Finding pH of a buffer solution on addition of hydrogen iodide to it Question:


  A buffer with $\mathrm{pH} = 4.88$ was prepared by dissolving $\pu{0.10 mol}$ of benzoic acid ($K_\mathrm{a} = \pu{6.3E-5}$) and $\pu{0.50 mol}$ of sodium benzoate in sufficient pure water to form $\pu{1.00 L}$ solution. To a $\pu{70.0  mL}$ aliquot of this solution was added $\pu{2.00 mL}$ of $\pu{2.00 M}$ $\ce{HI}$ solution. What was the $\mathrm{pH}$ of the new $\pu{72.0 mL}$ solution?


My solution:

Using the Henderson–Hasselbalch equation

$$4.88 = -\log(6.3 \cdot 10^{-5}) + \log \left( \frac{0.10-x}{x} \right),$$

$$x = \pu{0.0173 mol}$$

Amount of conjugate base in $\pu{70.0 mL}$ is 

$$0.0173 \times 70/1000 = \pu{1.211e-3 mol}$$

Amount of $\ce{HI}$ added is $\pu{4e-3 mol}$.

I'm stuck here having more $\ce{H+}$ than my base.
Amount of acid added: 
$$0.002\ \mathrm L \times 0.2\ \mathrm{mol/L} = 0.004\ \mathrm{mol}$$

This will neutralize $0.004\ \mathrm{mol}$ of the benzoate: $$0.50\ \mathrm{mol} - 0.004\ \mathrm{mol} = 0.496\ \mathrm{mol}$$ of benzoate

This will increase benzoic acid by $0.004\ \mathrm{mol}$: $$0.10\ \mathrm{mol} + 0.004\ \mathrm{mol} = 0.104\ \mathrm{mol}$$ of benzoic acid

$$[\ce{H+}] = K_\mathrm a\left(\frac{n(\ce{HA})}{n(\ce{A-})}\right)$$ (yes, it's supposed to be $[\ce{HA}]/[\ce{A-}]$, but volume cancels out leaving amount of substance so long as your numbers leave you inside the titration curve between the starting $\mathrm{pH}$ and the equivalence point $\mathrm{pH}$)

$$[\ce{H+}] = 6.3 \times 10^{-5} \left(\frac{0.104\ \mathrm{mol}}{0.496\ \mathrm{mol}}\right) = 1.32 \times 10^{-5}$$

$$-\log\left(1.32 \times 10^{-5}\right) = \mathrm{pH} = 4.88$$
(actually $4.8794$, so you have a small change in $\mathrm{pH}$, as you would expect with a buffer with so much of the conjugate base in solution, but not enough to make a significant difference.)

I'm not at all clear why people are married to the Henderson–Hasselbalch equation for buffer problems. The equilibrium expression covers just about any situation I can think of, there's no formula to remember if you know how to do an equilibrium expression.