Question

100ml of 0.1 M NaOH is added to 100 ml of a 0.2 M CH₃COOH solution. The pH of the resulting solution will be (pk_a=4.74)

Answer 1

Millimole of 100ml of 0.1 M NaOH=100x0.1=10 millimole.

Millimole of 100 ml of a 0.2 M CH₃COOH =100x 0.2=20 millimole.

$$\begin{gathered} \underset{\mathrm{Sodium}\mathrm{Hydroxide}}{\mathrm{NaOH}(\mathrm{aq}.)}+\underset{\mathrm{Acetic}\mathrm{acid}}{{\mathrm{CH}}_3\mathrm{COOH}(\mathrm{aq}.)}⟶\underset{\mathrm{Sodium}\mathrm{Acetate}}{{\mathrm{CH}}_3\mathrm{COONa}(\mathrm{aq}.)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)} \\ 10\mathrm{millimole}20\mathrm{millimole} \\ 010\mathrm{millimole}10\mathrm{millimole} \end{gathered}$$

Now it is a Buffer.

$\mathrm{pH}={\mathrm{pk}}_{\mathrm{a}}+\log \left[\frac{\mathrm{Salt}}{\mathrm{Acid}}\right]$

$\mathrm{pH}=4.74+\log \left[\frac{10}{10}\right]=4.74+\log \left[1\right]=4.74+0=4.74$

100ml of 0.1 M NaOH is added to 100 ml of a 0.2 M CH₃COOH solution. The pH of the resulting solution will be 4.74.