Find the pH of a 0.350 M aqueous benzoic acid solution. For benzoic acid, Ka = 6.5
×
10–5.

Question

Find the pH of a 0.350 M aqueous benzoic acid solution. For benzoic acid, Ka = 6.5 
×
 10–5.

BlackTom AI · Accepted Answer

We are given a 0.350 M solution of a weak acid, benzoic acid, with Ka = 6.5 × 10^-5. To find the pH, we treat it as a monoprotic weak acid and use the approximation [H+] ≈ sqrt( Ka × C ) when the dissociation is small (x << C).

Option-by-option analysis:
- Option: 4.64
This would correspond to [H+] ≈ 10^-4.64 ≈ 2.29 × 10^-5 M. Such a small [H+] would imply a very weak acidity and an effectively higher pH, not matching the given Ka and concentration. At the observed C and Ka, the acid should yield a considerably larger [H+], so this option is inconsistent with the data.

- Option: 2.32 (the correct answer)
Compute [H+] ≈ sqrt( Ka × C ) = sqrt( (6.5 × 10^-5) × 0.350 ) = sqrt(2.275 × 10^-5) ≈ 4.77 × 10^-3 M. Then pH = -log10(4.77 × 10^-3) ≈ 2.32. This aligns with the expected result for a 0.350 M solution of a moderately weak acid with Ka in the 10^-5 range.

- Option: 11.68
pH values around 11–12 would indicate a strongly basic solution, not an acidic benzoic acid solution. Clearly, a benzoic acid solution cannot have a pH in the acidic/basic high range like 11.68, so this is erroneous.

- Option: 4.19
A pH of 4.19 corresponds to [H+] ≈ 6.5 × 10^-5 M. If we plug this into the acid dissociation expression for a weak acid, Ka ≈ [H+]^2 / (C − [H+]) ≈ (6.5 × 10^-5)^2 / (0.350 − 6.5 × 10^-5) ≈ 4.2 × 10^-9 / 0.349935 ≈ 1.2 × 10^-8, which is far smaller than the given Ka of 6.5 × 10^-5. This indicates that pH 4.19 would not be consistent with the provided Ka and concentration, so this choice is incorrect.

In summary, the calculation using the standard weak-acid approximation yields pH ≈ 2.32, consistent with option 2.32 and inconsistent with the remaining choices due to mismatches with Ka and the resulting [H+].

Find the pH of a 0.350 M aqueous benzoic acid solution. For benzoic acid, Ka = 6.5 × 10–5.

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