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Concept 06 / 08Equilibrium

Weak Acids and Bases: Ka, Kb, Ostwald's Dilution Law and Polyprotic Acids

Ionisation constants of weak acids and bases, Ostwald's dilution law, calculating degree of ionisation and pH, the relation Ka Kb = Kw for conjugate pairs, factors affecting acid strength, polyprotic acids with successive ionisation constants, and the common ion effect on weak electrolytes.

Figure 1
Weak acids. A weak acid HA ionises only partly: HA+H2O⇌H3O++A−\mathrm{HA + H_{2}O \rightleftharpoons H_{3}O^{+} + A^{-}}, Ka=[H+][A−][HA]K_{a} = \frac{[\mathrm{H^{+}}][\mathrm{A^{-}}]}{[\mathrm{HA}]}, pKa=−log⁡Ka\text{p}K_{a} = -\log K_{a}. A larger KaK_{a} (smaller pKa\text{p}K_{a}) means a stronger acid. Examples: HF\mathrm{HF} 3.5×10−43.5 \times 10^{-4}, HCOOH\mathrm{HCOOH} 1.8×10−41.8 \times 10^{-4}, CH3COOH\mathrm{CH_{3}COOH} 1.8×10−51.8 \times 10^{-5}, HCN\mathrm{HCN} 4.9×10−104.9 \times 10^{-10}. Ostwald's dilution law. For concentration cc and degree of ionisation α\alpha: [H+]=[A−]=cα[\mathrm{H^{+}}] = [\mathrm{A^{-}}] = c\alpha, [HA]=c(1−α)[\mathrm{HA}] = c(1 - \alpha), so Ka=cα21−αK_{a} = \frac{c\alpha^{2}}{1 - \alpha}. If α≪1\alpha \ll 1: α≈Kac\alpha \approx \sqrt{\frac{K_{a}}{c}}, [H+]=Kac[\mathrm{H^{+}}] = \sqrt{K_{a}c}, pH=12(pKa−log⁡c)\text{pH} = \frac{1}{2}(\text{p}K_{a} - \log c). Dilution increases α\alpha (more molecules ionise), even though [H+][\mathrm{H^{+}}] decreases. When α\alpha comes out larger than about 5%5\%, solve the quadratic instead (see figure: the approximation overshoots at low concentration). Weak bases. B+H2O⇌BH++OH−\mathrm{B + H_{2}O \rightleftharpoons BH^{+} + OH^{-}}, Kb=[BH+][OH−][B]K_{b} = \frac{[\mathrm{BH^{+}}][\mathrm{OH^{-}}]}{[\mathrm{B}]}. Similarly [OH−]=Kbc[\mathrm{OH^{-}}] = \sqrt{K_{b}c}. NH3\mathrm{NH_{3}}: Kb=1.8×10−5K_{b} = 1.8 \times 10^{-5}. Conjugate pairs. Ka×Kb=KwK_{a} \times K_{b} = K_{w}, so pKa+pKb=14\text{p}K_{a} + \text{p}K_{b} = 14 at 298 K298\,\mathrm{K}. For NH4+\mathrm{NH_{4}^{+}}, Ka=10−141.8×10−5=5.6×10−10K_{a} = \frac{10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}. Factors affecting acid strength. Along a period, acid strength of hydrides rises with electronegativity (CH4<NH3<H2O<HF\mathrm{CH_{4} < NH_{3} < H_{2}O < HF}). Down a group, it rises with atomic size, because the H-X bond weakens (HF<HCl<HBr<HI\mathrm{HF < HCl < HBr < HI}). For oxoacids, more oxygen atoms on the central atom stabilise the conjugate base (HClO<HClO2<HClO3<HClO4\mathrm{HClO < HClO_{2} < HClO_{3} < HClO_{4}}). Polyprotic acids. Ionise in steps with Ka1≫Ka2≫Ka3K_{a1} \gg K_{a2} \gg K_{a3}, because removing a proton from a