Q.The ionization constant of HF is 3.2 × 10⁻⁴. Calculate the degree of dissociation of HF in its 0.02 M solution. Calculate the concentration of all species present (H3O+, F– and HF) in the solution and its pH.
Concept understanding — Hydrogen Ion Concentration pH
Hydrogen Ion Concentration and pH
Imagine you have a glass of pure water. It looks simple, but inside, a tiny fraction of water molecules are constantly splitting apart and re-forming. This splitting creates two kinds of charged particles: a hydrogen ion (H+) and a hydroxide ion (OH−). In pure water, these two are perfectly balanced — there are exactly as many H+ as OH−.
Now, if you add something like lemon juice (an acid), you increase the number of hydrogen ions. The balance tips: more H+ than OH−. If you add baking soda (a base), you decrease H+ or increase OH−, and the balance tips the other way.
The question is: how do we measure this imbalance in a simple, practical way? The numbers of H+ ions are incredibly tiny — in pure water, only about 1 in every 10 million water molecules is split at any moment. Writing these numbers directly (like 0.0000001 moles per litre) is clumsy. That's where pH comes in.
The Precise Definition
pH is a mathematical shortcut. It stands for "power of hydrogen" (from the French puissance d'hydrogène).
pH=−log10[H+]
where [H+] is the concentration of hydrogen ions in moles per litre (mol/L).
The logarithm base 10 does two things at once:
- It compresses a huge range of numbers (from 10−14 to 100) into a manageable scale of 0 to 14.
- The negative sign flips the direction: higher [H+] gives a lower pH, and lower [H+] gives a higher pH.
What the Numbers Mean
| [H+] (mol/L) | pH | Example |
|---|---|---|
| 10−1 | 1 | Stomach acid |
| 10−3 | 3 | Lemon juice |
| 10−7 | 7 | Pure water (neutral) |
| 10−9 | 9 | Baking soda solution |
| 10−13 | 13 | Household bleach |
Notice the pattern: each step of 1 in pH means a tenfold change in [H+]. A solution of pH 3 has 10 times more H+ than pH 4, and 100 times more than pH 5.
The Key Insight
pH is not a measure of "how acidic" something is in a vague sense — it is a precise, logarithmic measure of the actual number of hydrogen ions present. The scale runs from 0 (most acidic, highest [H+]) to 14 (most basic, lowest [H+]), with 7 being neutral.
pH = 7 is neutral only at 25°C. At body temperature (37°C), neutral pH is about 6.8. The definition stays the same — only the reference point shifts.
A Quick Check
If a solution has [H+]=2.5×10−4 mol/L, what is its pH?
pH=−log10(2.5×10−4)=−(log102.5+log1010−4)=−(0.398−4)=3.602
So pH ≈ 3.6 — acidic, as expected from a 10−4 order concentration.
The beauty of pH is that it turns a microscopic, hard-to-grasp number into a simple, intuitive scale you can read on a meter or test with litmus paper. Once you understand that pH is just a clever way to write "how many hydrogen ions are floating around," the rest follows naturally.
If you've searched "Hydrogen Ion Concentration pH class 11 chemistry notes" or "Hydrogen Ion Concentration pH NCERT solutions", this page covers exactly that ground — the concept is a standard part of the Class 11 Chemistry NCERT/CBSE syllabus. It also carries real weight in JEE Main, NEET and state CET Chemistry papers, where questions on hydrogen ion concentration ph test both conceptual understanding and calculation speed.
Concept: Degree of ionization of a weak acid from Ka, using the exact quadratic (not the α≪1 shortcut, since Ka here is not small enough to justify it).
Step 1 -- Set up the ICE table for HF+H2O⇌H3O++F−, initial c=0.02 M:
[HF]=0.02(1−α),[H3O+]=[F−]=0.02α
Step 2 -- Substitute into Ka.
Ka=0.02(1−α)(0.02α)2=1−α0.02α2=3.2×10−4
Step 3 -- Solve the quadratic. This gives α2+0.016α−0.016=0, whose positive root is α≈0.12 (12%).
Step 4 -- Equilibrium concentrations and pH.
[H3O+]=[F−]=0.02×0.12=2.4×10−3 M
[HF]=0.02(1−0.12)=17.6×10−3 M
pH=−log(2.4×10−3)=2.62
The degree of dissociation is α=0.12 (12%); [H3O+]=[F−]=2.4×10−3 M, [HF]=17.6×10−3 M, and pH =2.62.
Because Ka is not tiny relative to the initial concentration, the α≪1 shortcut isn't safe here -- solving the full quadratic gives α=0.12, [H3O+]=[F−]=2.4×10−3 M, [HF]=17.6×10−3 M, and pH =2.62.
