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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 $4.74$.

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

$$\pu{pH} = 4.74$$

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