Question

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

• A. $pH=3.56$, Volume = 5.86 mL
• B. $pH=3.56$, Volume = 4.86 mL
• C. $pH=4.56$, Volume = 4.86 mL
• D. $pH=4.56$, Volume = 3.86 mL

Answer 1

The correct option is C $pH=4.56$, Volume = 4.86 mL

Number of millimoles = Molarity $\times$ Volume (mL)

$\therefore$ millimoles of $NaOH=0.2\times 20=4$

Millimoles of $C{H}_{3}COOH=0.2\times 50=10$

Millimoles of $C{H}_{3}COONa$ produced $=4$

$\therefore$ millimoles of $C{H}_{3}COOH$ remained $=10-4=6$

Now,
$pH=p{K}_a+log\frac{\text{millimoles of salt}}{\text{millimoles of acid}}$$=-log(1.8\times {10}^{-5})+log\frac{4}{6}$ $=4.56$

Let, the additional volume of $NaOH$ be $\upsilon$ mL to make the pH 4.74.
Millimoles of $NaOH$ added $=0.2\upsilon$

Since, $0.2\upsilon$ mmoles of $NaOH$ will neutralise $0.2\upsilon$ mmoles of $C{H}_{3}COOH$ and form $0.2\upsilon$ mmoles of $C{H}_{3}COONa$.

millimoles of $C{H}_{3}COOH=(6-0.2\upsilon )$

millimoles of $C{H}_{3}COONa=(4+0.2\upsilon )$

Hence,
$pH=p{K}_a+log\frac{4+0.2\upsilon }{6-0.2\upsilon }$

$4.74=-log(1.86\times {10}^{-5})+log\frac{4+0.2\upsilon }{6-0.2\upsilon }$

$4.74=4.7447+log\frac{4+0.2\upsilon }{6-0.2\upsilon }$

$-0.0047=log\frac{4+0.2\upsilon }{6-0.2\upsilon }$

$log\frac{6-0.2\upsilon }{4+0.2\upsilon }=0.0047$

Taking antilog, $\frac{6-0.2\upsilon }{4+0.2\upsilon }=1.011⟹\upsilon =4.86mL.$

Hence, option C is correct.