Question

A weak acid HA after treatment with 12 mL of 0.1 M strong base BOH has a pH of 5. At the end point, the volume of same base required is 26.6 mL. K_a of acid is:

• A. 1.8 × 10^-5
• B. 8.12×10^-6
• C. 1.8×10^-6
• D. 8.2×10^-5

Answer 1

The correct option is B 8.12×10^-6

for neutralization,
Total Meq. of acid$=$Meq of base$=26\times 0.1=2.66$
now, for partial neutralization of acid
$HA+BOH\to BA+H_{2}O$
Meq. before reaction $2.66$ $1.22$ $0$ $0$
Meq. after reaction $1.46$ $0$ $1.2$ $1.2$
the resultant mixture acts as a buffer and $\left[HA\right]$ and $\left[BA\right]$ may be placed in terms of Meq. since volume of mixture is constant
$\therefore pH=-\log k_a+\log \frac{\left[Conjugatebase\right]}{\left[Acid\right]}$
or
$5=-\log K_a+\log \frac{\left[1.2\right]}{\left[1.46\right]}$
$\therefore K_a=8.129\times {10}^{-6}$