temperature:
- Endothermic Dissolution: For salts where dissolution is endothermic, solubility typically increases with temperature, and thus \(K_{sp}\) increases.
- Exothermic Dissolution: For salts where dissolution is exothermic, solubility may decrease with temperature, resulting in a lower \(K_{sp}\) at higher temperatures.
2. Common Ion Effect:
- Adding a common ion (an ion already present in solution) decreases the solubility of a compound. This occurs because the increased ion concentration shifts the equilibrium towards the solid phase, reducing solubility. For example, adding \(NaCl\) to a \(PbCl_2\) solution reduces the solubility of \(PbCl_2\) due to the increase in \(Cl^-\) ions.
3. pH of the Solution:
- The solubility of certain salts, especially those with basic anions (like \(OH^-\) or \(CO_3^{2-}\)), increases in acidic solutions. For example, \(CaCO_3\) becomes more soluble in acidic conditions as \(CO_3^{2-}\) reacts with \(H^+\) ions to form \(HCO_3^-\) or \(H_2CO_3\), effectively removing \(CO_3^{2-}\) ions from the equilibrium.
Applications of \(K_{sp}\)
1. Predicting Precipitation Reactions:
- \(K_{sp}\) helps predict whether a precipitate will form when two solutions are mixed. By calculating the ion product (Q) (similar in form to \(K_{sp}\)) and comparing it with \(K_{sp}\):
- If \(Q > K_{sp}\): The solution is supersaturated, and a precipitate will form.
- If \(Q < K_{sp}\): The solution is unsaturated, and no precipitate will form.
- If \(Q = K_{sp}\): The solution is at equilibrium, and no further precipitation occurs.
2. Qualitative Analysis:
- In qualitative inorganic analysis, \(K_{sp}\) values help separate ions in a mixture. Selective precipitation can be induced by adjusting conditions to precipitate one ion while keeping others in solution.
3. Environmental Science:
- \(K_{sp}\) is used to understand the solubility of minerals and salts in natural waters. For example, calcium carbonate (\(CaCO_3\)) saturation in oceans is vital for marine life, affecting coral formation and shell strength.
4. Pharmaceuticals and Medicine:
- \(K_{sp}\) concepts are used in drug formulation, especially for drugs that require specific solubility characteristics to be effective. Understanding the solubility product helps in designing compounds that dissolve at appropriate rates.
Example Calculation Using \(K_{sp}\)
For \(AgCl\), a sparingly soluble salt:
\[
AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)
\]
Given \(K_{sp} = 1.8 \times 10^{-10}\) at 25°C, we can calculate the solubility (\(s\)):
1. Set up the \(K_{sp}\) expression:
\[
K_{sp} = [Ag^+][Cl^-] = s \times s = s^2
\]
2. Solve for \(s\):
\[
s = \sqrt{1.8 \times 10^{-10}} \approx 1.34 \times 10^{-5} \, \text{mol/L}
\]
Thus, the solubility of \(AgCl\) in water at 25°C is approximately \(1.34 \times 10^{-5}\) mol/L.
Limitations of \(K_{sp}\)
1. Only for Sparingly Soluble Compounds:
- \(K_{sp}\) applies to compounds with low solubility in water, not to highly soluble salts where complete dissociation occurs.
2. Temperature Dependence:
- \(K_{sp}\) values are highly temperature-dependent, and calculations are accurate only if the temperature is specified.
3. Assumption of Ideal Behavior:
- \(K_{sp}\) assumes ideal behavior, which may not hold in highly concentrated solutions or those with significant ionic interactions.
Conclusion
The solubility product constant (\(K_{sp}\)) is a valuable tool in understanding the solubility of sparingly soluble salts in water. It allows chemists to predict precipitation, design separation processes, and calculate solubility in various conditions. By applying \(K_{sp}\) concepts, chemists and environmental scientists can control and predict the behavior of ionic compounds in solutions, making it essential in fields such as qualitative analysis, pharmaceuticals, and environmental science.
Factors Affecting \(K_{sp}\) Calculation
The solubility product constant (\(K_{sp}\)) is influenced by several factors, each of which can affect the accuracy of \(K_{sp}\) calculations and the solubility of sparingly soluble compounds. Key factors include temperature, the common ion effect, pH of the solution, ionic strength, and the presence of complexing agents. Here’s a closer look at each factor and how it impacts \(K_{sp}\) and solubility:
1. Temperature
- Effect: Temperature has a significant impact on \(K_{sp}\), as most solubility equilibria are temperature-dependent.
- For endothermic dissolution reactions (where heat is absorbed), an increase in temperature increases the solubility of the salt, raising the \(K_{sp}\) value.
- For exothermic dissolution reactions (where heat is released), an increase in temperature decreases the solubility of the salt, lowering the \(K_{sp}\) value.
- Reason: According to Le Chatelier’s Principle, increasing the temperature of an endothermic process shifts the equilibrium towards more dissolved ions, thus increasing \(K_{sp}\), while it shifts the equilibrium towards the undissolved solid in exothermic processes, reducing \(K_{sp}\).
- Implication: \(K_{sp}\) values are typically tabulated at a specific temperature (e.g., 25°C). Using the wrong temperature in calculations can lead to inaccurate predictions of solubility or precipitation.
2. Common Ion Effect
- Effect: Adding a common ion (an ion that is already present in the equilibrium) reduces the solubility of a sparingly soluble salt.
- Reason: When a common ion is added, the concentration of that ion in the solution increases, which shifts the equilibrium toward the solid form, according to Le Chatelier’s Principle. This decreases the concentration of the other ions in solution, effectively reducing the solubility.
