Question

$100mL$ of a $0.5MC{H}_{3}COONa\left(aq\right)$ is titrated against a $0.5MHCl\left(aq\right)$. Calculate the $pH$ of the solution initially i.e. when no $HCl$ has been added.
Given: $p{K}_a\left(C{H}_{3}COOH\right)=4.74$

• A. 6.58
• B. 8.24
• C. 7.96
• D. 9.22

Answer 1

The correct option is D 9.22

When only $100mL$ of a $0.5MC{H}_{3}COONa\left(aq\right)$ is present. It will show simple salt hydrolysis.
The formula to calculate the $pH$ of a salt of strong base and weak acid is :
$pH=\frac{1}{2}(p{K}_w+p{K}_a+\log C)pH=\frac{1}{2}(14+4.74+\log 0.5)pH=\frac{1}{2}(18.74-1+\log 5)pH=\frac{1}{2}\left(18.44\right)pH=9.22$