What is the pH of 0.010 M aqueous hypochlorous acid? (K_a of HOCl = 3.5 × 10^-8) A) 2.00 B) 4.50 C) 4.73 D) 6.54 E) 7.45

Question

Quizplus · Accepted Answer

The equation for the dissociation of hypochlorous acid (HOCl) in water is:

HOCl + H2O ⇌ H3O+ + OCl-

From this equation, we can see that when HOCl dissociates, it produces one hydronium ion (H3O+) and one hypochlorite ion (OCl-).

The acid dissociation constant (Ka) for this reaction is given as:

Ka = [H3O+][OCl-] / [HOCl]

We are given that the Ka value for HOCl is 3.5 × 10-8. Since we are dealing with a weak acid, we can assume that the equilibrium concentration of the acid (HOCl) is equal to the initial concentration of the acid ([HOCl]eq = [HOCl]0). Thus:

Ka = [H3O+][OCl-] / [HOCl]0

We are also given that the initial concentration of HOCl is 0.010 M. At equilibrium, some of this will have reacted to form H3O+ and OCl-. Let x be the equilibrium concentration of H3O+. Then:

[H3O+] = x

[OCl-] = x

[HOCl]eq = [HOCl]0 - x

Substituting these values into the Ka expression and solving for x gives:

3.5 × 10-8 = x2 / (0.010 - x)

Assuming x is small compared to 0.010, we can simplify this expression to:

3.5 × 10-8 ≈ x2 / 0.010

Solving for x gives:

x ≈ 1.3 × 10-4

So the concentration of H3O+ (and also OCl-) at equilibrium is 1.3 × 10-4 M.

The pH of the solution is given by:

pH = -log[H3O+]

pH = -log(1.3 × 10-4)

pH ≈ 4.73

Therefore, the correct answer is C: 4.73.

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