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Acids and Bases - Calculations involving acids and bases (HL)

Grade 11IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The ionic product of water, Kw=[H+][OH−]K_w = [H^+][OH^-], is temperature-dependent. At 298 K298\text{ K}, Kw=1.0×10−14K_w = 1.0 \times 10^{-14}. In pure water, [H+]=[OH−]=Kw[H^+] = [OH^-] = \sqrt{K_w}, which means pH=7.0pH = 7.0 only at 298 K298\text{ K}.

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For weak acids (HAHA), the acid dissociation constant is Ka=[H+][A−][HA]K_a = \frac{[H^+][A^-]}{[HA]}. For weak bases (BB), the base dissociation constant is Kb=[BH+][OH−][B]K_b = \frac{[BH^+][OH^-]}{[B]}.

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The relationship between KaK_a and KbK_b for a conjugate acid-base pair is Ka×Kb=KwK_a \times K_b = K_w, which implies pKa+pKb=pKwpK_a + pK_b = pK_w.

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Buffer solutions consist of a weak acid and its conjugate base (or a weak base and its conjugate acid). They maintain a relatively constant pHpH when small amounts of acid or base are added.

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During a titration of a weak acid with a strong base, at the half-equivalence point, [HA]=[A−][HA] = [A^-]. At this point, pH=pKapH = pK_a.

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Salt hydrolysis occurs when ions of a salt react with water. Salts from a strong acid and weak base produce acidic solutions (pH<7pH < 7), while salts from a weak acid and strong base produce basic solutions (pH>7pH > 7).

📐Formulae

pH=−log⁡10[H+]pH = -\log_{10}[H^+]

pOH=−log⁡10[OH−]pOH = -\log_{10}[OH^-]

Kw=[H+][OH−]=1.0×10−14 (at 298 K)K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \text{ (at 298 K)}

pKa=−log⁡10Ka and pKb=−log⁡10KbpK_a = -\log_{10}K_a \text{ and } pK_b = -\log_{10}K_b

pH+pOH=pKw=14.00 (at 298 K)pH + pOH = pK_w = 14.00 \text{ (at 298 K)}

[H+]=Ka×[HA]initial (for weak acids where dissociation is small)[H^+] = \sqrt{K_a \times [HA]_{initial}} \text{ (for weak acids where dissociation is small)}

pH=pKa+log⁡10([base][acid]) (Henderson-Hasselbalch equation)pH = pK_a + \log_{10}\left(\frac{[base]}{[acid]}\right) \text{ (Henderson-Hasselbalch equation)}

💡Examples

Problem 1:

Calculate the pHpH of a 0.100 mol dm−30.100\text{ mol dm}^{-3} solution of ethanoic acid (CH3COOHCH_3COOH) at 298 K298\text{ K}. Given Ka=1.8×10−5K_a = 1.8 \times 10^{-5}.

Solution:

[H+]=Ka×[CH3COOH]=1.8×10−5×0.100=1.8×10−6≈1.34×10−3 mol dm−3[H^+] = \sqrt{K_a \times [CH_3COOH]} = \sqrt{1.8 \times 10^{-5} \times 0.100} = \sqrt{1.8 \times 10^{-6}} \approx 1.34 \times 10^{-3}\text{ mol dm}^{-3}. pH=−log⁡10(1.34×10−3)≈2.87pH = -\log_{10}(1.34 \times 10^{-3}) \approx 2.87.

Explanation:

Because ethanoic acid is a weak acid, we assume the equilibrium concentration of the acid is approximately equal to its initial concentration (0.100−x≈0.1000.100 - x \approx 0.100).

Problem 2:

A buffer solution is prepared using 0.20 mol dm−30.20\text{ mol dm}^{-3} propanoic acid (pKa=4.87pK_a = 4.87) and 0.10 mol dm−30.10\text{ mol dm}^{-3} sodium propanoate. Calculate the pHpH of the buffer.

Solution:

pH=pKa+log⁡10([A−][HA])pH = pK_a + \log_{10}\left(\frac{[A^-]}{[HA]}\right) pH=4.87+log⁡10(0.100.20)=4.87+log⁡10(0.5)=4.87−0.30=4.57pH = 4.87 + \log_{10}\left(\frac{0.10}{0.20}\right) = 4.87 + \log_{10}(0.5) = 4.87 - 0.30 = 4.57.

Explanation:

The Henderson-Hasselbalch equation is used to find the pHpH of a buffer. Here, the concentration of the salt (conjugate base) is half that of the acid, making the pHpH lower than the pKapK_a.