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:
$\ce{CH3COOH &lt;=&gt; H+ + CH3COO-}$
$K_\mathrm{a} = \frac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \frac{x\cdot x}{0.2} = 1.8 \cdot 10^{-5} \rightarrow \mathrm{pH} = 2.72$
Then
$\ce{CH3COONa -&gt; Na+ + CH_3COO-}$
$\ce{CH3COO- + H2O &lt;=&gt; CH3COOH + OH-}$
\begin{align}
K_\mathrm{b} = \frac {K_\mathrm{w}}{K_\mathrm{a}} &amp;= 5.56 \cdot 10^{-10} \\
 &amp;= \frac{[\ce{OH-}][\ce{CH3COOH}]}{[\ce{CH3COO-}]}  = \frac{x^2}{0.2} \\
 \implies \mathrm{pOH} &amp;= 4.97 \\ 
\implies \mathrm{pH} &amp;= 9.02
\end{align}
Hence, $\mathrm{pH}_\text{solution}= (9.02+2.72)/2 = 5.87$
However, the result should be $4.74$.
Because the acid and conjugate bases are equimolar - and because the equilibrium constant is small, obviating the need to solve a quadratic if you approached this in another way - the Henderson-Hasselbalch equation is:
$$\pu{pH} = {\rm p}K_\mathrm{a} + {\rm log}_{10}\left({[A^-]\over[HA]}\right)$$
$$\pu{pH} = 4.74 + {\rm log}_{10}\left({0.2\ {\rm M}\over 0.2\ {\rm M}}\right)$$
which reduces to
$$\pu{pH} = 4.74$$