The pH of a solution of a 0.15 M solution of HOCl is 4.14. What is the Ka for HOCl?

Question

BlackTom AI · Accepted Answer

To tackle this problem, I first extract the key data: a monoprotic weak acid HOCl with initial concentration C = 0.15 M, and the solution pH = 4.14.

Option 1: 5.7 × 10−2 (0.057). This value is far too large for an acid of this strength at 0.15 M; such a large Ka would produce a much stronger acid than HOCl under these conditions, and would yield a pH much lower than 4.14. Therefore this is not plausible.

Option 2: 4.8 × 10−4 (0.00048). This is several orders of magnitude larger than the calculated Ka for HOCl at this concentration. If Ka were this big, the H+ concentration would be around sqrt(Ka*C) ≈ sqrt(0.00048×0.15) ≈ sqrt(7.2e-5) ≈ 0.0085 M, giving a pH around 2.07, not 4.14. Hence this is incorrect.

Option 3: 8.8 × 10−3 (0.0088). Again, this is too large. Using the same estimation as above would give a pH significantly lower than 4.14, so this Ka value does not fit the given pH.

Option 4: 3.5 × 10−8. This Ka is very small, consistent with a weak acid like HOCl. To verify, compute [H+] from the pH: [H+] = 10^−4.14 ≈ 7.2×10^−5 M. For a weak acid, [H+] ≈ sqrt(Ka × C) when x ≪ C, so Ka ≈ [H+]^2 / C ≈ (7.2×10^−5)^2 / 0.15 ≈ 5.18×10^−9 / 0.15 ≈ 3.5×10^−8. This aligns with the provided Ka value, supporting its correctness.

Option 5: 7.2 × 10−5 (0.000072). This is simply the numeric value of [H+] derived from the pH, not the acid dissociation constant Ka. It represents the hydrogen ion concentration, not the equilibrium constant, so it cannot be the Ka for HOCl.

In summary, the pH tells us [H+] ≈ 7.2×10^−5 M, and using Ka = [H+]^2 / (C − [H+]) with C = 0.15 M yields Ka ≈ 3.5×10^−8, which matches Option 4. The other options either overestimate or misinterpret the quantities involved, leading to inconsistent pH predictions with the given data.

The pH of a solution of a 0.15 M solution of HOCl is 4.14. What is the Ka for HOCl?单项选择题

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