Q.The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pK a of bromoacetic acid.
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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 HA, the degree of ionization α gives the fraction of molecules that dissociate. The equilibrium concentrations are derived from initial concentration C and α.
Step 1 – Equilibrium concentrations
For HA ⇌ H⁺ + A⁻, with C=0.1 M and α=0.132:
[H+]=Cα=0.1×0.132=0.0132 M
[A−]=0.0132 M, [HA]=C(1−α)=0.1×0.868=0.0868 M
Step 2 – pH of the solution
pH=−log[H+]=−log(0.0132)
0.0132=1.32×10−2, so pH=2−log1.32≈2−0.1206=1.879
Step 3 – Ka and pKa …
[H+]=Cα gives pH=1.88; the Ostwald dilution law Ka=1−αCα2 gives Ka=2.0×10−3, so pKa=2.70.
1. Hydrogen-ion concentration and pH. For HA⇌H++A− with degree of ionization α:
[H+]=Cα=0.1×0.132=1.32×10−2 M
pH=−log(1.32×10−2)=1.88
2. Acid dissociation constant. Equilibrium concentrations are [HA]=C(1−α) and [H+]=[A−]=Cα, so: …
- 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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