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Problems · Problem 6.18

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.

Uttarakhand UbseTextbookSubjective· 3mImportance★★★★★est
12% · 18/155 Questions
✓ Free question

Because KaK_a is not tiny relative to the initial concentration, the α≪1\alpha \ll 1 shortcut isn't safe here -- solving the full quadratic gives α=0.12\alpha = 0.12, [H3O+]=[F−]=2.4×10−3[\text{H}_3\text{O}^+]=[\text{F}^-]=2.4\times10^{-3} M, [HF]=17.6×10−3[\text{HF}]=17.6\times10^{-3} M, and pH =2.62= 2.62.

1. Write the equilibrium and its ICE table.

HF+H2O⇌H3O++F−\text{HF} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{F}^-

Let c=0.02c = 0.02 M be the initial concentration and α\alpha the degree of ionization:

HFH3_3O+^+F−^-
Initial0.0200
Change−0.02α-0.02\alpha+0.02α+0.02\alpha+0.02α+0.02\alpha
Equilibrium0.02(1−α)0.02(1-\alpha)0.02α0.02\alpha0.02α0.02\alpha

2. Substitute into KaK_a.

Ka=[H3O+][F−][HF]=(0.02α)20.02(1−α)=0.02 α21−α=3.2×10−4K_a = \frac{[\text{H}_3\text{O}^+][\text{F}^-]}{[\text{HF}]} = \frac{(0.02\alpha)^2}{0.02(1-\alpha)} = \frac{0.02\,\alpha^2}{1-\alpha} = 3.2\times10^{-4}

3. Do NOT drop (1−α)≈1(1-\alpha)\approx1 here. Check first: if we (wrongly) assumed α≪1\alpha \ll 1, α≈Ka/c=3.2×10−4/0.02≈0.126\alpha \approx \sqrt{K_a/c} = \sqrt{3.2\times10^{-4}/0.02} \approx 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=00.02\alpha^2 = 3.2\times10^{-4}(1-\alpha) \implies \alpha^2 + 0.016\alpha - 0.016 = 0

α=−0.016+0.0162+4(0.016)2=−0.016+0.0642562≈−0.016+0.25352≈0.12\alpha = \frac{-0.016 + \sqrt{0.016^2 + 4(0.016)}}{2} = \frac{-0.016+\sqrt{0.064256}}{2} \approx \frac{-0.016+0.2535}{2} \approx 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[\text{H}_3\text{O}^+] = [\text{F}^-] = c\alpha = 0.02 \times 0.12 = 2.4\times10^{-3}\ \text{M}

[HF]=c(1−α)=0.02(0.88)=17.6×10−3 M[\text{HF}] = c(1-\alpha) = 0.02(0.88) = 17.6\times10^{-3}\ \text{M}

6. pH.

pH=−log⁡[H3O+]=−log⁡(2.4×10−3)=2.62\text{pH} = -\log[\text{H}_3\text{O}^+] = -\log(2.4\times10^{-3}) = 2.62

Watch out

A common mistake here is applying the α=Ka/c\alpha=\sqrt{K_a/c} shortcut without first checking that α≪1\alpha \ll 1 actually holds. With Ka=3.2×10−4K_a=3.2\times10^{-4} and c=0.02c=0.02 M, the shortcut's own output (≈12.6%\approx 12.6\%) is already too large to trust itself -- that's the signal to go back and solve the exact quadratic instead.

✓Final answer

α=0.12\alpha = \boxed{0.12} (12%); [H3O+]=[F−]=2.4×10−3 M[\text{H}_3\text{O}^+]=[\text{F}^-]=\boxed{2.4\times10^{-3}\ \text{M}}; [HF]=17.6×10−3 M[\text{HF}]=\boxed{17.6\times10^{-3}\ \text{M}}; pH =2.62=\boxed{2.62}.

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