Questions: The pH of a* 0.88 M solution of hydrocyanic acid (HCN) is measured to be 4.63. Calculate the acid dissociation constant Ka of hydrocyanic acid. Be sure your answer has the correct number of significant digits.

Solution Steps

The pH of the solution is given as 4.63. We can use the pH to find the concentration of hydrogen ions \([H^+]\) using the formula: \[ \text{pH} = -\log[H^+] \] Rearranging the formula to solve for \([H^+]\): \[ [H^+] = 10^{-\text{pH}} \] Substituting the given pH value: \[ [H^+] = 10^{-4.63} \approx 2.3442 \times 10^{-5} \, \text{M} \]

The dissociation of hydrocyanic acid (HCN) in water can be represented as: \[ \text{HCN} \leftrightarrow \text{H}^+ + \text{CN}^- \] The acid dissociation constant \(K_a\) is given by: \[ K_a = \frac{[H^+][\text{CN}^-]}{[\text{HCN}]} \]

Assume the initial concentration of HCN is 0.88 M. At equilibrium, the concentration of \(\text{H}^+\) and \(\text{CN}^-\) will both be equal to \([H^+]\), and the concentration of HCN will be reduced by \([H^+]\): \[ [\text{HCN}]_{\text{eq}} = 0.88 - [H^+] \approx 0.88 - 2.3442 \times 10^{-5} \approx 0.88 \, \text{M} \] Since \([H^+]\) is very small compared to the initial concentration of HCN, the change in \([\text{HCN}]\) is negligible.

Using the equilibrium concentrations in the expression for \(K_a\): \[ K_a = \frac{[H^+][\text{CN}^-]}{[\text{HCN}]_{\text{eq}}} = \frac{(2.3442 \times 10^{-5})(2.3442 \times 10^{-5})}{0.88} \] \[ K_a = \frac{(2.3442 \times 10^{-5})^2}{0.88} \approx \frac{5.4943 \times 10^{-10}}{0.88} \approx 6.2435 \times 10^{-10} \]

Final Answer