K_a for HCN is 4.9 × 10^-10. What is the pH of a 0.068 M aqueous solution of sodium cyanide?

Question

Quizplus · Accepted Answer

Sodium cyanide (NaCN) dissociates in water to form Na+ and CN-. CN- is the conjugate base of the weak acid HCN. The Kb for CN- can be calculated using the relation Kw = Ka * Kb, where Kw = 1.0 × 10^-14. Solving for Kb gives Kb = Kw / Ka = 1.0 × 10^-14 / 4.9 × 10^-10 = 2.04 × 10^-5. Using the Kb expression and the initial concentration of CN-, the hydroxide ion concentration [OH-] can be calculated, and then the pOH can be found. Finally, the pH is calculated as 14 - pOH, which is approximately 11.07.

Ka for HCN Is 4

K_a for HCN Is 4