Calculate the percent ionization of nitrous acid in a solution that is 0.210 mol L^-1 in nitrous acid (HNO₂)and 0.290 mol L^-1 in potassium nitrite (KNO₂). The acid dissociation constant of nitrous acid is 4.50 × 10^-4. A)58.0 B)0.154 C)16.6 D)2.74 × 10^-3 E)1.55

Question

Quizplus · Accepted Answer

To solve this problem, we need to use the equilibrium constant expression for the ionization of nitrous acid:

K = [H+][NO2-]/[HNO2]

We also know that the initial concentration of nitrous acid is 0.210 mol and its acid dissociation constant is 4.50 × 10^-4. Let's assume that x mol of nitrous acid ionizes, so the equilibrium concentrations of the species are:

[HNO2] = 0.210 - x
[H+] = x
[NO2-] = 0.290 + x

Substituting these values into the equilibrium constant expression, we get:

4.50 × 10^-4 = x*(0.290 + x)/(0.210 - x)

Simplifying the equation and solving for x, we get:

x = 0.0154 mol

The percent ionization is:

(0.0154 mol/[HNO2]0) x 100% = 0.154%

Therefore, the correct answer is B.

Calculate the Percent Ionization of Nitrous Acid in a Solution