- Example: For a saturated solution of \(AgCl\):
\[
AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)
\]
Adding \(NaCl\) introduces more \(Cl^-\) ions, decreasing the solubility of \(AgCl\) as more \(AgCl\) precipitates to maintain the constant \(K_{sp}\) value.
- Implication: The common ion effect does not change \(K_{sp}\) itself but affects the ion concentrations in solution, thus affecting how solubility is calculated in the presence of other sources of the ions involved.
3. pH of the Solution
- Effect: The pH of the solution can alter the solubility of salts, especially those containing anions that can react with \(H^+\) ions (e.g., salts with \(OH^-\), \(CO_3^{2-}\), or \(PO_4^{3-}\) ions).
- Reason: In acidic solutions, anions like \(OH^-\), \(CO_3^{2-}\), or \(PO_4^{3-}\) can react with \(H^+\), removing them from the equilibrium. This shifts the equilibrium toward more dissolution to replenish the ions, increasing the solubility.
- Example: The solubility of \(CaCO_3\) in water increases in acidic conditions:
\[
CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_3^{2-}(aq)
\]
If \(H^+\) is added, it reacts with \(CO_3^{2-}\) to form \(HCO_3^-\), effectively reducing \(CO_3^{2-}\) concentration and increasing \(CaCO_3\) solubility.
- Implication: Calculations of \(K_{sp}\) and solubility need to consider the pH, especially for salts that contain anions prone to reacting with \(H^+\).
4. Ionic Strength of the Solution
- Effect: The presence of additional ions in solution (even if they are not common ions) can impact the solubility of sparingly soluble salts.
- Reason: Ionic strength affects activity coefficients, which measure the “effective concentration” of ions in non-ideal solutions. In solutions with high ionic strength, ions are shielded by each other, reducing interactions and effectively increasing the solubility of some salts.
- Example: Adding a neutral salt (like \(NaNO_3\)) to a solution with \(AgCl\) can increase the solubility of \(AgCl\) by increasing the ionic strength, which decreases the electrostatic interactions between \(Ag^+\) and \(Cl^-\), thus affecting their activities.
- Implication: In concentrated solutions, using concentration values without correcting for ionic strength (using activity coefficients) can lead to inaccurate \(K_{sp}\) calculations. For accurate results, especially in complex or highly ionic solutions, activities should be used instead of concentrations.
5. Presence of Complexing Agents
- Effect: Complexing agents or ligands, such as \(NH_3\), \(CN^-\), or \(OH^-\), can significantly increase the solubility of sparingly soluble salts by forming complex ions with the metal cations.
- Reason: When a complex ion forms, it reduces the concentration of the free metal ion, shifting the equilibrium towards dissolution to maintain \(K_{sp}\).
- Example: For \(AgCl\), adding \(NH_3\) results in the formation of the complex ion \([Ag(NH_3)_2]^+\):
\[
AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)
\]
\[
Ag^+(aq) + 2NH_3(aq) \rightarrow [Ag(NH_3)_2]^+(aq)
\]
The formation of \([Ag(NH_3)_2]^+\) reduces \(Ag^+\) concentration, thus increasing the solubility of \(AgCl\).
- Implication: The presence of complexing agents should be taken into account in \(K_{sp}\) calculations, as they can dramatically alter the apparent solubility of the salt.
Summary of Factors Affecting \(K_{sp}\) Calculations
| Factor | Effect on \(K_{sp}\) or Solubility |
|--||
| Temperature | Changes \(K_{sp}\) based on endothermic/exothermic nature of dissolution. |
| Common Ion Effect | Reduces solubility by shifting equilibrium towards the solid phase. |
| pH of the Solution | Alters solubility of salts with anions that react with \(H^+\). |
| Ionic Strength | Increases solubility by affecting ion activities, especially in concentrated solutions. |
| Complexing Agents | Increases solubility by forming complex ions, lowering the concentration of free metal ions. |
Understanding these factors is crucial for accurately calculating \(K_{sp}\) and predicting the solubility of ionic compounds under various conditions. Adjusting for these factors allows chemists to control and manipulate precipitation, dissolution, and separation processes in both laboratory and industrial settings.
