What is the hydronium ion concentration of a 0.150 M hypochlorous acid solution with
K_a = 3.5 × 10^-8? The equation for the dissociation of hypochlorous acid is:
HOCl(aq) + H₂O(l) ⇌ H₃O⁺(aq) + OCl^-(aq) .

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the dissociation of hypochlorous acid is:
KS1U1B1aS1U1B0 = [HS1U1B13S1U1B0OS1U1P1+S1S1P0][OClS1U1P1-S1S1P0]/[HOCl]
Since the initial concentration of HOCl is 0.150 M, and the concentration of the products (HS1U1B13S1U1B0OS1U1P1+S1S1P0 and OClS1U1P1-S1S1P0) is negligible in comparison, we can assume that the concentration of HOCl at equilibrium is approximately 0.150 M. Therefore, we can simplify the expression to:
KS1U1B1aS1U1B0 = [HS1U1B13S1U1B0OS1U1P1+S1S1P0][OClS1U1P1-S1S1P0]/0.150
Substituting the given value of KS1U1B1aS1U1B0 = 3.5 × 10S1U1P1-8S1S1P0, and since the concentrations of HS1U1B13S1U1B0OS1U1P1+S1S1P0 and OClS1U1P1-S1S1P0 are equal at equilibrium (let's call this concentration x), we can solve for x:
3.5 × 10S1U1P1-8S1S1P0 = xS1U1B1S1S1P0S1S1P1/0.150
x = 7.2 × 10S1U1P1-5S1S1P0M
Since x represents the concentration of both HS1U1B13S1U1B0OS1U1P1+S1S1P0 and OClS1U1P1-S1S1P0, we can use the definition of acidity to find the concentration of hydronium ions:
[H3O+] = [HS1U1B13S1U1B0OS1U1P1+S1S1P0] + [H3O+] from the dissociation of water
Assuming that [H3O+] from the dissociation of water is negligible compared to the concentration from the dissociation of HOCl, we can approximate:
[H3O+] ≈ [HS1U1B13S1U1B0OS1U1P1+S1S1P0] = x
Therefore, the hydronium ion concentration of the solution is approximately 7.2 × 10S1U1P1-5S1S1P0 M, which corresponds to choice D.

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