Question

Calculate the pH when 60 mL of 1.25 M NaOH is added to 977 mL of 1.80 M of acetic acid.

Please show full calculations. Thanks! :)

Answer 1

The acid-base reaction between NaOH and CH₃COOH is given by the equation:

$NaOH(aq) + CH_3COOH(aq) \rightarrow CH_3COONa(aq) + H_2O(l)$

The pH of the medium depend on the volume of NaOH added to the solution:

- :

$pH = \frac{pK_a- log[CH_3COOH]_0}{2}$

:

$pH = pK_a+ log\left ( \frac{[CH_3COO^-]}{{CH_3COOH}} \right )= pK_a+ log\left ( \frac{M_{NaOH}\times V_{NaOH}^{added}}{M_{CH_3COOH}\times V_{CH_3COOH} -M_{NaOH}\times V_{NaOH}^{added}} \right )$

$pH = \frac{14+pK_a+ log \frac{M_{CH_3COOH}\times V_{CH_3COOH}}{V_{CH_3COOH}+ V_{NaOH}^{Equivalence}}}{2}$

$pH = 14 + log[OH^-]= 14+ log\left ( \frac{M_{NaOH}\times V_{NaOH}^{added}-M_{NaOH}\times V_{NaOH}^{Equivalence}}{V_{CH_3COOH} +V_{NaOH}^{added}} \right )$

At the equivalence pH:

$n_{NaOH}=n_{CH_3COOH}$

$\Rightarrow M_{NaOH}\times V_{NaOH}^{Equivalence}=M_{CH_3COOH}\times V_{CH_3COOH}$

$\Rightarrow V_{NaOH}^{Equivalence}=\frac{M_{CH_3COOH}\times V_{CH_3COOH}}{ M_{NaOH}}=\frac{1.80 M\times 977 mL}{1.25 M}=1406.88 mL$

$\Rightarrow V_{NaOH}^{Added}=60 mL< V_{NaOH}^{Equivalence}$

$\Rightarrow pH = pK_a+ log\left ( \frac{M_{NaOH}\times V_{NaOH}^{added}}{M_{CH_3COOH}\times V_{CH_3COOH} -M_{NaOH}\times V_{NaOH}^{added}} \right )=4.75 + log\frac{1.25\times 60}{(1.80\times 977-1.25\times 60)}=3.4$