Question 12.EP.14: The Ka of acetic acid, CH3COOH, is 1.8 × 10^−5. Calculate th...

The initial concentration of acetic acid is 0.10 M and those of the ions are zero, leading to the following equilibrium table:

CH_{3}COOH(aq)+H_{2}O(l)\rightleftarrows H_{3}O^{+}(aq)+CH_{3}COO^{-}(aq)

Write the equilibrium constant expression, and substitute values from the table:

\                                K_{a}=\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]}

K_{a}=\frac{(x)(x)}{(0.10-x)}=\frac{x^{2}}{(0.10-x)}=1.8\times 10^{-5}

We should expect the extent of dissociation to be small for a weak acid. So we can simplify the calculation if we assume that x is small enough that (0.10 − x) ≈ 0.10. With that assumption, we can rewrite the equation above so that we will not need to use the quadratic formula:

K_{a}\cong \frac{x^{2}}{0.10}=1.8\times 10^{-5}

Solve this for x:

x² = 1.8 × 10^{−5}(0.10) = 1.8 × 10^{−6}

\                                       x = 1.3 × 10^{−3} M

Therefore, [H_{3}O^{+}] = [CH_{3}COO^{−}] = 1.3 × 10^{−3} M. Because we now know [H_{3}O^{+}], it is easy to find the pH:

pH = −log (1.3 × 10^{−3}) = 2.9

Analyze Your Answer
We should confirm that the assumption we made was reasonable. We assumed that (0.10 − x) ≈ 0.10. So we can test this by subtracting the value obtained for x from the initial concentration of CH_{3}COOH.

0.10 M − 1.3 × 10^{−3} M = 0.0987 ≈ 0.10 M

Our initial concentration was given only to two significant figures, so our simplifying approximation is reasonably valid.