|
Q 1. If the solubility of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) is \(\mathrm{S}\) moles/litre, its solubility product will be; |
|
(A) \(S^{2}\); |
|
(B) \(S^{3}\); |
|
(C) \(4 \mathrm{~S}^{3}\); |
|
(D) \(2 \mathrm{~S}^{3}\); |
|
Q 2. The solubility of \(\mathrm{CdSO}_{4}\) in water is \(8.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\). Its solubility in \(0.01 \mathrm{MH}_{2} \mathrm{SO}_{4}\) solution is \(\times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}\) (Round off to the Nearest integer) (Assume that solubility is much less than \(0.01 \mathrm{M}\)); |
|
(A)\(64 \times 10^{-6} \mathrm{M}\); |
|
(B)\(10 \times 10^{-6} \mathrm{M}\); |
|
(C)\(32 \times 10^{-6} \mathrm{M}\); |
|
(D)\(26 \times 10^{-6} \mathrm{M}\); |
|
Q 3. If the solubility is represented as ' \(S\) ' molL and the solubility product as \(\mathrm{K}_{\mathrm{sp}}\), the relation between the two for \(\mathrm{Zr}_{3}\left(\mathrm{PO}_{4}\right)_{4}\) is; |
|
(A) \(\mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}\right)^{1 / 7}\) ; |
|
(B) \(\mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{27}\right)^{1 / 3}\); |
|
(C) \(\mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{m}}}{6912}\right)^{1 / 7}\); |
|
(D) \(\mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}\right)^{1 / 5}\); |
|
Q 4. If \(\mathrm{S}_{1}, \mathrm{~S}_{2}, \mathrm{~S}_{3}\) and \(\mathrm{S}_{4}\) are the solubility of \(\mathrm{AgCl}\) in water, in \(0.01 \mathrm{MCaCl}_{2}\) in \(0.01 \mathrm{MNaCl}\) and in\n\(0.05 \mathrm{MAgNO}_{3}\) respectively at a certain temperature, the correct order of solubility is; |
|
(A) \(S_{1}>S_{2}>S_{3}>S_{4}\); |
|
(B) \(\mathrm{S}_{1}>\mathrm{S}_{3}>\mathrm{S}_{2}>\mathrm{S}_{4}\); |
|
(C) \(\mathrm{S}_{1}>\mathrm{S}_{2}=\mathrm{S}_{3}>\mathrm{S}_{4}\); |
|
(D) \(\mathrm{S}_{1}>\mathrm{S}_{3}>\mathrm{S}_{4}>\mathrm{S}_{2}\); |
|
Q 5. If the solubility of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) is \(\mathrm{S}\) moles/litre, its solubility product will be; |
|
(A) \(S^{2}\); |
|
(B) \(S^{3}\); |
|
(C) \(4 \mathrm{~S}^{3}\); |
|
(D) \(2 \mathrm{~S}^{3}\); |
|
Q 6. The solubility of \(\mathrm{CdSO}_{4}\) in water is \(8.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\). Its solubility in \(0.01 \mathrm{MH}_{2} \mathrm{SO}_{4}\) solution is \(\times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}\) (Round off to the Nearest integer) (Assume that solubility is much less than \(0.01 \mathrm{M}\)); |
|
(A)\(64 \times 10^{-6} \mathrm{M}\); |
|
(B)\(10 \times 10^{-6} \mathrm{M}\); |
|
(C)\(32 \times 10^{-6} \mathrm{M}\); |
|
(D)\(26 \times 10^{-6} \mathrm{M}\); |
|
Q 7. If the solubility is represented as ' \(S\) ' molL and the solubility product as \(\mathrm{K}_{\mathrm{sp}}\), the relation between the two for \(\mathrm{Zr}_{3}\left(\mathrm{PO}_{4}\right)_{4}\) is; |
|
(A) \(\mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}\right)^{1 / 7}\) ; |
|
(B) \(\mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{27}\right)^{1 / 3}\); |
|
(C) \(\mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{m}}}{6912}\right)^{1 / 7}\); |
|
(D) \(\mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}\right)^{1 / 5}\); |
|
Q 8. If \(\mathrm{S}_{1}, \mathrm{~S}_{2}, \mathrm{~S}_{3}\) and \(\mathrm{S}_{4}\) are the solubility of \(\mathrm{AgCl}\) in water, in \(0.01 \mathrm{MCaCl}_{2}\) in \(0.01 \mathrm{MNaCl}\) and in\n\(0.05 \mathrm{MAgNO}_{3}\) respectively at a certain temperature, the correct order of solubility is; |
|
(A) \(S_{1}>S_{2}>S_{3}>S_{4}\); |
|
(B) \(\mathrm{S}_{1}>\mathrm{S}_{3}>\mathrm{S}_{2}>\mathrm{S}_{4}\); |
|
(C) \(\mathrm{S}_{1}>\mathrm{S}_{2}=\mathrm{S}_{3}>\mathrm{S}_{4}\); |
|
(D) \(\mathrm{S}_{1}>\mathrm{S}_{3}>\mathrm{S}_{4}>\mathrm{S}_{2}\); |
|
Q 9. The solubility of AgI in NaI solution is less than that in pure water because :; |
|
(a) the temperature of the solution decreases ; |
|
(b) solubility product to \(\mathrm{AgI}\) is less than that of \(\mathrm{NaI}\); |
|
(c) of common ion effect; |
|
(d) AgI forms complex with \(\mathrm{NaI}\); |
|
Q 10. The \(K_{s p}\) for \(\mathrm{Cr}(\mathrm{OH})_{3}\) is \(1 6 \times 10^{-30}\) The solubility of this compound in water is :; |
|
(a) \(\sqrt[4]{1.6 \times 10^{-30}}\); |
|
(b) \(\sqrt[4]{1.6 \times 10^{-30} / 27}\); |
|
(c) \(1.6 \times 10^{-30 / 27}\); |
|
(d) \(\sqrt{1.6 \times 10^{-30}}\); |
Equilibrium Constant Expressions (\(K_c\) and \(K_p\))
Definition of the Equilibrium Constant
The equilibrium constant is a value that expresses the relationship between the concentrations or partial pressures of reactants and products in a chemical reaction at equilibrium. It quantifies the extent of a reaction and is a key tool for predicting reaction behavior. Equilibrium constants vary depending on whether the reaction components are measured in concentration (\(K_c\)) or partial pressure (\(K_p\)).
For a general reversible reaction:
\[
aA + bB \rightleftharpoons cC + dD
\]
The equilibrium constant expression can be written in terms of:
- \(K_c\): Based on molar concentrations (mol/L),
- \(K_p\): Based on partial pressures (typically in atmospheres, atm) for gaseous reactions.
Equilibrium Constant Expression in Terms of Concentration (\(K_c\))
The equilibrium constant in terms of concentration, \(K_c\), is calculated using the molar concentrations of the reactants and products at equilibrium. For the general reaction above, the \(K_c\) expression is:
\[
K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}
\]
Where:
- \([A]\), \([B]\), \([C]\), and \([D]\) represent the equilibrium concentrations of the reactants and products,
- \(a\), \(b\), \(c\), and \(d\) are the stoichiometric coefficients.