1. Write the equilibrium and its ICE table.
HF+H2O⇌H3O++F−
Let c=0.02 M be the initial concentration and α the degree of ionization:
| HF | H3O+ | F− | |
|---|---|---|---|
| Initial | 0.02 | 0 | 0 |
| Change | −0.02α | +0.02α | +0.02α |
| Equilibrium | 0.02(1−α) | 0.02α | 0.02α |
2. Substitute into Ka.
Ka=[HF][H3O+][F−]=0.02(1−α)(0.02α)2=1−α0.02α2=3.2×10−4
3. Do NOT drop (1−α)≈1 here. Check first: if we (wrongly) assumed α≪1, α≈Ka/c=3.2×10−4/0.02≈0.126 -- over 5%, so the approximation is NOT valid and the full quadratic must be solved.
4. Solve the quadratic exactly. Rearranging:
0.02α2=3.2×10−4(1−α)⟹α2+0.016α−0.016=0
α=2−0.016+0.0162+4(0.016)=2−0.016+0.064256≈2−0.016+0.2535≈0.12
(the negative root is rejected -- a degree of ionization cannot be negative).
5. Equilibrium concentrations.
[H3O+]=[F−]=cα=0.02×0.12=2.4×10−3 M
[HF]=c(1−α)=0.02(0.88)=17.6×10−3 M
6. pH.
pH=−log[H3O+]=−log(2.4×10−3)=2.62
A common mistake here is applying the α=Ka/c shortcut without first checking that α≪1 actually holds. With Ka=3.2×10−4 and c=0.02 M, the shortcut's own output (≈12.6%) is already too large to trust itself -- that's the signal to go back and solve the exact quadratic instead.
α=0.12 (12%); [H3O+]=[F−]=2.4×10−3 M; [HF]=17.6×10−3 M; pH =2.62.
- KEAM 2026Set eng-2026-04194 marksMCQQ.The concentration of hydrogen ions in a hydrochloric acid solution is 3 x 10−3 M. Its pH value is about ( log3 = 0.4771) (A) 2.32 (B) 2.52 (C) 2.47 (D) 3.47 (E) 5.52
›Reveal solutionSolution
pH=−log[H+]=3−log3=2.52.
Given [H+]=3×10−3M:
pH=−log(3×10−3)=−(log3+log10−3)=−log3+3.
pH=3−0.4771=2.52.
✓Final answerThe correct option is (B).
- KEAM 2026Set eng-2026-04214 marksMCQQ.Match the correct pH value with the following substances:(a) Milk of magnesia(i) pH = 6.8(b) Black coffee(ii) pH = 7.8(c) Egg white(iii) pH = 5(d) Milk(iv) pH = 10 (A) (a)-(iv), (b)-(ii), (c)-(i). (d)-(ii) (B) (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i) (C) (a)-(i), (b)-(iv), (c)-(iii). (d)-(ii) (D) (a)-(i), (b)-(iii), (c)-(ii), (d)-(iv) (E) (a)-(iv), (b)-(iii), (c)-(ii). (d)-(i)
›Reveal solutionSolution
(a) MoM→10, (b) coffee→5, (c) egg white→7.8, (d) milk→6.8, matching (iv),(iii),(ii),(i).
Match each substance to its characteristic pH:
- (a) Milk of magnesia — a base (Mg(OH)2 suspension), pH ≈10 ⇒ (iv).
- (b) Black coffee — acidic, pH ≈5 ⇒ (iii).
- (c) Egg white — slightly basic, pH ≈7.8 ⇒ (ii).
- (d) Milk — nearly neutral/slightly acidic, pH ≈6.8 ⇒ (i).
Thus the correct matching is (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i).
✓Final answerThe correct option is (E).
- KEAM 2026Set pha-2026-0418F4 marksMCQQ.The pH of an aqueous solution of weak mono acidic base is 11. What is the [H+] of the solution? [pKw = 14] (A) 1×10−11 M (B) 1×10−13 M (C) 1×10−14 M (D) 1×10−3 M (E) 1×10−10 M
›Reveal solutionSolution
[H+]=10−11 M when pH = 11.
Reasoning
By definition pH=−log[H+], so
[H+]=10−pH=10−11 M.
(The 'weak base' detail is a distractor — pH alone fixes [H+].)
✓Final answerThe correct option is (A).
- KEAM 2025Set eng-2025-04234 marksMCQQ.The concentration of hydrogen ions in a sample of soft drink is 2×10−4 mol lit−1. Its pH value is (log 2 = 0.3010) (A) 4.369 (B) 3.699 (C) 2.369 (D) 5.301 (E) 3.301
›Reveal solutionSolution
pH = -log(2×10^-4) = 3.699.
