Q.The ionization constant of propanoic acid is 1.32 × 10⁻⁵. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?
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.
Important
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)
For a weak acid, the degree of ionization α is found from Ka=Cα2/(1−α); in pure 0.05 M solution, α≈1.62×10−2 and pH ≈ 3.09. In the presence of 0.01 M HCl, the common ion effect suppresses ionization drastically, giving α′≈1.32×10−3.
Concept First: Weak Acid Ionization
Propanoic acid (CH3CH2COOH) is a weak monoprotic acid. Its ionization in water is an equilibrium:
CH3CH2COOH⇌CH3CH2COO−+H+
The equilibrium constant Ka=1.32×10−5 tells us the acid is weak — only a tiny fraction of molecules actually donate a proton. The degree of ionizationα is the fraction of acid molecules that have ionized. If the initial concentration is C, then at equilibrium:
[HA]=C(1−α)
[A−]=Cα
[H+]=Cα (from the acid alone)
The key formula is:
Ka=[HA][H+][A−]=C(1−α)(Cα)2=1−αCα2
When α is very small (typically α<0.05), we can approximate 1−α≈1, giving α≈Ka/C. But we must check the approximation — if it fails, we solve the quadratic.
1. Pure 0.05 M solution — find α
Given C=0.05 M, Ka=1.32×10−5.
First, test the approximation: α≈0.051.32×10−5=2.64×10−4=1.62×10−2.
Is this small enough? 1.62% is borderline — the approximation 1−α≈1 introduces about 1.6% error. For exam accuracy (usually 2–3 significant figures), this is acceptable. But let's be thorough and solve exactly.
Write the exact equation:
1−α0.05α2=1.32×10−5
Multiply through:
0.05α2=1.32×10−5(1−α)
0.05α2=1.32×10−5−1.32×10−5α
Bring all terms to one side:
0.05α2+1.32×10−5α−1.32×10−5=0
This is a quadratic in α. Using the quadratic formula:
The approximation gave 1.62×10−2 — identical to three significant figures. For weak acids where Ka/C<10−3, the approximation is safe. Here Ka/C=2.64×10−4, so it's fine.
So α=1.62×10−2 (or 1.62%).
2. pH of the pure solution
[H+]=Cα=0.05×1.62×10−2=8.10×10−4 M.
pH=−log10(8.10×10−4)=3.09
(Check: log8.1≈0.908, so 4−0.908=3.092.)
3. In presence of 0.01 M HCl — common ion effect
Now the solution already contains H+ from a strong acid (HCl) at 0.01 M. This shifts the weak acid equilibrium to the left — ionization is suppressed.
Let the new degree of ionization be α′. The initial concentration of propanoic acid is still C=0.05 M. At equilibrium: …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
COMEDK 2025Set 2025-A1 markMCQ
Q.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).
Q.In a 0.2M 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.
Q.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 out
A 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.