Question

A weak acid HA after treatment with $1.2\, mL$ of $0.1\, M$ strong base has a pH of 5. At the end point, the volume of same base required is $26.6\, mL$. The value of $ {{\text{K}}_{\text{a}}} $ is

• A. $ 8.2\times {{10}^{-6}} $
• B. $ 6.4\times {{10}^{-6}} $
• C. $ 5.3\,\times {{10}^{-5}} $
• D. $ 2.4\times {{10}^{-6}} $

Answer 1

Answer: (A)

For complete neutralisation,
Total milliequivalent of acid = milliequivalent of base $=26.6 \times 0.1=2.66$
For partial neutralisation;
$\begin{array}{llll} HA + BOH & \rightarrow BA + H _{2} O \\ 2.66 & 1.2 & 0 & C \text {-- before reaction } \\ 1.46 & 0 & 1.2 & ( C +1.2) \text { - after reaction. }\end{array}$
The resultant mixture has HA and BA and thus acts as a buffer.
$\therefore pH =-\log K _{ a }+\log \frac{[\text { Salt }]}{[\text { Acid }]}$
$5=-\log K _{ a }+\log \frac{1.2}{1.46}=-\log K _{ a }-0.085$
$\therefore \log K _{ a }=-5.085$
or $K _{ a }=$ anti $\log (-5.085)=8.21 \times 10^{-6}$