Questions: What is the pH of a 0.0620 M solution of hydrocyanic acid, HCN(Ka=4.9 × 10^-10)?

Solution Steps

Hydrocyanic acid (HCN) dissociates in water according to the following equation: \[ \mathrm{HCN} \leftrightarrow \mathrm{H}^+ + \mathrm{CN}^- \]

The acid dissociation constant \( K_a \) for hydrocyanic acid is given by: \[ K_a = \frac{[\mathrm{H}^+][\mathrm{CN}^-]}{[\mathrm{HCN}]} \] Given \( K_a = 4.9 \times 10^{-10} \) and the initial concentration of HCN is 0.0620 M.

Let \( x \) be the concentration of \(\mathrm{H}^+\) and \(\mathrm{CN}^-\) ions at equilibrium. The change in concentration of HCN will be \( -x \). Therefore, at equilibrium: \[ [\mathrm{H}^+] = x, \quad [\mathrm{CN}^-] = x, \quad [\mathrm{HCN}] = 0.0620 - x \]

Substitute the equilibrium concentrations into the \( K_a \) expression: \[ 4.9 \times 10^{-10} = \frac{x \cdot x}{0.0620 - x} \] Since \( K_a \) is very small, we can assume \( x \ll 0.0620 \), so \( 0.0620 - x \approx 0.0620 \). This simplifies the equation to: \[ 4.9 \times 10^{-10} = \frac{x^2}{0.0620} \]

Solve for \( x \): \[ x^2 = 4.9 \times 10^{-10} \times 0.0620 \] \[ x^2 = 3.038 \times 10^{-11} \] \[ x = \sqrt{3.038 \times 10^{-11}} \] \[ x = 5.514 \times 10^{-6} \] Thus, the concentration of \(\mathrm{H}^+\) ions is \( 5.514 \times 10^{-6} \) M.

The pH is calculated using the formula: \[ \text{pH} = -\log [\mathrm{H}^+] \] \[ \text{pH} = -\log (5.514 \times 10^{-6}) \] \[ \text{pH} \approx 5.258 \]

Final Answer