Characteristics of \(K_c\):
- Unit Dependency: The units of \(K_c\) depend on the reaction's stoichiometry. If there is a change in the number of moles of gases between reactants and products, the units of \(K_c\) will reflect this.
- Temperature Dependence: Like all equilibrium constants, \(K_c\) is only valid at a specific temperature and changes if the temperature changes.
Equilibrium Constant Expression in Terms of Partial Pressure (\(K_p\))
For gaseous reactions, the equilibrium constant in terms of partial pressure, \(K_p\), is used. It is defined similarly to \(K_c\), but it uses the partial pressures of gases instead of concentrations:
\[
K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}
\]
Where:
- \(P_A\), \(P_B\), \(P_C\), and \(P_D\) are the partial pressures of the reactants and products at equilibrium,
- \(a\), \(b\), \(c\), and \(d\) are the stoichiometric coefficients.
Characteristics of \(K_p\):
- Units: Like \(K_c\), the units of \(K_p\) depend on the reaction's stoichiometry and the number of moles of gaseous reactants and products.
- Applicability: \(K_p\) is specifically used for gaseous reactions, where pressure provides a convenient way to express the concentration of gases.
Relationship Between \(K_c\) and \(K_p\)
For reactions involving gases, \(K_c\) and \(K_p\) are related by the equation:
\[
K_p = K_c (RT)^{\Delta n}
\]
Where:
- \(R\) is the gas constant (0.0821 L·atm·K\(^{-1}\)·mol\(^{-1}\)),
- \(T\) is the temperature in Kelvin,
- \(\Delta n\) is the change in moles of gas (moles of gaseous products - moles of gaseous reactants).
This relationship shows that:
- If \(\Delta n = 0\), then \(K_p = K_c\), as no change in the number of gas molecules occurs.
- If \(\Delta n \neq 0\), \(K_p\) and \(K_c\) will differ based on the temperature and the number of gas molecules.
Example:
For the reaction:
\[
N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)
\]
\(\Delta n = 2 - (1 + 3) = -2\), so \(K_p\) and \(K_c\) can be related as:
\[
K_p = K_c (RT)^{-2}
\]
Interpreting \(K_c\) and \(K_p\) Values
1. Magnitude of \(K\):
- \(K \gg 1\): The reaction strongly favors the formation of products; equilibrium lies far to the right.
- \(K \ll 1\): The reaction favors the reactants, and equilibrium lies far to the left.
- \(K \approx 1\): Neither reactants nor products are strongly favored; the reaction reaches a balanced equilibrium.
2. Predicting Reaction Direction:
- The reaction quotient (Q) is used to determine if a reaction is at equilibrium:
- If \(Q = K\), the system is at equilibrium.
- If \(Q < K\), the reaction will proceed forward to produce more products.
- If \(Q > K\), the reaction will shift toward the reactants.
Factors Affecting \(K_c\) and \(K_p\)
1. Temperature:
- \(K_c\) and \(K_p\) are temperature-dependent. An increase or decrease in temperature will change the equilibrium constant.
- For exothermic reactions (\(\Delta H < 0\)), increasing temperature decreases \(K\).
- For endothermic reactions (\(\Delta H > 0\)), increasing temperature increases \(K\).
2. Concentration and Pressure Changes:
- Changing concentrations or partial pressures affects the position of equilibrium but does not change the values of \(K_c\) or \(K_p\). The equilibrium constant remains the same at a given temperature.
3. Catalysts:
- Catalysts speed up the rate at which equilibrium is achieved but do not alter the value of \(K_c\) or \(K_p\).
Applications of \(K_c\) and \(K_p\)
1. Predicting Reaction Yields:
- \(K_c\) and \(K_p\) help determine the extent to which a reaction will proceed, which is essential in calculating yields in industrial processes.
2. Chemical Engineering:
- Knowledge of equilibrium constants allows chemical engineers to optimize reaction conditions in processes like the Haber process for ammonia synthesis, where adjusting temperature and pressure influences equilibrium.
3. Environmental Chemistry:
- \(K_c\) and \(K_p\) are used to model atmospheric reactions, such as the equilibrium between different nitrogen oxides, which helps understand air pollution dynamics.
4. Biological Systems:
- In biochemistry, equilibrium constants are critical for understanding enzyme-catalyzed reactions and metabolic pathways, where equilibrium helps determine the direction of biochemical reactions.
Example Calculation of \(K_c\) and \(K_p\)
For the reaction:
\[
H_2(g) + I_2(g) \rightleftharpoons 2HI(g)
\]
If the equilibrium concentrations are \([H_2] = 0.1\) M, \([I_2] = 0.1\) M, and \([HI] = 0.2\) M, we can calculate \(K_c\) as follows:
\[
K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(0.2)^2}{(0.1)(0.1)} = 4
\]
If the reaction involves gases, we can find \(K_p\) using the relationship between \(K_c\) and \(K_p\), given \(\Delta n = 0\), so \(K_p = K_c = 4\) in this case.
