Question

What will be the $pH$ at the equivalent point during the titration of a $100$ ml of $0.2M$ solution of $C{H}_{3}COONa$ with $0.2M$ solution of $HCl$ ?

[Given : $K_a=2\times {10}^{-5}$]

• A. $3-\log \sqrt{2}$
• B. $3+\log \sqrt{2}$
• C. $3-\log 2$
• D. $3+\log 2$

Answer 1

The correct option is A $3-\log \sqrt{2}$

The balanced chemical equation for the titration reaction between sodium acetate and $HCl$ is : $C{H}_{3}COONa+HCl⟶C{H}_{3}COOH+NaCl$

$100$ ml of $0.2M$ sodium acetate will react with $100$ ml of $0.2M$ $HCl$ to form $200$ ml of $0.1M$ sodium acetate.

Note that the concentration of acetic acid will be one-half the concentration of acetic acid or $HCl$.

For acetic acid, $K_a$ is $2\times {10}^{-5}$.

Hence, $p{K}_a$ is $-log{K}_a$ which is $-log(2\times {10}^{-5})$ or $4.69$.

The $pH$ of the solution is $pH=\frac{1}{2}[p{K}_a-log[C{H}_{3}COOH\left]\right]$

$pH=\frac{1}{2}[4.69-log0.1]=2.849=3-log\sqrt{2}$

Hence, option A is correct.