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  • 0.1 M acetic acid solution having pH =3 is titrated against 0.05 M NaOH solution. Calculate pH at 1/4th stage of neutralization of acid.Option: 1</stro

0.1 M acetic acid solution having pH =3 is titrated against 0.05 M NaOH solution. Calculate pH at 1/4th stage of neutralization of acid.

Option: 1

4.5


Option: 2

5


Option: 3

9.3


Option: 4

8.2


Answers (1)

best_answer

It is given that the pH of 0.1 M CH_{3}COOH = 3

\therefore [H^{+}] = 10^{-3}

Also, [H^{+}] = c\alpha

\Rightarrow \alpha =\frac{[H^{+}]}{c} =\frac{10^{-3}}{0.1} =10^{-2}

\therefore K_{a}=c\alpha ^{2} = 10^{-5}

At 1/4 neutralization of acid

                                {\qquad \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NaOH} \longrightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{H}_{2} \mathrm{O}}

before addition of NaOH 0.1                                                        0

after addition of NaOH    0.1 *(3/4)                                           0.1*(1/4)

 

\mathrm{pH}=-\log K_{a}+\log \frac{[\text { Conjugate base }]}{[\text { Acid }]}

\mathrm{pH}=-\log 10^{-5}+\log \frac{[0.1/4]}{[0.3/4]}

\mathrm{pH}=4.523

Posted by

Ritika Jonwal

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