Question

$500 \,ml$ of $0.2 \,M \,HCl$ is mixed with $500 \,ml$ of $0.2\, M \,CH _{3} COOH .25 \,ml$ of the mixture is titrated with $0.1 \,M\, NaOH$ solution. By how many units does the $pH$ change from the start to the stage when $HCl$ is just completely neutralized. $K _{ a }$ for acetic acid $=2.0 \times 10^{-5}$.

• A. 3.7
• B. 4.4
• C. 2.0
• D. 3.0

Answer 1

Answer: (C)

$[ HCl ]_{\text {mix }}=0.1$ and $\left[ CH _{3} COOH \right]_{\text {mix }}=0.1 M$
Initial $pH$ of the mixture $=1.0$
Volume of $NaOH$ required just to neutralize $HCl$ in $25 ml \, \text{mix} =25 \,ml$
Concentration of $CH _{3} COH$ after $HCl$ in neutralized
$=\frac{25 \times 0.1}{50}=0.05 M$
$\left[ H ^{+}\right] =\sqrt{ K _{ a } \times C }=\sqrt{2 \times 10^{-5} \times 0.05} $
$=10^{-3} M $
$ pH =3.0$
Change in $pH =3.0-1.0=2.0$