Calculate the percent ionization of nitrous acid in a solution that is 0.222 M in nitrous acid (HNO₂)and 0.278 M in potassium nitrite (KNO₂). The acid dissociation constant of nitrous acid is 4.50 × 10^-4. A)55.6 B)0.162 C)15.5 D)2.78 × 10^-3 E)3.448

Question

Quizplus · Accepted Answer

Step 1: Write the balanced equation for the ionization of nitrous acid: 
HNOS1U1B12S1U1B0 + H2OS1U1L1S1U1B0 ⇌ H3OS1U1L1S1U1B+ + NOS1U1L1S1U1B2-

Step 2: Write out the initial concentrations and set up an ICE (Initial, Change, Equilibrium) table:

|         | HNOS1U1B12S1U1B0 | H2OS1U1L1S1U1B0 |  H3OS1U1L1S1U1B+ | NOS1U1L1S1U1B2- |
|---------|-------------|-------------|-----------------|-------------|
|Initial  |     0.222 M |     --      |   --            |     0.278 M |
|Change   |    -x       |     +x      |   +x             |     +x      |
|Equilibrium|0.222 - x  |     x      | x             |0.278 + x |

Step 3: Write the expression for the acid dissociation constant, Ka:

Ka = [H3O+][NO2-] / [HNOS1U1B12S1U1B0]

Step 4: Plug in the equilibrium concentrations into the expression for Ka:

4.50 × 10S1U1P1-4S1S1P0 = x^2 / (0.222 - x)

Since x is much smaller than 0.222, we can assume that x ≈ [H3O+] ≈ [NO2-] ≈ 0.

4.50 × 10S1U1P1-4S1S1P0 = (0)^2 / (0.222 - 0) 
x = sqrt(Ka*[HNOS1U1B12S1U1B0]) = sqrt(4.50 × 10S1U1P1-4S1S1P0 * 0.222) = 1.62 × 10S1U1P1-3S1S1P0 M

Step 5: Calculate the percent ionization:

% Ionization = (x / [HNOS1U1B12S1U1B0]) * 100% 
% Ionization = (1.62 × 10S1U1P1-3S1S1P0 M / 0.222 M) * 100% = 0.162 * 100% = 16.2%

Therefore, the answer is B.

Calculate the Percent Ionization of Nitrous Acid in a Solution