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: …
- CBSE 2026Set ANNUAL1 markMCQQ.Ostwald's dilution law is not obeyed by :(a) CH3COOH(b) NH4OH(c) HCl(d) H2CO3
›Reveal solutionSolution
Ostwald's dilution law applies only to weak electrolytes; HCl, being a strong electrolyte, does not obey it.
Ostwald's dilution law gives the relationship between the dissociation (ionisation) constant Ka of a weak electrolyte, its degree of dissociation α, and its concentration C:
Ka=1−αCα2
This law assumes that the electrolyte is only partially/weakly dissociated in solution, so that an equilibrium exists between the undissociated molecules and the ions, and α can meaningfully vary with dilution. CH3COOH, NH4OH and H2CO3 are all weak electrolytes that only partially ionise in water, so they obey this law.
…
- CBSE 2026Set ANNUAL1 markMCQQ.The heat of neutralization of a strong acid and strong base is(a) 13.7 kcal(b) Greater than 13.7 kcal(c) Less than 13.7 kcal(d) None of these
›Reveal solutionSolution
Heat of neutralisation of a strong acid + strong base ≈ 13.7 kcal (57.1 kJ) per mole.
Strong acids and strong bases are completely ionised, so neutralisation is essentially H+ + OH- → H2O. This releases a constant amount of energy, about 57.1 kJ (≈ 13.7 kcal) per mole of water formed, regardless of which strong …
- CBSE 2026Set ANNUAL1 markMCQQ.An aqueous solution of NH4Cl is(a) Acidic(b) Alkaline(c) Neutral(d) None of these
›Reveal solutionSolution
NH4Cl solution is acidic.
NH4Cl is formed from a strong acid (HCl) and a weak base (NH4OH). In water the cation hydrolyses: NH4+ + H2O ⇌ NH4OH + H+, releasing H+ …
- CBSE 2025Set ANNUAL1 markMCQQ.Degree of dissociation of 0.1 N CH3COOH is (Kacid = 1 x 10^-5)(a) 10^-5(b) 10^-4(c) 10^-3(d) 10^-2
›Reveal solutionSolution
0.1 N acetic acid (Ka = 1x10^-5) dissociates to the extent of α = 10^-2 (1%).
For a weak monobasic acid, the degree of dissociation (for small α, using the approximation valid when α << 1) is:
α = √(Ka / C)
…
- CBSE 2024Set ANNUAL1 markMCQQ.The hydrogen ion concentration of a weak acid of dissociation constant Ka and concentration C is equal to(a) √(Ka/C)(b) C/Ka(c) KaC(d) √(KaC)
›Reveal solutionSolution
[H⁺] of a weak acid = √(Ka·C).
For a weak acid HA⇌H++A− with initial concentration C and degree of dissociation α (small, so 1−α≈1):
…
- CBSE 2022Set ANNUAL1 markMCQQ.Which of the following is weak electrolytic?(a) NaCl(b) HCl(c) CH3COOH(d) K2SO4
›Reveal solutionSolution
Acetic acid ionises partially, so it is the weak electrolyte — option (c).
From NCERT Class 11 Chemistry (Equilibrium):
- NaCl, K₂SO₄: strong (ionic) electrolytes — fully dissociated.
- HCl: strong acid — essentially completely ionised. …
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