Conclusion
The equilibrium constant expressions \(K_c\) and \(K_p\) are essential in understanding the balance between reactants and products in chemical reactions at equilibrium. While \(K_c\) uses concentration and applies to both aqueous and gaseous reactions, \(K_p\) is specific to gases and uses partial pressures. These constants help chemists predict reaction yields, calculate equilibrium concentrations, and understand how temperature changes influence reaction behavior. Both constants are central to fields like industrial chemistry, environmental science, and biochemistry, making them invaluable tools in predicting and controlling chemical processes.
|
Q 1. \n\(Assertion\): With change in concentration of the reactant, the value of equilibrium constant changes. \n\(Reason\) : More the value of \(K_{C}\) lesser is the extent of reaction in the forward direction.; |
|
(D) If both Assertion and Reason are false statements; |
|
(C) If Assertion is true statement but Reason is false; |
|
(B) If both Assertion \& Reason are true but the reason is not the correct explanation of the assertion; |
|
(A) If both Assertion \& Reason are true and the reason is the correct explanation of the assertion; |
|
Q 2. \n\(Assertion\): With change in concentration of the reactant, the value of equilibrium constant changes. \n\(Reason\) : More the value of \(K_{C}\) lesser is the extent of reaction in the forward direction.; |
|
(B) If both Assertion \& Reason are true but the reason is not the correct explanation of the assertion; |
|
(C) If Assertion is true statement but Reason is false; |
|
(D) If both Assertion and Reason are false statements; |
|
(A) If both Assertion \& Reason are true and the reason is the correct explanation of the assertion; |
|
Q 3. \(\mathrm{K}_{1}\) and \(\mathrm{K}_{2}\) are equilibrium constant for reactions (1) and (2) \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) \(\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) Then,; |
|
(a) \(\mathrm{K}_{1}=\left(\frac{1}{\mathrm{~K}_{2}}\right)^{2}\); |
|
(b) \(\mathrm{K}_{1}=\mathrm{K}_{2}^{2}\); |
|
(c) \(\mathrm{K}_{1}=\frac{1}{\mathrm{~K}_{2}}\); |
|
(d) \(\mathrm{K}_{1}=\left(\mathrm{K}_{2}\right)^{0}\); |
|
Q 4. The equilibrium constant for the reversible reaction \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) is \(K\) and for reaction \(\frac{1}{2} \mathrm{~N}_{2}+\frac{3}{2} \mathrm{H}_{2} \rightleftharpoons \mathrm{NH}_{3}\), the equilibrium constant is \(K^{\prime}\) The \(K\) and \(K^{\prime}\) will be related as:; |
|
(a) \(K \times K^{\prime}=1\); |
|
(b) \(K=K^{\prime}\); |
|
(c) \(K^{\prime}=\sqrt{K}\); |
|
(d) \(K=\sqrt{K^{\prime}}\); |
|
Q 5. A reaction is \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}\) Initially we start with equal concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) At equilibrium we find that the moles of \(\mathrm{C}\) is two times of \(\mathrm{A}\) What is the equilibrium constant of the reaction?; |
|
(d) 2; |
|
(c) 4; |
|
(b) \(\frac{1}{2}\); |
|
(a) \(\frac{1}{4}\); |
|
Q 6. In \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}\) The unit of equilibrium constant is :; |
|
(d) No unit; |
|
(c) Mole litre \({ }^{-1}\); |
|
(b) Mole litre; |
|
(a) Litre mole \({ }^{-1}\); |
|
Q 7. Steam reacts with iron at high temperature to give hydrogen gas and \(\mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{~s})\) The correct expression for the equilibrium constant is; |
|
(d) \(\frac{\left[\mathrm{Fe}_{3} \mathrm{O}_{4}\right]}{[\mathrm{Fe}]}\); |
|
(c) \(\frac{\left(\mathrm{P}_{\mathrm{H}_{2}}\right)^{4}\left[\mathrm{Fe}_{3} \mathrm{O}_{4}\right]}{\left(\mathrm{P}_{\mathrm{H}_{2} \mathrm{O}}\right)^{4}[\mathrm{Fe}]}\); |
|
(b) \(\frac{\left(\mathrm{P}_{\mathrm{H}_{2}}\right)^{4}}{\left(\mathrm{P}_{\mathrm{H}_{2} \mathrm{O}}\right)^{4}}\); |
|
(a) \(\frac{\mathrm{P}_{\mathrm{H}_{2}}^{2}}{\mathrm{P}_{\mathrm{H}_{2} \mathrm{O}}^{2}}\); |
|
Q 8. The rate constant for forward and backward reaction of hydrolysis of ester are \(1 1 \times 10^{-2}\) and \(1 5 \times 10^{-3}\) per minute respectively Equilibrium constant for the reaction \(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}^{+} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) is; |
|
(d) 7.33; |
|
(c) 6.33; |
|
(b) 5.33; |
|
(a) 4.33; |
|
Q 9. Unit of equilibrium constant for the given reaction is \(\mathrm{Ni}(\mathrm{s})+4 \mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(\mathrm{~g})\); |
|
(d) \((\mathrm{mol} / \mathrm{l})^{4}\); |
|
(c) \(\left(\mathrm{mol} / \mathrm{l}^{-4}\right.\); |
|
(b) \((\mathrm{mol} / \mathrm{l})^{3}\); |
|
(a) \((\mathrm{mol} / \mathrm{l})^{-3}\); |
|
Q 10. In a reversible chemical reaction having two reactants in equilibrium, if the concentration of the reactants are doubled then the equilibrium constant will; |
|
(a) Also be doubled; |
|
(b) Be halved; |
|
(c) Become one-fourth; |
|
(d) Remain the same; |
Le Chatelier’s Principle
Definition of Le Chatelier’s Principle
Le Chatelier’s Principle states that if a system at equilibrium is subjected to a change (stress) in concentration, temperature, or pressure, the system will adjust its position of equilibrium to counteract that change and restore a new equilibrium. This principle is essential in understanding how equilibrium systems respond to external influences and is widely used to predict the outcome of changes in chemical reactions.
