Question

Calculate the $pH$ of 0.2 M $NaOCl$ solution.
$K_a\left(HOCl\right)=3.5\times {10}^{-8}$

• A. $10.01$
• B. $12.53$
• C. $8.63$
• D. $9.53$

Answer 1

The correct option is A $10.01$

$NaOCl$ is a salt of weak acid $HOCl$ and strong base $NaOH$
$pKa=-log\left(K_a\right)p{K}_a=-log(3.5\times {10}^{-8})p{K}_a=(8-0.54)=7.46pH=7+\frac{1}{2}(p{K}_a+logC)pH=7+\frac{1}{2}(7.46+log(0.2\left)\right)pH=7+\frac{1}{2}(7.46-0.7)pH=10.38$