A 0.10 M solution of acetic acid (Ka = 1.8×10^-5) is prepared. What is its pH?

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Multiple Choice

A 0.10 M solution of acetic acid (Ka = 1.8×10^-5) is prepared. What is its pH?

Explanation:
Weak acids only partially ionize in water, so the hydrogen ion concentration is small and can be estimated from the acid’s Ka and its initial concentration. For a monoprotic weak acid HA with initial concentration C, Ka = x^2/(C − x) where x = [H+]. If the dissociation is small (x ≪ C), you can use x ≈ sqrt(Ka × C). Here, C = 0.10 M and Ka = 1.8 × 10^-5. So [H+] ≈ sqrt(1.8 × 10^-5 × 0.10) = sqrt(1.8 × 10^-6) ≈ 1.34 × 10^-3 M. The pH is −log10([H+]) ≈ −log10(1.34 × 10^-3) ≈ 2.87. Using the exact equation Ka = x^2/(C − x) gives a very similar value (about 0.00133 M for [H+]), yielding pH ≈ 2.87.

Weak acids only partially ionize in water, so the hydrogen ion concentration is small and can be estimated from the acid’s Ka and its initial concentration. For a monoprotic weak acid HA with initial concentration C, Ka = x^2/(C − x) where x = [H+]. If the dissociation is small (x ≪ C), you can use x ≈ sqrt(Ka × C).

Here, C = 0.10 M and Ka = 1.8 × 10^-5. So [H+] ≈ sqrt(1.8 × 10^-5 × 0.10) = sqrt(1.8 × 10^-6) ≈ 1.34 × 10^-3 M. The pH is −log10([H+]) ≈ −log10(1.34 × 10^-3) ≈ 2.87.

Using the exact equation Ka = x^2/(C − x) gives a very similar value (about 0.00133 M for [H+]), yielding pH ≈ 2.87.

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