Question

$100$ml of $0.01$M $C{H}_{3}COOH$ reacts with $40$ml of $0.01$M $NaOH$ $(K_a={10}^{-5})$. Find $pH$.

Answer 1

Molarity of ${CH}_{3}COOH$ are $M_1$ and volume $V_2$

$M_1{V}_1=100\times 0.01=1$

Molarity of $NaOH$ are $M_2$ $+$ volume $V_2$

$M_2{V}_2=40\times 0.01=0.4$

Clearly $M_1{V}_1>M_2{V}_2$

$\therefore$ Resulting solution will be acidic.

Molarity of resulting solution $M=\frac{M_1{V}_1-M_2{V}_2}{V_1+V_2}=\frac{1-0.4}{140}=0.0042$

$pH=-log\left(0.0042\right)$

$=2.37$

$\therefore$ $pH$ of solution $=2.37$