Q.The ionization constant of chloroacetic acid is 1.35 × 10⁻³. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?
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Weak Acid Ionization: From Intuition to Precision
Imagine you drop a spoonful of sugar into a glass of water. Some sugar dissolves, but a lot just sits at the bottom. Now imagine you drop a spoonful of salt — it all dissolves completely. Acids behave the same way. Some acids, like hydrochloric acid (HCl), dissolve completely in water — every single molecule breaks apart. Others, like acetic acid (vinegar), only partially break apart. Most of the acid molecules stay intact, and only a few actually ionize.
That's the core idea: weak acids are shy about giving away their hydrogen ion. They don't fully commit.
The Precise Statement
A weak acid (HA) in water establishes an equilibrium between the intact acid molecule and its ions:
HA(aq)+H2O(l)⇌H3O(aq)++A(aq)−
The double arrow (⇌) is the key. It tells you the reaction happens in both directions simultaneously. Some HA molecules break apart to form H3O+ and A−, while some H3O+ and A− recombine back into HA. At equilibrium, both processes happen at the same rate — so the concentrations stop changing.
For a weak acid, most of the acid remains as HA at equilibrium. Only a tiny fraction exists as ions. This is the opposite of a strong acid, where the forward reaction goes to completion (single arrow: →).
The Quantitative Measure: Ka
Every weak acid has a number that tells you exactly how "shy" it is — the acid dissociation constant, Ka:
Ka=[HA][H3O+][A−]
Ka=[HA][H3O+][A−]
The smaller the Ka, the weaker the acid. For acetic acid (vinegar), Ka≈1.8×10−5. That tiny number means the numerator (ions) is very small compared to the denominator (intact acid). For a strong acid like HCl, Ka is effectively infinite — the denominator is essentially zero because all the acid has ionized.
A Concrete Example
Suppose you dissolve 0.10 mol of acetic acid (CH3COOH) in 1 L of water. At equilibrium, you'll find:
- [CH3COOH]≈0.0998 M (almost all of it is still intact)
- [H3O+]≈0.0013 M (only about 1.3% has ionized)
- [CH3COO−]≈0.0013 M
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---------|-------------|------------|-----------------|
| CH3COOH | 0.10 | −x | 0.10−x |
| H3O+ | 0 | +x | x |
| CH3COO− | 0 | +x | x |
Plugging into Ka=0.10−xx2=1.8×10−5 and solving gives x≈0.0013 M.
Why This Matters …
Concept: Weak Acid Ionization — For a weak acid and its salt, the pH of the acid alone is found via the ionization equilibrium, while the pH of the salt solution is governed by the hydrolysis of the conjugate base.
Step 1: pH of 0.1 M chloroacetic acid
Let HA represent the acid. For HA⇌H++A−, with Ka=1.35×10−3 and initial concentration C=0.1 M. Since Ka is not very small, use the exact quadratic:
Ka=C−xx2, where x=[H+].
Solving: x2=1.35×10−3(0.1−x)
⇒x2+1.35×10−3x−1.35×10−4=0
x=2−1.35×10−3+(1.35×10−3)2+4×1.35×10−4
x≈0.0111 M.
Thus, pH=−log(1.16×10−2)=1.94.
Step 2: pH of 0.1 M sodium salt (NaA) solution
The salt NaA fully dissociates, giving A− (conjugate base) at 0.1 M. A− hydrolyzes: A−+H2O⇌HA+OH−. …
For 0.1 M chloroacetic acid (Ka=1.35×10−3) the pH=1.94; for its 0.1 M sodium salt the ion hydrolyses to give pH=7.94.
Part 1 - The acid (0.1 M chloroacetic acid)
Chloroacetic acid is a weak acid: ClCH2COOH⇌ClCH2COO−+H+.
