Question

$20mL$ of $0.2M$ sodium hydroxide is added to $50mL$ of $0.2M$ acetic acid to give $70mL$ of the solution. What is the pH of the solution? Calculate the additional volume of $0.2M$ $NaOH$ required to make the pH of solution $4.74$. The ionisation constant of acetic acid is $1.8\times {10}^{-5}$.

Answer 1

No. of moles of $NaOH$ in $20mL=\frac{0.2}{1000}\times 20=0.004$
No. of moles of acetic acid in $50mL=\frac{0.2}{1000}\times 50=0.01$
When $NaOH$ is added, ${CH}_{3}COONa$ is formed
${CH}_{3}COOH+NaOH\rightleftharpoons {CH}_{3}COONa+H_{2}O$
$1mole$ $1mole$ $1mole$ $1mole$
No. of moles of ${CH}_{3}COONa$ in $70mL$ solution $=0.004$
No. of moles of ${CH}_{3}COOH$ in $70mL$ solution
$=(0.01-0.004)=0.006$
Applying Henderson's equation
$pH=\log \frac{\left[Salt\right]}{\left[Acid\right]}-\log K_a$
$=\log \frac{0.004}{0.006}-\log 1.8\times {10}^{-5}=4.5687$
On further addition of $NaOH$, the pH becomes $4.74$
$pH=\log \frac{\left[Salt\right]}{\left[Acid\right]}-\log K_a$
$pH=\log \frac{\left[Salt\right]}{\left[Acid\right]}-\log 1.8\times {10}^{-5}$
or $\log \frac{\left[Salt\right]}{\left[Acid\right]}=pH+\log 1.8\times {10}^{-5}=-0.0048$
so, $\log \frac{\left[Salt\right]}{\left[Acid\right]}=\overline{1}.9952$
$\frac{\left[Salt\right]}{\left[Acid\right]}=0.9891$
Let $x$ moles of $NaOH$ be added
$\frac{\left[Salt\right]}{\left[Acid\right]}=\frac{0.004+x}{0.006-x}=0.9891$
or $0.004+x=0.9891\times 0.006-0.9891x$
$x=0.000972mole$
Volume of $0.2M$ $NaOH$ solution having $0.000972moles$
$=\frac{1000}{0.2}\times 0.000972=4.86mL$