Question

Bromophenol blue is an acidic indicator with $K_a$ value of $6\times {10}^{-5}$. What percentage of this indicator is in its basic form at a pH of $5$? (Given that log $6=0.78$)

• A. $82.5\%$
• B. $85.7\%$
• C. $83.6\%$
• D. $87.2\%$

Answer 1

The correct option is B $85.7\%$

The equilibrium reaction for an acidic indicator $\left(HIn\right)$ is given below:
$HIn\rightleftharpoons I{n}^{-}+H^{+}$

The expression for the pH of the indicator is as given below:
$pH=p{K}_{In}+log\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}$

$p{K}_{In}=-log\left(K_{In}\right)=-log(6\times {10}^{-5})=4.22$

$pH=5$

Substituting values in the above expression, we get,

$5=4.22+log\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}$ or $\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}=\frac{100}{16.6}$

The percentage of indicator in basic form in the solution $=\frac{100}{100+16.6}\times 100=85.7\%$

Option B is correct.