Question

For the acid indicator thymol blue, pH is 2.0 when half the indicator is in unionised form. Find the $\%$ of indicator in unionised form in the solution with $\left[H^{+}\right]=4\times {10}^{-3}M$.

• A. $25.4\%$
• B. $27.2\%$
• C. $28.6\%$
• D. $29.4\%$

Answer 1

The correct option is C $28.6\%$

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

When half the indicator is in unionised form,
$2.0=p{K}_{In}+log\frac{1}{1}$

Hence, $p{K}_{In}=2.0$

For $\left[H^{+}\right]=4\times {10}^{-3}M$, $pH=-log\left[H^{+}\right]=-log(4\times {10}^{-3})=2.4$

Hence, $2.4=2.0+log\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}$ or $\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}=\frac{100}{39.8}$

The % of indicator in unionised form in the solution $=\frac{39.8}{100+39.8}\times 100=28.6\%$.