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.


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.