Question

The indicator phenol red is half in the ionic form when $pH$ is $7.2$. If the ration of the undissociated from to the ionic form is $1:5$, find the $pH$ of the solution. With the same $pH$ for solution, if indicator is altered such that the ratio of undissociated form to dissociated form becomes $1:4$, find the $pH$ when $50$% of the new indicator is in ionic form.

• A. $pH=7.3;7.9$
• B. $pH=7.9;7.3$
• C. $pH=6.9;8.3$
• D. $pH=8.3;6.9$

Answer 1

The correct option is B $pH=7.9;7.3$

pH of indicator,

$pH=pKIn+log\frac{\left[In\right]}{\left[HIn\right]}$

When indicator is half in ionic form $pH=p{K}_a=7.2$
$pH=7.2+\log 5=7.898$

Now with this $pH$
$7.898=p{K}_{a_1}+\log 4=p{K}_{a_1}=7.2959$

When $50$% of new indicator is in ionic form
$pH=p{K}_{a_1}=7.2959$