Question

The $pH$ of a solution obtained by mixing $100ml$ of $0.2MC{H}_{3}COOH$ is with $100ml$ of $0.2MNaOH$ would be:

[Note : $p{K}_a$ for $C{H}_{3}COOH=4.74$ and $\log 2=0.301)$].

• A. $4.74$
• B. $8.87$
• C. $9.10$
• D. $8.57$

Answer 1

The correct option is B $8.87$

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

Moles of $C{H}_{3}COOH=0.2\times 100\times 1=20mmol$

Moles of $NaOH=0.2\times 100\times 1=20mmol$

So there is complete neutralisation

$\left[salt\right]=C=\frac{20}{100+100}=\frac{20}{200}=0.1M$

$pH=7+\frac{1}{2}p{K}_a+\frac{1}{2}{\log }_{10}C$

$=7+\frac{1}{2}\left(4.74\right)+\frac{1}{2}\log \left(0.1\right)$

$=7+2.37-\frac{1}{2}{\log }_{10}10$

$pH=9.37-0.5=8.87$