Question

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

• A. $1.95$
• B. $2.95$
• C. $0.95$
• D. $3.95$

Answer 1

The correct option is C $0.95$

$HIn\rightleftharpoons H^{+}+I{n}^{-}$
$K_{In}=K_a=\frac{\left[H^{+}\right]\left[I{n}^{-}\right]}{\left[HIn\right]}=\frac{\left[H^{+}\right]\left[base\right]}{\left[acid\right]}\left[H^{+}\right]=K_a\frac{acid}{base}$

For $75\%$ red,
$\left[H^{+}\right]=\frac{(3\times {10}^{-5})\left(75\right)}{25}=9\times {10}^{-5}$
$\therefore pH=4.05$

For $75\%$ blue,
$\left[H^{+}\right]=\frac{(3\times {10}^{-5})\left(25\right)}{75}=1\times {10}^{-5}$
$\therefore pH=5$

$\text{The change in}pH=5-4.05=0.95$