Question

What fraction of an indicator $HIn$ is in basic form at a $pH$ of $6$ if the $p{K}_a$ of the indicator is $5$ ?

• A. $\frac{1}{2}$
• B. $\frac{1}{11}$
• C. $\frac{10}{11}$
• D. $\frac{1}{10}$

Answer 1

The correct option is C $\frac{10}{11}$

Given: $pH$ of $HIn=6$
$p{K}_a=5$
We know,
$pH=p{K}_a+\text{log}\frac{\text{[Ionised]}}{\text{[un Ionised]}}$
$6=5+\text{log}\frac{\text{[Ionised]}}{\text{[un Ionised]}}$
$1=\text{log}\frac{\text{[Ionised]}}{\text{[un Ionised]}}$
$\frac{\text{[Ionised]}}{\text{[un Ionised]}}=10$
we get,
$\left[\text{ionized}\right]=10$ and $\left[\text{unionized}\right]=1$
$\frac{\text{[Ionised]}}{\text{Ionised}+\text{[un Ionised]}}=\left[\frac{10}{11}\right]$