Le Chatelier’s Principle is applicable to various types of chemical equilibria, including gaseous reactions, aqueous solutions, and industrial processes.
Types of Changes That Affect Equilibrium
1. Changes in Concentration
- Adding or removing reactants or products will shift the equilibrium to minimize the effect of the concentration change.
- Adding Reactants: Shifts the equilibrium towards the products to consume the added reactants.
- Removing Reactants: Shifts the equilibrium towards the reactants to produce more reactants.
- Adding Products: Shifts the equilibrium towards the reactants to consume the added products.
- Removing Products: Shifts the equilibrium towards the products to replace the removed products.
Example:
For the reaction:
\[
N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)
\]
Adding more \(N_2\) or \(H_2\) shifts the equilibrium towards the production of \(NH_3\).
2. Changes in Pressure (for Gaseous Equilibria)
- Changing the pressure of a system affects only reactions involving gases. According to Le Chatelier’s Principle:
- Increasing Pressure: The equilibrium shifts towards the side with fewer moles of gas to reduce pressure.
- Decreasing Pressure: The equilibrium shifts towards the side with more moles of gas to increase pressure.
Example:
For the reaction:
\[
N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)
\]
Increasing the pressure shifts the equilibrium toward the product side (fewer moles), favoring \(NH_3\) formation.
Note: If the number of moles of gases is the same on both sides, pressure changes have no effect on the equilibrium position.
3. Changes in Temperature
- Temperature changes affect the equilibrium constant (\(K\)) and can shift the equilibrium position depending on whether the reaction is exothermic or endothermic.
- Increasing Temperature:
- For an endothermic reaction (\(\Delta H > 0\)), the equilibrium shifts towards the products (right) to absorb the added heat.
- For an exothermic reaction (\(\Delta H < 0\)), the equilibrium shifts towards the reactants (left) to release heat.
- Decreasing Temperature:
- For an endothermic reaction, the equilibrium shifts towards the reactants.
- For an exothermic reaction, the equilibrium shifts towards the products.
Example:
For the exothermic reaction:
\[
N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) + \text{heat}
\]
Increasing the temperature shifts the equilibrium towards the reactants (left), reducing \(NH_3\) formation.
Applications of Le Chatelier’s Principle
1. Industrial Processes
- Le Chatelier’s Principle is widely used in industry to optimize reaction conditions for maximum product yield.
- Haber Process: The synthesis of ammonia (\(NH_3\)) from nitrogen and hydrogen is influenced by pressure, temperature, and concentration adjustments. High pressure and moderate temperature favor ammonia production.
- Contact Process: In the production of sulfuric acid, adjusting temperature and pressure is crucial for maximizing sulfur trioxide (SO\(_3\)) yield.
2. Chemical Analysis
- Le Chatelier’s Principle helps predict the outcomes of reactions in analytical chemistry. For example, shifting equilibria is used in complexometric titrations to ensure complete reaction with the analyte.
3. Biological Systems
- Biological systems rely on equilibrium adjustments to maintain homeostasis. For example, blood uses the bicarbonate buffer system (HCO\(_3^-\)/H\(_2\)CO\(_3\)) to maintain pH levels. Changes in \(CO_2\) levels (such as during respiration) cause the equilibrium to shift, helping regulate blood pH.
4. Environmental Science
- In natural water bodies, changes in temperature and concentration can affect equilibrium systems. For example, increased \(CO_2\) levels in water can shift carbonate equilibria, affecting the solubility of minerals and the health of aquatic ecosystems.
Le Chatelier’s Principle and Reaction Quotient (Q)
Le Chatelier’s Principle is closely related to the reaction quotient (Q):
- Reaction Quotient (Q): Q is calculated the same way as the equilibrium constant (K) but with initial concentrations or partial pressures instead of equilibrium values.
- Comparison of Q and K:
- If \(Q < K\): The reaction will proceed in the forward direction to reach equilibrium.
- If \(Q > K\): The reaction will proceed in the reverse direction to reach equilibrium.
- If \(Q = K\): The system is already at equilibrium, and no shift occurs.
Understanding \(Q\) and \(K\) allows us to predict the direction of a reaction when conditions are altered, aligning with Le Chatelier’s Principle.
Examples of Le Chatelier’s Principle in Action
1. CO\(_2\) Solubility in Oceans:
- When atmospheric \(CO_2\) levels increase, more \(CO_2\) dissolves in oceans, forming carbonic acid (H\(_2\)CO\(_3\)). This dissociates into \(H^+\) and \(HCO_3^-\), affecting ocean pH and carbonate equilibria. By Le Chatelier’s Principle, the increase in \(CO_2\) shifts the equilibrium toward the formation of \(HCO_3^-\), acidifying the water.
2. Carbon Monoxide Poisoning:
- Hemoglobin binds oxygen in the blood, following an equilibrium process. In the presence of carbon monoxide (CO), the equilibrium shifts in favor of CO binding to hemoglobin over oxygen, causing CO poisoning. This shift can be mitigated by increasing oxygen concentration, which shifts the equilibrium to displace CO from hemoglobin.
3. Titration Curves in Chemistry:
- During titrations, adding acid or base causes shifts in pH equilibrium, which Le Chatelier’s Principle helps explain. In buffer solutions, adding small amounts of H\(^+\) or OH\(^-\) does not cause drastic pH changes, as the buffer adjusts by neutralizing added ions.
