Question

A solution of weak acid was titrated with base $NaOH$. The equivalence point was reached when $36.12$ $mL$ of $0.1M$ $NaOH$ have been added. Now, $18.06mL$ of $0.1M$ $HCl$ were added to titrated solution, the $pH$ was found to be $4.92$. $K_a$ of acid is

• A. $K_a=1.3\times {10}^{-5}$
• B. $K_a=1.0\times {10}^{-5}$
• C. $K_a=1.2\times {10}^{-5}$
• D. $K_a=1.1\times {10}^{-5}$

Answer 1

The correct option is C $K_a=1.2\times {10}^{-5}$

For complete neutralisation,
Meq. of acid = Meq. of $NaOH=36.12\times 0.1=3.612$

$\therefore$ $\underset{3.6120}{HA}+\underset{3.6120}{NaOH}⟶\underset{03.612}{NaA}+\underset{03.612}{H_{2}O}$

Now $1.806$ Meq. of $HCl$ ($18.06\times 0.1$) are added to this solution containing $3.612$ Meq. of $NaA$

$\therefore$ $\underset{Meqbeforereaction3.612Meq.afterreaction1.806}{NaA}+\underset{1.8060}{H}Cl⟶\underset{01.806}{NaCl}+\underset{01.806}{HA}$

The solution has $HA$ and $NaA$ and thus, acts as buffer.

$\therefore$ $pH=-\log K_a+log\frac{1.806}{1.806}$

$4.92=-\log K_a$

$\therefore$ $K_a=1.2\times {10}^{-5}$