# pH of an aqueous solution with acetic acid and sodium acetate

An aqueous solution of $$\pu{1L}$$ contains $$\pu{0.2 mol}$$ of acetic acid and $$\pu{0.2 mol}$$ of sodium acetate ($$K_\mathrm{a} = 1.8 \times 10^{-5}$$). Compute the $$\mathrm{pH}$$ .

I'm trying to help my little sister in solving this exercise... however, a lot of years passed away since the time I attended the school...

Our (unsuccessful) attempt:

$$\ce{CH3COOH <=> H+ + CH3COO-}$$

$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \frac{x\cdot x}{0.2} = 1.8 \cdot 10^{-5} \rightarrow \mathrm{pH} = 2.72$$

Then

$$\ce{CH3COONa -> Na+ + CH_3COO-}$$

$$\ce{CH3COO- + H2O <=> CH3COOH + OH-}$$

\begin{align} K_\mathrm{b} = \frac {K_\mathrm{w}}{K_\mathrm{a}} &= 5.56 \cdot 10^{-10} \\ &= \frac{[\ce{OH-}][\ce{CH3COOH}]}{[\ce{CH3COO-}]} = \frac{x^2}{0.2} \\ \implies \mathrm{pOH} &= 4.97 \\ \implies \mathrm{pH} &= 9.02 \end{align}

Hence, $$\mathrm{pH}_\text{solution}= (9.02+2.72)/2 = 5.87$$

However, the result should be $$4.74$$.

• It is much simpler, if you realize that the ratio of concentrations of acetic acid and acetate anion is 1, so Ka=[H+]. Apr 30 at 17:44

$$\pu{pH} = {\rm p}K_\mathrm{a} + {\rm log}_{10}\left({[A^-]\over[HA]}\right)$$
$$\pu{pH} = 4.74 + {\rm log}_{10}\left({0.2\ {\rm M}\over 0.2\ {\rm M}}\right)$$
$$\pu{pH} = 4.74$$