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)? A) 11.11 B) 2.89 C) 1.3 × 10^-3 D) 7.8 × 10^-12 E) 3.9 × 10^-11

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Quizplus · Accepted Answer

Sodium cyanide (NaCN) dissociates in water to form Na+ and CN-. The CN- ion is the conjugate base of HCN, and it will hydrolyze (react with water) to form HCN and OH-. The pH of the solution can be calculated using the Kb of CN- (which can be found using Kw = Ka * Kb) and the initial concentration of NaCN. The calculation involves finding the concentration of OH- produced and then converting that to pH. Given the acid-dissociation constant (Ka) of HCN is 4.9 × 10^-10, the Kb for CN- can be calculated and used to find the pH of the solution, which is 11.11.

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