Question

$50ml0.1MNaOH$ is added to $50ml$ of $0.1M$ $C{H}_{3}COOH$ solution, The $pH$ will be: (pKa ${CH}_{3}COOH=4.7447$, Given $\log 20=1.3$)

• A. $4.7447$
• B. $9.2553$
• C. $8.7218$
• D. $noneoftheabove$.

Answer 1

The correct option is D $noneoftheabove$.

$NaOH+C{H}_{3}COOH\to C{H}_{3}COONa+H_{2}O$

no. of mili equivalents = molarity $\times$ volume $\times$ n-factor

Meq. of $NaOH=0.1\times 60\times 1=6$

Meq. of $C{H}_{3}COOH=0.1\times 50\times 1=5$

$NaOH+C{H}_{3}COOH\to C{H}_{3}COONa+H_{2}O$

initial $6$ $5$

final $1$ $0$ $5$ $5$

In solution, we have strong base. so salt $C{H}_{3}COONa$ is neglected.

(SO) $BOH=-\log \left[O{H}^{-}\right]$

no. of mili equivalent of $NaOH=1$

so, $1=normality\times volume$

$1=N\times 110$

$1=N$

_____

$110$

$pOH=-\log \left[1/110\right]=2.04$

So, $pH=11.96$