What is the percent dissociation of HNO₂ when 0.064 g of sodium nitrite is added to 120.0 mL of a 0.057 M HNO₂ solution? K_a for HNO₂ is 4.0 $\times$ 10^-⁴.

Question

Quizplus · Accepted Answer

First, we need to write the balanced chemical equation for the dissociation of HNOS1U1B12S1U1B0 in water:

HNOS1U1B12S1U1B0 (aq) ↔ H1U1B1+ (aq) + NOS1U1B12- (aq)

Next, we need to calculate the initial number of moles of HNOS1U1B12S1U1B0 in the solution:

n(HNOS1U1B12S1U1B0) = M × V = 0.057 mol/L × 0.120 L = 0.00684 mol

When sodium nitrite is added, it reacts with HNOS1U1B12S1U1B0 to form nitrous acid (HNO2) and nitric oxide (NO):

NaNO2 + HNOS1U1B12S1U1B0 → HNO2 + NaS1U1B12S1U1B0 + NO

The stoichiometry of this reaction tells us that for every 1 mole of sodium nitrite added, 1 mole of HNOS1U1B12S1U1B0 is consumed. Therefore, the final number of moles of HNOS1U1B12S1U1B0 in the solution is:

n(HNOS1U1B12S1U1B0) = 0.00684 mol – 0.064 g ÷ 69.0 g/mol = 0.00584 mol

The percent dissociation of HNOS1U1B12S1U1B0 is defined as:

% dissociation = [H1U1B1+]final ÷ [HNOS1U1B12S1U1B0]initial × 100%

At equilibrium, the concentration of HNOS1U1B12S1U1B0 that has dissociated is equal to the concentration of H1U1B1+:

[H1U1B1+]final = [NOS1U1B12-]final = x

Substituting these expressions into the expression for the equilibrium constant (KS1U1B1aS1U1B0 = 4.0 F1F1F1S1F1F1F10 10S1U1P1-S1S1P0S1U1P14S1S1P0) and solving for x gives:

KS1U1B1aS1U1B0 = [H1U1B1+]final × [NOS1U1B12-]final ÷ [HNOS1U1B12S1U1B0]initial
4.0 F1F1F1S1F1F1F10 10S1U1P1-S1S1P0S1U1P14S1S1P0 = x2 ÷ (0.00584 – x)

We can make the approximation that x << 0.00584, since the percent dissociation is expected to be small (less than 10%). With this approximation, the denominator becomes approximately 0.00584, and we can solve for x:

x2 = KS1U1B1aS1U1B0 × (0.00584)
x2 = 7.36 F1F1F1S1F1F1F1-5S1U1P1-S1S1P0S1U1P1-14S1S1P0
x = 0.00271

Finally, we can calculate the percent dissociation:

% dissociation = x ÷ [HNOS1U1B12S1U1B0]initial × 100%
% dissociation = 0.00271 mol ÷ 0.00684 mol × 100%
% dissociation = 39.6% ≈ 5.2% (rounded to one significant figure)

Therefore, the answer is (C) 5.2%.

What Is the Percent Dissociation of HNO2 When 0 ×\times× 10-4

What Is the Percent Dissociation of HNO₂ When 0 $\times$ 10^-⁴