Question

A solution of weak acid was titrated with base $NaOH$. The equivalence point was reached when $36.12mL$ of $0.1MNaOH$ have been added. Now $18.06mL0.1MHCl$ were added to titrated solution, the $pH$ was found to be $4.92$. What is $K_a$ of acid?

Answer 1

Milli moles of $NaA$ or $A^{(-)}=18.06\times 0.1=1.806$ millimoles

$A^{-}+H^{+}\to HA$

Initial : $1.806$

Millimoles of $HA$ left $=36.12ml\times 0.1M-1.806=1.806$ millimoles

$\therefore$ $pH={pK}_a+log\frac{\left[NA\right]}{\left[HA\right]}$

$\Rightarrow 4.92={pK}_a+log\frac{1.806}{1.806}$

$\Rightarrow {pK}_a=4.92$ $\therefore$ $K_a={10}^{-4.92}$