Concept and Intuition
pH is the negative logarithm of the hydrogen-ion concentration. Splitting the log of a product lets us use the given log 2.
Step-by-Step Solution
- pH = -log[H+] = -log(2 × 10^-4).
- = -(log 2 + log 10^-4) = -(0.3010 - 4).
- = 4 - 0.3010 = 3.699.
Common Mistakes
- Writing pH = 4 - 0.3 as 3.7 without the given precise log value.
- Sign errors giving 4.301 instead of 3.699.
✓Final answerThe correct option is (B) — 3.699.
ANSWER: B
- KEAM 2025Set eng-2025-04284 marksMCQQ.The pH of a solution is 4 then it's OH− ion concentration (in mol dm−3) is (A) 10−4 (B) 10−10 (C) 10−2 (D) 10−12 (E) 10−6
›Reveal solutionSolution
Use [H+][OH−]=Kw=10−14 at 25 ∘C.
pH =4 means [H+]=10−pH=10−4 mol dm−3.
Since [H+][OH−]=10−14,
[OH−]=10−410−14=10−10 mol dm−3.
✓Final answerThe correct option is (B).
- KEAM 2025Set pha-2025-0424A4 marksMCQQ.An aqueous solution of which of the following has the highest pH value? (A) 0.10 M HCl (B) 0.50 M H2SO4 (C) 0.10 M NaOH (D) 0.5 M HCl (E) 0.01 M NaOH
›Reveal solutionSolution
0.10 M NaOH has the highest pH (≈13).
Strong bases give high pH; strong acids give low pH. Evaluate each:
- 0.10 M HCl: [H+]=0.10, pH =1.
- 0.50 M H2SO4: [H+]≈1.0, pH ≈0.
- 0.10 M NaOH: [OH−]=0.10⇒pOH=1⇒pH=13.
- 0.5 M HCl: pH ≈0.3.
- 0.01 M NaOH: pOH=2⇒pH=12.
The largest pH is from 0.10 M NaOH.
✓Final answerThe correct option is (C).
- KEAM 2024Set eng-2024-06064 marksMCQQ.Equal volumes of pH 3, 4 & 5 are mixed in a container. The concentration of H+ in the mixture is (Assume there is no change in the volume during mixing) (A) 1×10−3 M (B) 3.7×10−4 M (C) 1×10−4 M (D) 3.7×10−5 M (E) 3×10−5 M
›Reveal solutionSolution
Average the three [H+] values (equal volumes): (10−3+10−4+10−5)/3=3.7×10−4 M.
The hydrogen-ion concentrations are 10−3, 10−4 and 10−5 M. Mixing equal volumes and conserving moles of H+:
[H+]=3VV(10−3)+V(10−4)+V(10−5)=31.11×10−3=3.7×10−4 M.
✓Final answerThe correct option is (B).
- KEAM 2024Set pha-2024-06104 marksMCQQ.0.1 M HCl and 0.1 M H_{2}SO_{4} each of volume 2 mL are mixed and the volume is made up to 6 mL by adding 2 mL of 0.01 N NaCl solution. The pH of the resulting mixture is (A) $1.17$ (B) $1.0$ (C) $0.3$ (D) $\log 2 - \log 3$ (E) $\log 3 - \log 2$
›Reveal solutionSolution
Sum the H⁺ moles from both acids and divide by the final 6 mL.
Moles of H⁺:
- HCl: 0.1 M×2 mL=0.2 mmol.
- H2SO4: 0.1 M×2 mL=0.2 mmol ×2 (diprotic) =0.4 mmol.
Total H⁺ =0.6 mmol. NaCl is a neutral salt and contributes no H⁺. Final volume =6 mL:
[H+]=6 mL0.6 mmol=0.1 M⇒pH=−log(0.1)=1.0.
✓Final answerThe correct option is (B). [H+]=0.1 M gives pH = 1.0.
- KEAM 2024Set pha-2024-06104 marksMCQQ.In which of the following aqueous solutions of salt, is pH independent of the concentration of the salt? (A) Ammonium chloride (B) Ferric chloride (C) Ammonium acetate (D) Sodium acetate (E) Ammonium sulphate
›Reveal solutionSolution
For a salt of a weak acid and weak base, pH depends only on pKa and pKb.
For a salt of a weak acid and a weak base such as ammonium acetate,
pH=7+21(pKa−pKb),
which contains no concentration term — the pH is independent of salt concentration. The other salts (from a strong acid + weak base, or weak acid + strong base) have pH that varies with concentration.
✓Final answerThe correct option is (C). Ammonium acetate's pH is concentration-independent.
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