Conclusion
Le Chatelier’s Principle provides a framework for predicting how equilibrium systems respond to changes in concentration, pressure, or temperature. It plays a critical role in various applications, from industrial chemical synthesis to environmental science and biological processes. By applying Le Chatelier’s Principle, chemists can optimize reaction conditions to maximize yields, control pH in buffer solutions, and understand natural equilibria in ecosystems.
|
Q 1. When pressure is applied to the equilibrium system Ice \(\rightleftharpoons\) Water Which of the following phenomenon will happen?; |
|
(a) More ice will be formed; |
|
(b) Water will evaporate; |
|
(c) More water will be formed; |
|
(d) Equilibrium will not be formed; |
|
Q 2. Value of \(\mathrm{K}_{\mathrm{P}}\) in the reaction \(\mathrm{MgCO}_{3(\mathrm{~s})} \rightarrow \mathrm{MgO}_{(\mathrm{s})}+\mathrm{CO}_{2(\mathrm{~g})}\) is; |
|
(a) \(\mathrm{K}_{\mathrm{P}}=\mathrm{P}_{\mathrm{CO}_{2}}\); |
|
(b) \(\mathrm{K}_{\mathrm{P}}=\mathrm{P}_{\mathrm{CO}_{2}} \times \frac{\mathrm{P}_{\mathrm{CO}_{2}} \times \mathrm{P}_{\mathrm{MgO}}}{\mathrm{P}_{\mathrm{MgCO}_{3}}}\); |
|
(c) \(\mathrm{K}_{\mathrm{P}}=\frac{\mathrm{P}_{\mathrm{CO}_{2}} \times \mathrm{P}_{\mathrm{MgO}}}{\mathrm{P}_{\mathrm{MgCO}_{3}}}\); |
|
(d) \(\mathrm{K}_{\mathrm{P}}=\frac{\mathrm{P}_{\mathrm{MgCO}_{3}}}{\mathrm{P}_{\mathrm{CO}_{2}} \times \mathrm{P}_{\mathrm{MgO}}}\); |
|
Q 3. The reaction quotient \(\mathrm{Q}\) is used to; |
|
(a) predict the extent of a reaction on the basis of its magnitude; |
|
(b) predict the direction of the reaction; |
|
(c) calculate equilibrium concentrations; |
|
(d) calculate equilibrium constant; |
|
Q 4. Which one of the following information can be obtained on the basis of Le Chatelier principle?; |
|
(a) Dissociation constant of a weak acid; |
|
(b) Entropy change in a reaction; |
|
(c) Equilibrium constant of a chemical reaction; |
|
(d) Shift in equilibrium position on changing value of a constraint; |
|
Q 5. For the manufacture of ammonia by the reaction \[ \mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}+2 \mathrm{kcal} \] the favourable conditions are; |
|
(a) Low temperature, low pressure and catalyst; |
|
(b) Low temperature, high pressure and catalyst; |
|
(c) High temperature, low pressure and catalyst; |
|
(d) High temperature, high pressure and catalyst; |
|
Q 6. Which of the following reaction will be favoured at low pressure?; |
|
(a) \(\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI}\); |
|
(b) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\); |
|
(c) \(\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3}+\mathrm{Cl}_{2}\); |
|
(d) \(\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}\); |
|
Q 7. The equilibrium which remains unaffected by pressure change is; |
|
(a) \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\); |
|
(b) \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\); |
|
(c) \(2 \mathrm{O}_{3}(\mathrm{~g}) \rightleftharpoons 3 \mathrm{O}_{2}(\mathrm{~g})\); |
|
(d) \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\); |
|
Q 8. Suitable conditions for melting of ice :; |
|
(a) high temperature and high pressure; |
|
(b) high temperature and low pressure; |
|
(c) low temperature and low pressure; |
|
(d) low temperature and high pressure; |
|
Q 9. In a two-step exothermic reaction \[ \mathrm{A}_{2}(\mathrm{~g})+\mathrm{B}_{2}(\mathrm{~g}) \underset{\text { Step 1 }}{\rightleftharpoons} 3 \mathrm{C}(\mathrm{g}) \underset{\text { Step 2 }}{\rightleftharpoons} \mathrm{D}(\mathrm{g}) \] Steps 1 and 2 are favoured respectively by; |
|
(a) high pressure, high temperature and low pressure, low temperature; |
|
(b) high pressure, low temperature and low pressure, high temperature; |
|
(c) low pressure, high temperature and high pressure, high temperature; |
|
(d) low pressure, low temperature and high pressure, low temperature; |
|
Q 10. What happens when an inert gas is added to an equilibrium keeping volume unchanged?; |
|
(a) More product will form; |
|
(b) Less product will form; |
|
(c) More reactant will form; |
|
(d) Equilibrium will remain unchanged; |
Ionization of Water and pH Scale
Ionization of Water
Water (\(H_2O\)) can undergo a process called self-ionization (or autoionization) where two water molecules react to form hydronium ions (H\(_3\)O\(^+\)) and hydroxide ions (OH\(^-\)):
\[
H_2O(l) + H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)
\]
This reaction is often simplified to show that water dissociates into hydrogen ions (H\(^+\)) and hydroxide ions:
\[
H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)
\]
The equilibrium constant for this reaction is called the ion-product constant for water, denoted by \(K_w\). At 25°C (298 K), the value of \(K_w\) is:
\[
K_w = [H^+][OH^-] = 1.0 \times 10^{-14}
\]
Since the concentration of \(H^+\) and \(OH^-\) ions is equal in pure water, each concentration is:
\[
[H^+] = [OH^-] = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7} \, \text{M}
\]
This balance gives water a neutral pH at 25°C, where \([H^+]\) and \([OH^-]\) are equal.
pH Scale
The pH scale is a logarithmic scale that measures the concentration of hydrogen ions (H\(^+\)) in a solution, providing a way to quantify acidity or basicity. The formula for calculating pH is:
\[
pH = -\log[H^+]
\]
- Acidic Solutions: Solutions with a pH less than 7 have a higher concentration of H\(^+\) ions than OH\(^-\) ions.
