Question

The $pH$ of a solution obtained by mixing $100mL$ of $0.2M$ $C{H}_{3}COOH$ with $100mL$ of $0.2MNaOH$ will be: ($p{K}_{a}forC{H}_{3}COOH=4.74$)

• A. 4.74
• B. 8.87
• C. 9.1
• D. 8.87

Answer 1

The correct option is D 8.87

$0.2MC{H}_{3}COOH0.2MNaOH$ each $100mL$ is taken here.

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

$C{H}_{3}COOH+NaOH\rightleftharpoons C{H}_{3}COONa+H_{2}O$
$20+20\rightleftharpoons 0+0$
$0+0\rightleftharpoons 20+20$
So there is complete neutralisation
$\left[salt\right]=\frac{20}{200}=0.1$

Using the formula,
$pH=7+\frac{1}{2}logp{K}_a+logC$
$pH=7+\frac{1}{2}logp{K}_a+\frac{1}{2}log\left[salt\right]$


putting the values,
$pH=7+\frac{1}{2}log\left(4.74\right)+\frac{1}{2}log\left(0.1\right)$
$pH=7+2.37-0.5$
$pH=8.87$