Question

A $20mL$ of $0.02MC{H}_{3}COOH$ is titrated with $0.2MNaOH$. Find the $pH$ of the solution when $80mL$ of $NaOH$ has been added.
Given $p{K}_a\left(C{H}_{3}COOH\right)=4.74$

• A. 13.2
• B. 12.2
• C. 0.8
• D. 1.8

Answer 1

The correct option is A 13.2

mmol of $C{H}_{3}COOH=20\times 0.02=0.4$
mmol of $NaOH=80\times 0.2=16$

$C{H}_{3}COOH\left(aq\right)+NaOH\left(aq\right)\to C{H}_{3}COONa\left(aq\right)+H_{2}O\left(l\right)Initially:0.41600Final:015.60.40.4$
$\text{Final Concentration}=\frac{\text{Moles}}{\text{Total Volume}}\left[NaOH\right]=\frac{15.6}{100}=0.156M\left[C{H}_{3}COONa\right]=\frac{0.4}{100}=0.004M$
Since concentration of $NaOH$ is fairly high in comparsion to $\left[C{H}_{3}COONa\right]$, we can neglect the concentration of $O{H}^{-}$ that comes from hydrolysis of the salt $\left[C{H}_{3}COONa\right]$
All the concentration of $O{H}^{-}$ can be assumed from $NaOH$ only.
$\left[NaOH\right]=\left[O{H}^{-}\right]=0.156MpOH=-log\left[O{H}^{-}\right]pOH=-log\left(0.156\right)=0.80pH=14-0.8=13.2$