Question

When $100\text{mL}$ of $0.4MC{H}_{3}COOH$ is mixed with $100\text{mL}$ of $0.2MNaOH$, the approximate value of $\left[H_3{O}^{+}\right]$ in the solution mixture is? $\left[K_a\right(C{H}_{3}COOH)=1.8\times {10}^{-5}]$

• A. $1.8\times {10}^{-6}$
• B. $1.8\times {10}^{-5}$
• C. $9\times {10}^{-6}$
• D. $9\times {10}^{-5}$

Answer 1

The correct option is B $1.8\times {10}^{-5}$

When volume of solution is doubled, the concentration reduced to half.
Thus 100 mL of $C{H}_{3}COOH$ and 100 mL of $NaOH$ added to give 200mL of solution of 0.2 M $C{H}_{3}COOH$ and 0.1 M $NaOH$.

$\begin{array}{ccccccc}C{H}_{3}COOH& +& NaOH& \to & C{H}_{3}COONa& +& H_{2}O\\ 40mmol& & 20mmol& & 0& & \\ Afterreaction,& & & & & & \\ 20mmol& & 0& & 20mmol& & \end{array}$

$\left[C{H}_{3}COOH\right]=\frac{20}{200}=0.1M,\left[C{H}_{3}COONa\right]=\frac{20}{200}=0.1M$
$pH=p{K}_a+\text{log}\frac{\text{[salt]}}{\text{[acid]}}$
$pH=p{K}_a$
$\left[H^{+}\right]=K_a=1.8\times {10}^{-5}$