negative ion is harder. For H2S\mathrm{H_{2}S}: Ka1≈1.0×10−7K_{a1} \approx 1.0 \times 10^{-7}, Ka2≈1.3×10−13K_{a2} \approx 1.3 \times 10^{-13}. Practically all H+\mathrm{H^{+}} comes from the first step, and in a solution of the acid alone, [A2−]≈Ka2[\mathrm{A^{2-}}] \approx K_{a2}. Common ion effect. Adding a strong electrolyte that supplies an ion already produced by a weak electrolyte suppresses the weak one's ionisation. With CH3COONa\mathrm{CH_{3}COONa} present at concentration ss, [H+]=Kacs[\mathrm{H^{+}}] = \frac{K_{a}c}{s}. Adding HCl\mathrm{HCl} to H2S\mathrm{H_{2}S} lowers [S2−][\mathrm{S^{2-}}], which is used in qualitative analysis to precipitate only the least soluble sulfides (group II). Key Takeaways: - Ka=cα21−α≈cα2K_{a} = \frac{c\alpha^{2}}{1 - \alpha} \approx c\alpha^{2}; α=Kac\alpha = \sqrt{\frac{K_{a}}{c}}; [H+]=Kac[\mathrm{H^{+}}] = \sqrt{K_{a}c}. - pH=12(pKa−log⁡c)\text{pH} = \frac{1}{2}(\text{p}K_{a} - \log c) for a weak acid. - KaKb=KwK_{a}K_{b} = K_{w} for a conjugate pair. - Ka1≫Ka2K_{a1} \gg K_{a2}; [A2−]≈Ka2[\mathrm{A^{2-}}] \approx K_{a2} for H2A\mathrm{H_{2}A} alone. - Pitfall: dilution increases α\alpha but decreases [H+][\mathrm{H^{+}}] and raises pH. Brain-Teaser 1: Find the pH and percent ionisation of 0.1 M0.1\,\mathrm{M} acetic acid (Ka=1.8×10−5K_{a} = 1.8 \times 10^{-5}). Solution: α=1.8×10−50.1=0.0134\alpha = \sqrt{\frac{1.8 \times 10^{-5}}{0.1}} = 0.0134 (1.34%1.34\%). [H+]=0.1×0.0134=1.34×10−3[\mathrm{H^{+}}] = 0.1 \times 0.0134 = 1.34 \times 10^{-3}, pH =2.87= 2.87. (The approximation is valid since α<5%\alpha < 5\%.) Brain-Teaser 2: What is the degree of ionisation of 0.001 M0.001\,\mathrm{M} acetic acid? Is the approximation valid? Solution: The approximation gives α=0.018=0.134\alpha = \sqrt{0.018} = 0.134, over 5%5\%, so solve cα21−α=Ka\frac{c\alpha^{2}}{1 - \alpha} = K_{a}: 0.001α2+1.8×10−5α−1.8×10−5=00.001\alpha^{2} + 1.8 \times 10^{-5}\alpha - 1.8 \times 10^{-5} = 0, giving α=0.125\alpha = 0.125 (12.5%12.5\%). The approximation overestimates by about 7%7\%. Brain-Teaser 3: Find [H+][\mathrm{H^{+}}] in a solution that is 0.1 M0.1\,\mathrm{M} in acetic acid and 0.1 M0.1\,\mathrm{M} in sodium acetate, and compare with Brain-Teaser 1. Solution: The acetate ion suppresses ionisation: [H+]=Ka[HA][A−]=1.8×10−5[\mathrm{H^{+}}] = \frac{K_{a}[\mathrm{HA}]}{[\mathrm{A^{-}}]} = 1.8 \times 10^{-5}, pH =4.74= 4.74. The H+\mathrm{H^{+}} concentration has fallen about 7575-fold, from 1.34×10−31.34 \times 10^{-3}.

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Weak Acids and Bases: Ka, Kb, Ostwald's Dilution Law and Polyprotic Acids — DPP

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