- Neutral Solutions: Solutions with a pH of 7 (such as pure water at 25°C) have equal concentrations of H\(^+\) and OH\(^-\) ions.
- Basic (Alkaline) Solutions: Solutions with a pH greater than 7 have a higher concentration of OH\(^-\) ions than H\(^+\) ions.
Example:
For a solution with \([H^+] = 1.0 \times 10^{-3}\) M, the pH is:
\[
pH = -\log(1.0 \times 10^{-3}) = 3
\]
pOH Scale
Similarly, the pOH scale is used to measure the concentration of hydroxide ions (OH\(^-\)):
\[
pOH = -\log[OH^-]
\]
The relationship between pH and pOH is:
\[
pH + pOH = 14 \quad \text{(at 25°C)}
\]
This means that if the pH of a solution is known, the pOH can be calculated, and vice versa. This relationship holds as long as \(K_w\) remains constant, which is true at a given temperature (25°C for \(K_w = 1.0 \times 10^{-14}\)).
Example:
For a solution with pH = 5:
\[
pOH = 14 - 5 = 9
\]
Applications of the pH Scale
1. Biological Systems:
- pH is crucial in biological systems, where enzymes and cellular processes function optimally within narrow pH ranges (e.g., human blood pH is maintained around 7.4 by buffers).
2. Environmental Science:
- pH helps assess the acidity or alkalinity of natural water bodies. Acid rain, for example, lowers the pH of lakes and rivers, affecting aquatic life.
3. Industrial Applications:
- In industries like pharmaceuticals and food production, pH control is vital to ensure product stability, quality, and safety.
4. Soil Chemistry:
- Soil pH affects nutrient availability for plants. Acidic or alkaline soils require treatment to bring the pH to levels optimal for crop growth.
Importance of the Ionization Constant \(K_w\)
The value of \(K_w\) provides a foundation for understanding the pH scale, and it depends on temperature. As temperature increases, \(K_w\) also increases, leading to a higher concentration of H\(^+\) and OH\(^-\) ions. This temperature dependence means that the neutral point (where \([H^+]\) = \([OH^-]\)) may have a different pH at temperatures other than 25°C.
For example:
- At temperatures above 25°C, \(K_w\) increases, making pure water slightly acidic as \([H^+]\) increases.
- At temperatures below 25°C, \(K_w\) decreases, and pure water becomes slightly basic as \([OH^-]\) predominates.
This variability underscores the importance of considering temperature when using pH as a measure of neutrality.
Calculating pH in Different Scenarios
1. Strong Acids and Bases:
- For strong acids like HCl, which dissociates completely, \([H^+]\) is equal to the acid’s molarity.
\[
\text{pH of 0.01 M HCl} = -\log(0.01) = 2
\]
- For strong bases like NaOH, \([OH^-]\) is equal to the base’s molarity, and pH can be calculated via pOH:
\[
\text{pOH of 0.01 M NaOH} = -\log(0.01) = 2
\]
\[
\text{pH} = 14 - 2 = 12
\]
2. Weak Acids and Bases:
- Weak acids and bases do not fully dissociate, so calculating pH involves their dissociation constant (\(K_a\) for acids and \(K_b\) for bases) and equilibrium expressions.
- Example for Acetic Acid (\(K_a = 1.8 \times 10^{-5}\)):
\[
CH_3COOH \rightleftharpoons H^+ + CH_3COO^-
\]
Set up an ICE table and solve for \([H^+]\) to calculate pH.
3. Water Ionization:
- In pure water at 25°C:
\[
[H^+] = [OH^-] = 1.0 \times 10^{-7} \, \text{M}
\]
\[
\text{pH} = -\log(1.0 \times 10^{-7}) = 7
\]
- Thus, pure water has a neutral pH of 7 at 25°C.
Conclusion
The ionization of water forms the basis of the pH scale, a vital tool in chemistry for measuring the acidity or basicity of solutions. The pH scale ranges from 0 to 14, with pH values below 7 indicating acidity, above 7 indicating basicity, and exactly 7 being neutral. Both the ionization constant (\(K_w\)) and the temperature dependence of pH are critical in various scientific and industrial fields, influencing everything from biological processes to environmental and industrial applications.
|
Q 1. The ionization constant of ammonium hydroxide is \(1 77 \times 10^{-5}\) at \(298 \mathrm{~K}\) Hydrolysis constant of ammonium chloride is: [2009]; |
|
(a) \(6.50 \times 10^{-12}\); |
|
(b) \(5.65 \times 10^{-13}\); |
|
(c) \(5.65 \times 10^{-12}\); |
|
(d) \(5.65 \times 10^{-10}\); |
Boric acid acts as a Lewis acid by accepting OH- from water molecules, forming B(OH)4-. |
The equilibrium constant (K) for a dissociation reaction depends on temperature and degree of dissociation. |
pH and equilibrium in ammonium solutions depend on concentrations of NH4+ and OH- in equilibrium. |
Strong acids like HCl dissociate completely in water, affecting pH below 7. |
Buffer solutions resist changes in pH by neutralizing added acids or bases. |
Acid-Base Equilibria |
Law of Chemical Equilibrium |
Buffer Solutions and their Role |
Solubility Product Constant (Ksp) |
Equilibrium Constant Expressions (Kc, Kp) |
Le Chatelier’s Principle |
Ionization of Water and pH Scale |