Question

$100$ml of $0.01$M $C{H}_{3}COOH$ reacts with $40$ml, $0.01$M $NaOH$ $(K_a={10}^{-5})$.
Find $pH$ when $1$ ml of $0.01$M $HCl$ is added.

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 $=2.37$

$M_1{V}_1$ are for mixture.

$M_1=0.0042$, $V_1=140$ $\Rightarrow M_1{V}_1=0.588M$

$M_2{V}_2$ are $HCl$

$H_1=0.01M$ $V_2=1ml\Rightarrow M_2{V}_2=0.01M$

$M=\frac{0.588-0.01}{140+1}=\frac{0.0578}{141}=4.099\times {10}^{-3}$

$pH=-log(4.099\times {10}^{-3})=2.387$

$pH$ of solution $=2.387$.