Question

The equivalent point in a titration of $40.0mL$ of a solution of a weak monoprotic acid occurs when $35.0mL$ of a $0.10MNaOH$ solution has been added. The $pH$ of the solution is $5.75$ after the addition of $20.0mL$ of $NaOH$ solution. What is the dissociation constant of the acid?

Answer 1

$V_a=40ml$, $V_b=35ml$, $NaOH$ molarity$=0.1M$ $pH=5.5$

$H_1{V}_1=M_2{V}_2$

$M_1=\frac{0.1\times 35}{40}=0.0875M$

$pH=-log\left[H^{+}\right]$

$\left[H^{+}\right]={10}^{-45}$

$\left[H^{+}\right]=3.16\times {10}^{-6}$

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

$K_a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]}$

$\left[H^{+}\right]=\left[A^{-}\right]$

$K_a=\frac{{(3.16\times {10}^{-6})}^2}{0.0875}$

$K_a=1.1\times {10}^{-10}$

Dissociation Constant of acid$=1.1\times {10}^{-10}$