Question

$5.0mL$ of $0.1MNaOH$ solution is added to $50mL$ of the $0.1M$ acetic acid solution. Calculate the $pH$ of the resulting acetic acid solution.

$(Ka=1.8\times {10}^{-5})$

Answer 1

$C{H}_{3}COOH+H_{2}O⟷C{H}_{3}CO{O}^{-}+H_3{O}^{+}$
$K_a=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C{H}_{3}COOH\right]}$
Let $\left[H_3{O}^{+}\right]=\left[C{H}_{3}CO{O}^{-}\right]=x$
$K_a=\frac{x^2}{0.1M}{M}^2$
After adding $5㎖$ of $NaOH$
Moles of $C{H}_{3}COOH=0.050\ell \times 0.1㏖/\ell =0.005㏖$
Moles of $NaOH$ added $=0.0050\ell \times 0.1㏖/\ell =0.0005㏖$
$C{H}_{3}COOH+O{H}^{-}⟶C{H}_{3}CO{O}^{-}+H_{2}O$
After adding $NaOH,0.0005㏖$ of $C{H}_{3}CO{O}^{-}$
and $0.0045㏖$ of $C{H}_{3}COOH$ is left.
$K_a=1.8\times {10}^{-5}=\frac{\left[H_3{O}^{+}\right]\left[0.0005\right]}{\left[0.0045\right]}$
$\left[H_3{O}^{+}\right]=0.000162M$
$pH=-\log \left(0.000162\right)$
$=3.79$