Question

An indicator has $p{K}_{in}=\mathrm{5.3.}$ In a certain solution, this indicator is found to be $20\%$ ionised. Calculate the pH of the solution?

• A. $1.6$
• B. $6$
• C. $5.3$
• D. $4.7$

Answer 1

The correct option is D $4.7$

Consider $C$ to be the initial concentration of the indicator.
Since, indicator is $80\%$ ionized in acid form $\left(HIn\right)$:
$\therefore \left[HIn\right]=0.8C$
$0.2C=\left[H^{+}\right]=\left[I{n}^{-}\right]$
$\begin{array}{ccccc}HIn\left(aq\right)& \rightleftharpoons & H^{+}\left(aq\right)& +& I{n}^{-}\left(aq\right)\\ C& & 0& & 0\\ 0.8C& & 0.2C& & 0.2C\end{array}$

$pH=pK{}_a+lo{g}_{10}\frac{\left[l{n}^{-}\right]}{\left[Hln\right]}pH=5.3+lo{g}_{10}\left(\frac{0.2C}{0.8C}\right)pH=5.3+lo{g}_{10}\left(\frac{1}{4}\right)pH=5.3-2lo{g}_{10}\left(2\right)=5.3-(2\times 0.3)pH=4.7$