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Difficulty: HardRelative Strength and Ionization of Acids and Bases

An aqueous solution of hydrocyanic acid (HCN\text{HCN}), a weak monobasic acid, has a concentration of 0.40 mol dm−30.40\text{ mol dm}^{-3} at 25∘C25^\circ\text{C}. Given that the acid dissociation constant (KaK_a) for HCN\text{HCN} is 4.9×10−10 mol dm−34.9 \times 10^{-10}\text{ mol dm}^{-3}, calculate the hydrogen ion concentration, [H+][\text{H}^+], in mol dm−3\text{mol dm}^{-3}. Express your answer as the coefficient AA in the form A×10−5 mol dm−3A \times 10^{-5}\text{ mol dm}^{-3}.

Answer: 1.4

Answer

The coefficient A is 1.4, which corresponds to a hydrogen ion concentration of 1.4×10−5 mol dm−31.4 \times 10^{-5}\text{ mol dm}^{-3}.
For a weak monobasic acid, the hydrogen ion concentration is determined using the weak acid ionization relationship [H+]=Ka⋅c[\text{H}^+] = \sqrt{K_a \cdot c}. Substituting Ka=4.9×10−10 mol dm−3K_a = 4.9 \times 10^{-10}\text{ mol dm}^{-3} and c=0.40 mol dm−3c = 0.40\text{ mol dm}^{-3} yields [H+]=1.96×10−10=1.4×10−5 mol dm−3[\text{H}^+] = \sqrt{1.96 \times 10^{-10}} = 1.4 \times 10^{-5}\text{ mol dm}^{-3}, giving a coefficient of 1.4.

Step-by-Step Solution

1
Write the ionization reaction and equilibrium constant expression
HCN(aq)⇌H(aq)++CN(aq)−\text{HCN}_{(aq)} \rightleftharpoons \text{H}^+_{(aq)} + \text{CN}^-_{(aq)}, giving Ka=[H+][CN−][HCN]K_a = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]}
Hydrocyanic acid is a weak monobasic acid that ionizes partially in water.
2
Apply weak acid approximations
Since [H+]=[CN−][\text{H}^+] = [\text{CN}^-] and KaK_a is extremely small, [HCN]≈c=0.40 mol dm−3[\text{HCN}] \approx c = 0.40\text{ mol dm}^{-3}. Thus, Ka=[H+]2cK_a = \frac{[\text{H}^+]^2}{c}.
The negligible ionization degree allows the equilibrium concentration of un-ionized acid to be approximated as its initial concentration.
3
Substitute values and solve for [H+][\text{H}^+]
[H+]=Ka×c=4.9×10−10×0.40=1.96×10−10=1.4×10−5 mol dm−3[\text{H}^+] = \sqrt{K_a \times c} = \sqrt{4.9 \times 10^{-10} \times 0.40} = \sqrt{1.96 \times 10^{-10}} = 1.4 \times 10^{-5}\text{ mol dm}^{-3}
Multiplying KaK_a by the molar concentration gives the square of the hydrogen ion concentration.
4
Determine the coefficient A
A=1.4A = 1.4
Matching 1.4×10−5 mol dm−31.4 \times 10^{-5}\text{ mol dm}^{-3} to the requested standard scientific notation form A×10−5 mol dm−3A \times 10^{-5}\text{ mol dm}^{-3} yields A=1.4A = 1.4.

Key Concept

Weak Acid Ionization Equilibrium and Ka Calculations
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Topics

Relative Strength and Ionization of Acids and Bases

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