Question

An acid base indicator has $Ka=3x{10}^{-5}$ The acid form of indicator is red and the basic form of indicator is blue. By how much must the pH change in order to change the indicator from 75% red to 75% blue?

• A. $7.91x{10}^{-5}$
• B. 0.95
• C. $5x{10}^{23}$
• D. 9

Answer 1

The correct option is A $7.91x{10}^{-5}$

$pH=p{K}_a+\log \frac{\left[I{n}^{-}\right]}{\left[HIn\right]}$

Initial concentration of $\left[HIn\right]=75$%

Final concentration of $\left[HIn\right]=25$%

Initial concentration of $\left[I{n}^{-}\right]=25$%

Final concentration of $\left[I{n}^{-}\right]=75$%

$\therefore p{H}_1=-\log (3\times {10}^{-5})+\log \left[\frac{75}{25}\right]$

$\Rightarrow p{H}_1=5\Rightarrow [H_1{]}^{+}={10}^{-5}M$

$\therefore p{H}_2=-\log (3\times {10}^{-5}+\log \left(\frac{25}{75}\right)$

$\Rightarrow p{H}_2=-4.045\Rightarrow \left[H_2^{+}\right]=8.91\times {10}^{-5}M$

$\therefore \left[H_2^{+}\right]-\left[H_1^{+}\right]=7.91\times {10}^{-5}M$