0
$\begingroup$

Calculate the pH of the solution that results when $\pu{40.0 mL}$ of $\pu{0.100 M}$ $\ce{CH3COOH}$ is mixed with $\pu{30.0 mL}$ of $\pu{0.200 M}$ $\ce{NaCH3COO}.$ $K_\mathrm{a}(\ce{CH3COOH}) = \pu{1.8E-5}.$

Can I use Henderson–Hasselbalch equation to solve this? It would be

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{CH3COOH}]}{[\ce{NaCH3COO}]} \quad\implies\quad \mathrm{pH} = 4.44,$$

but if I set up the ICE table

\begin{array}{rrrrr} \ce{&CH3COOH &-> &H+ &+ &CH3COO-}\\ \text{I} & 0.100 && 0 && 0.200 \\ \text{C} & -x && +x && -x \\ \text{E} & 0.100 - x && x && 0.200 + x \end{array}

and solve for $x,$ it gives me a different $\mathrm{pH}.$

The volumes threw me off and make me think that I have to find the total volume and new molarity before solving for $\mathrm{pH}.$

Improve this question
$\endgroup$
4
  • $\begingroup$ Yes, you should take into consideration the effect of dilution when you mix the two. Note however that your could also consider the molar ratio of acid and conjugate base. This should become obvious if your write the HH equation out explicitly. $\endgroup$
    – Buck Thorn ♦
    Commented May 10, 2022 at 16:11
  • $\begingroup$ Calculate the amount of acid and of acetate in moles. And then divide by the total volume. The use the HH equation. $\endgroup$
    – Maurice
    Commented May 10, 2022 at 16:29
  • $\begingroup$ Thank you, Maurice. That actually makes a lot of sense. Does my equation look right? $\endgroup$
    – Nori
    Commented May 10, 2022 at 16:44
  • 1
    $\begingroup$ Shouldn’t it be $$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{NaCH3COO}]}{[\ce{CH3COOH}]}$$? $\endgroup$
    – Oleg Farenskiy
    Commented May 11, 2022 at 9:42

0

Reset to default

Browse other questions tagged or ask your own question.