Question

The equivalent point in a titration of $40.0mL$ of a solution of a weak monoprotic acid occurs when $35.0mL$ of a $0.10M$ $NaOH$ 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

Volume of Acid$=40ml$

Volume of Base$=35ml$

Molarity of $NaOH=0.1$

$pH=5.5$ $⟶pH=-log\left[H^{+}\right]$ $\left[H^{+}\right]=3.16\times {10}^{-6}$

$M_1{V}_1=M_2{V}_2$

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

$M_1=0.0875M$

Dissociation of weak monoportic acid.

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

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

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

$K_a\frac{(3.16\times {10}^{-9}{)}^2}{0.0875}$

$K_a=1.1\times {10}^{-10}⟶$ dissociation constant of acid.