Question

In an acidiic indicator $HIn$ has ionisation constant is ${10}^{-8}$. The acid form of indicator is yellow and alkaline form is red. Which is correct statement: (Given : $log2=0.3,log3=0.48$)

• A. The $pH$ range of indicator is $7\text{to}9$
• B. Change in $pH$ is $0.96$ when $75\%$ acidic form of the indicator changes to $75\%$ alkaline form
• C. This indicator is suitable for the titration of strong acid Vs strong base
• D. $pH$ of indicator is $8.3$ when ratio of acid form to alkaline form is $2$

Answer 1

Answer: (A), (B), (C)

The correct options are
A The $pH$ range of indicator is $7\text{to}9$
B Change in $pH$ is $0.96$ when $75\%$ acidic form of the indicator changes to $75\%$ alkaline form
C This indicator is suitable for the titration of strong acid Vs strong base

$p{K}_a=-\log \left({10}^{-8}\right)=8$
$pH$ indicator changes colour in solution over a range of $pH$ values. These narrow range of $pH$ values is known as $pH$ range.
Generally the $pH$ range of an indicator is falls between $p{K}_a\underset{-}{+}1.$
So, $pH$ range is $7\text{to}9$.

As the $pH$ range of the indicator is $7\text{to}9$, then it is suitable for the titration of strong acid and strong base.

$p{H}_1=8+log\frac{25}{75}=8+log\frac{1}{3}$
$p{H}_2=8+log\frac{75}{25}=8+log3$
$\text{Change in}pH=log3-log\frac{1}{3}=2log3=2\times 0.48=0.96$

When ratio of acid form to alkaline form is $2$, then
$pH=8+log(\frac{1}{2}=8-0.3=7.7)$