The acid- dissociation constant of hydrocyanic acid (HCN) at 25.0 °C is 4.9 × 10^-¹⁰. What is the pH of an aqueous solution of 0.080 M sodium cyanide (NaCN)?

Question

Quizplus · Accepted Answer

The correct answer is A because the pH of an aqueous solution of 0.080 M sodium cyanide (NaCN) can be calculated using the dissociation constant of hydrocyanic acid (HCN) and the concentration of NaCN. The dissociation constant of HCN is 4.9 × 10^-10, which means that the equilibrium constant for the dissociation of HCN is 4.9 × 10^-10. The concentration of NaCN is 0.080 M, which means that the concentration of HCN in the solution is 0.080 M. The pH of the solution can be calculated using the following equation:```pH = -log[H+],```where [H+] is the concentration of hydrogen ions in the solution. The concentration of hydrogen ions can be calculated using the following equation:```[H+] = Kw/[CN-],```where Kw is the dissociation constant of water (1.0 × 10^-14) and [CN-] is the concentration of cyanide ions in the solution. The concentration of cyanide ions can be calculated using the following equation:```[CN-] = [NaCN] - [HCN],```where [NaCN] is the concentration of sodium cyanide in the solution. Substituting the values of Kw, [NaCN], and [HCN] into the equation for [H+], we get:```[H+] = 1.0 × 10^-14/[0.080 M - 0.080 M] = 1.0 × 10^-14/0 = ∞,```which means that the pH of the solution is 0.

The Acid- Dissociation Constant of Hydrocyanic Acid (HCN) at 25