Question

Calculate $\left[H^{+}\right]$ at equivalent point between titration of $0.1M,25mL$ of weak acid $HA(K_{a\left(HA\right)}={10}^{-5})$ with $0.05MNaOH$ solution:

• A. $3\times {10}^{-9}$
• B. $1.732\times {10}^{-9}$
• C. $8$
• D. $10$

Answer 1

The correct option is B $1.732\times {10}^{-9}$

Given:
$K_{a\left(HA\right)}={10}^{-5}$
Solution:

• Titration occurs between a weak acid and a weak base i.e., $HCl$ vs $NaOH$ .The reaction can be written as,

$HA+NaOH\rightleftharpoons NaA+H_{2}O$

• It is given that 25ml of weal acid is titrated with 0.05M NaOH solution so the reactant concentration for HA = $25\times 0.1$, $NaOH=0.05\times 50,NaA=2.5$.
• The concentration of $HA$ can be calculated by ,

$\left[NA\right]=\frac{mole}{volume}$

$=\frac{2.5}{25+50}$

$=\frac{2.5}{75}$

$=\frac{1}{30}$

$H^{+}=\sqrt{\frac{Kw.Ka}{C}}$

$=\sqrt{\frac{{10}^{-14}{10}^{-5}\times 30}{1}}$

$=\sqrt{30\times {10}^{-19}}$

$=\sqrt{3\times {10}^{-18}}$

$=1.732\times {10}^{-9}$

Hence, the correct option is B