Calculate the pH of 0.10 M acetic acid (Ka = 1.8×10−5).

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Multiple Choice

Calculate the pH of 0.10 M acetic acid (Ka = 1.8×10−5).

Explanation:
Weak acids only partially ionize in water. For a monoprotic acid HA with Ka = [H+][A-]/[HA], if the initial concentration is C and x = [H+] = [A-], then [HA] ≈ C − x. The equality becomes Ka = x^2/(C − x). Since Ka is small compared to C, we can use the approximation x ≈ sqrt(Ka × C). Plugging in C = 0.10 M and Ka = 1.8 × 10^-5 gives Ka × C = 1.8 × 10^-6, and x ≈ sqrt(1.8 × 10^-6) ≈ 1.34 × 10^-3 M. The pH is −log10(1.34 × 10^-3) ≈ 2.87. The exact solution (solving the quadratic Ka = x^2/(C − x)) yields a value very close to this, confirming the approximation is valid for these numbers.

Weak acids only partially ionize in water. For a monoprotic acid HA with Ka = [H+][A-]/[HA], if the initial concentration is C and x = [H+] = [A-], then [HA] ≈ C − x. The equality becomes Ka = x^2/(C − x). Since Ka is small compared to C, we can use the approximation x ≈ sqrt(Ka × C).

Plugging in C = 0.10 M and Ka = 1.8 × 10^-5 gives Ka × C = 1.8 × 10^-6, and x ≈ sqrt(1.8 × 10^-6) ≈ 1.34 × 10^-3 M. The pH is −log10(1.34 × 10^-3) ≈ 2.87.

The exact solution (solving the quadratic Ka = x^2/(C − x)) yields a value very close to this, confirming the approximation is valid for these numbers.

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