Question

An acid-base indicator has $K_a$ of $3.0\times {10}^{-5}$. The acid form of the indicator is red and the basic form is blue. Then

• A. pH is $4.05$ when indicator is $75\%$ red
• B. pH is $5.00$ when indicator is $75\%$ blue
• C. pH is $5.00$ when indicator is $75\%$ red
• D. pH is $4.05$ when indicator is $75\%$ blue

Answer 1

Answer: (A), (B)

The correct options are
A pH is $4.05$ when indicator is $75\%$ red
B pH is $5.00$ when indicator is $75\%$ blue

The equilibrium reaction for the acidic indicator $HIn$ is given below:

$HIn\rightleftharpoons I{n}^{-}+H^{+}$

The expression for the pH of the indicator is as given below:

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

For 75% red color,

$p{H}_1=-log(3.0\times {10}^{-5})+log\frac{25}{75}=4.05$

For 75% blue color,

$p{H}_2=-log(3.0\times {10}^{-5})+log\frac{75}{25}=5.00$