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 0.87 mol L^-1 sodium cyanide (NaCN)? A)0.06 B)13.94 C)4.2 × 10^-3 D)2.38 E)11.62

Question

Quizplus · Accepted Answer

First, we need to write the chemical equation for the dissociation of NaCN in water:

NaCN + H2O ⇌ Na+ + CN- + H2O

NaCN is a strong electrolyte and will fully dissociate in water. Therefore, we can assume that the [CN-] concentration is equal to the initial NaCN concentration, which is 0.87 M.

The HCN produced by the reaction will react with water to form H3O+ and CN-, according to the following equation:

HCN + H2O ⇌ H3O+ + CN-

We can assume that x mol/L of the HCN will dissociate. Then, the equilibrium concentrations will be:

[HCN] = (0.87 - x) mol/L
[H3O+] = x mol/L
[CN-] = 0.87 mol/L + x mol/L

The acid dissociation constant for HCN is:

Ka = [H3O+][CN-]/[HCN]

Substituting the equilibrium concentrations and the Ka value, we get:

4.9 × 10^-10 = x^2 / (0.87 - x)

Solving for x using the quadratic formula, we get:

x = 2.8 × 10^-6 M

Therefore, [H3O+] = 2.8 × 10^-6 M and the pH of the solution is:

pH = -log[H3O+] = 11.62

The Acid-Dissociation Constant of Hydrocyanic Acid (HCN)at 25 10−1010 ^ { - 10 }10−10

The Acid-Dissociation Constant of Hydrocyanic Acid (HCN)at 25 $10 ^ { - 10 }$