[H+]=Kac=(1.35×10−3)(0.1)=1.35×10−4=1.16×10−2 M
pH=−log(1.16×10−2)=1.94
Part 2 - The sodium salt (0.1 M sodium chloroacetate)
The salt of a weak acid and strong base; the anion hydrolyses:
ClCH2COO−+H2O⇌ClCH2COOH+OH−
Hydrolysis constant
Kh=KaKw=1.35×10−31.0×10−14=7.41×10−12 …
- COMEDK 2025Set 2025-A1 markMCQQ.The ionisation constant of the weak acid HF whose concentration is 0.1 M is 3.5×10−4 The Equilibrium constant value for the reaction F−+H2O⇋HF+OH−is _________ and the pH of aqueous solution of the weak acid is __________ . (A) Keq=4.4×10−11&pH=4.25 (B) Keq=3.58×10−11&pH=4.19 (C) Keq=2.86×10−11&pH=3.23 (D) Keq=3.92×10−10&pH=4.07
›Reveal solutionSolution
The hydrolysis constant Keq=Kw/Ka=2.86×10−11 uniquely fixes option (C).
Hydrolysis of the conjugate base F−:
Keq=KaKw=3.5×10−41.0×10−14=2.86×10−11.
pH of the weak acid (0.1 M HF):
[H+]=KaC=3.5×10−4×0.1=5.9×10−3 M,
pH=−log(5.9×10−3)≈2.23. …
- COMEDK 2023Set 2023-M1 markMCQQ.In a 0.2 M aqueous solution, lactic acid is 6.9% dissociated. The value of dissociation constant is (A) 1.2×10−4 (B) 9.5×10−4 (C) 6.5×10−4 (D) 3.6×10−2
›Reveal solutionSolution
For a weak acid, Ka=Cα2/(1−α)≈Cα2. Substituting C=0.2M and α=0.069 gives Ka≈9.5×10−4.
Lactic acid is 6.9% dissociated, so degree of dissociation α=0.069, and C=0.2M.
Using Ostwald's dilution law:
Ka=1−αCα2
With the standard approximation (α≪1): …
- COMEDK 2021Set 2021-B1 markMCQQ.The degree of dissociation of 0.5 M NH3 at 25°C in a solution of pH = 12. (A) 12% (B) 1% (C) 4% (D) 2%
›Reveal solutionSolution
[OH−]=0.01 M from pH 12; dividing by 0.5 M gives α=0.02=2%.
From pH:
pOH=14−12=2⇒[OH−]=10−2=0.01 M
Degree of dissociation for NH3⇌NH4++OH−: …
- KCET 2020Set A-11 markMCQQ.When the same quantity of heat is absorbed by a system at two different temperatures T1 and T2, such that T1>T2, change in entropies are ΔS1 and ΔS2 respectively. Then : (A) ΔS2<ΔS1 (B) ΔS1<ΔS2 (C) ΔS1=ΔS2 (D) S2>S1
›Reveal solutionSolution
For a given amount of heat absorbed, the entropy change is larger at the lower temperature because entropy is inversely proportional to temperature: ΔS=Q/T. Hence ΔS2>ΔS1 when T1>T2.
The core idea here is the definition of entropy change for a reversible process: ΔS=TQrev. When a system absorbs a quantity of heat Q at a constant temperature T, the entropy change is simply Q/T. This tells us that the same heat causes a bigger entropy increase at a lower temperature.
Why? Think of entropy as a measure of disorder or the number of ways energy can be distributed. At a lower temperature, the system already has less thermal motion, so adding the same amount of heat creates a proportionally larger relative increase in the number of accessible microstates — hence a larger entropy change.
Let’s work through the reasoning step by step.
- Write the expression for entropy change For a system that absorbs heat Q reversibly at a constant temperature T,
ΔS=TQ
This is the defining relation from thermodynamics.
-
Apply it to the two cases
At temperature T1: ΔS1=T1Q
At temperature T2: ΔS2=T2Q
The same Q is absorbed in both cases.
-
Compare the denominators
We are given T1>T2. Since the numerator Q is the same, the fraction with the smaller denominator gives the larger value.
T2Q>T1Q
Therefore ΔS2>ΔS1.
Watch outA common mistake is to think that a higher temperature means a larger entropy change — but the opposite is true. Entropy change is inversely proportional to temperature for a fixed heat transfer. Don’t confuse ΔS with the absolute entropy S, which does increase with temperature.
- Examine the options …
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