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  • A weak acid HA after treatment with 12 mL of 0.1 M strong base has a pH of 5. At the end point, the volume of same base required is 26.6 mL. Calculate Ka of acid. (Given, log(1.2)

A weak acid HA after treatment with 12 mL of 0.1 M strong base has a pH of 5. At the end point, the volume of same base required is 26.6 mL. Calculate Ka of acid.
(Given, log(1.2) = 0.0791 and log(1.46) = 0.1643).

Option: 1

8.22 \times 10-6


Option: 2

8.22  \times 10+6


Option: 3

1.64  \times 10-6


Option: 4

-8.22  \times 10-6


Answers (1)

best_answer

\text { For complete neutralization: }
\begin{array}{l}{\mathrm{Total \: Meq.\: of acid = Meq.\: of \: base }} \\ {\text {Total Meq. of acid = } 26.6 \times 0.1=2.66} \\ {\mathrm{ Now\: for \: partial \: neutralization }}\end{array}
                                   \mathrm{HA}+\mathrm{BOH} \longrightarrow \mathrm{BA}+\mathrm{H}_{2} \mathrm{O}
Meq. before reaction      2.66        1.2                0          0
Meq. after reaction         1.46          0               1.2        1.2

\begin{array}{l}{\text { The resultant mixture has } \mathrm{HA} \text { and } B A \text { and thus acts as }} \\ {\text { buffer }} \\ {\qquad \begin{array}{c}{\mathrm{pH}=-\log K_{a}+\log \frac{[\text { Conjugate base }]}{[\text { Acid }]}} \\\\ {5=-\log K_{a}+\log \frac{1.2}{1.46}} \\\\ {\text { or } K_{a}=8.219 \times 10^{-6}}\end{array}}\end{array}

Therefore,option(1) is correct

Posted by

Divya Prakash Singh

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