Question

Indicator are weak acids or weak bases. For a basic indicator $pH=9.4$ when $80\%$ of indicator is in ionised form. Which of the following statements is correct if $\left[H^{+}\right]={10}^{-8}M$ in the solution.

• A. Almost complete indicator will be in ionised form
• B. Almost complete indicator will be in unionised form
• C. $50\%$ of indicator will be ionised
• D. $20\%$ of indicator will be ionised

Answer 1

The correct option is A Almost complete indicator will be in ionised form

Given $pH=9.4$ so $pOH=4.6$
$pOH=p{K}_{InOH}+\text{log}\frac{\left[I{n}^{+}\right]}{\left[InOH\right]}$
$4.6=p{K}_{InOH}+\text{log}\frac{80}{20}$
$(\because [l{n}^{+}]=Concentrationofionisedform=80\%$)
$\left(\right[lnOH]=Concenrtrationofunionisedform=20\%))$
$p{K}_{InOH}=4.6\times 2\text{log}2=4$
$K_{InOH}={10}^{-4}$
Again given that
$\left[H^{+}\right]={10}^{-8}$
so $\left[O{H}^{-}\right]={10}^{-6}\Rightarrow pOH=6$
$pOH=p{K}_{InOH}+log\frac{\left[I{n}^{+}\right]}{\left[InOH\right]}$
$6=4+\text{log}\frac{\left[I{n}^{+}\right]}{\left[InOH\right]}$
$\text{log}\frac{\left[I{n}^{+}\right]}{\left[InOH\right]}=2\Rightarrow \frac{\left[I{n}^{+}\right]}{\left[InOH\right]}={10}^2\Rightarrow [In{]}^{+}=100\times [InOH]$

% of Indicator ionised
$\frac{\left[I{n}^{+}\right]}{\left[I{n}^{+}\right]+\left[InOH\right]}\times 100=\frac{100\times \left[InOH\right]}{101\times \left[InOH\right]}\times 100=\frac{100}{101}\times 100=99\%\left(\%\text{of ionised form}\right)$

% of Indicator in unionised form
$\frac{\left[InOH\right]\times 100}{\left[I{n}^{+}\right]+\left[InOH\right]}=\frac{1}{101}\times 100=0.99\%=1\%$