Question

An acid-base indicator has $K_{HIn}=3.0\times {10}^{-5}$. The acid form of the indicator is red and the basic form is blue. The change in $\left[H^{+}\right]$ required to change the indicator from $75\text{\%}$ red to $75\text{\%}$ blue is:

• A. $8\times {10}^{-5}M$
• B. $9\times {10}^{-5}M$
• C. $1\times {10}^{-5}M$
• D. $3\times {10}^{-4}M$

Answer 1

Answer: (A)

$8\times {10}^{-5}M$

For a buffer solution,
$pH=p{K}_{HIn}+log\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}$
Given that $K_{HIn}=3.0\times {10}^{-5}$.
Substituting the values in the expression, we get
$p{H}_1=p{K}_{HIn}+log\frac{75}{25}$
$H_1^{+}=1\times {10}^{-5}$
$p{H}_2=p{K}_{HIn}+log\frac{25}{75}$
$H_2^{+}=9\times {10}^{-5}$
$H_2^{+}-H_1^{+}=(9-1)\times {10}^{-5}=8\times {10}^{-5}$
Hence, the change in $\left[H^{+}\right]$ required to change the indicator from $75\text{\%}$ red to $75\text{\%}$ blue is $8\times {10}^{-5}$.