Question

Choose the correct statement(s) regarding an indicator having the formula $HIn$ :

• A. When $50\%$ of the indicator is ionised, then $\log \left(\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}\right)=0.5$
• B. When $50\%$ of the indicator is ionised, then $\left[H^{+}\right]=K_{in}$
• C. When $50\%$ of the indicator is ionised, then $\left(\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}\right)=1$
• D. When $50\%$ of the indicator is ionised, then $pH=p{K}_{in}+0.5$

Answer 1

Answer: (B), (C)

The correct options are
B When $50\%$ of the indicator is ionised, then $\left[H^{+}\right]=K_{in}$
C When $50\%$ of the indicator is ionised, then $\left(\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}\right)=1$

$pH=p{K}_{in}+\log \left(\frac{\left[\text{Ionised form}\right]}{\left[\text{Unionised form}\right]}\right)$
When $50\%$ of the indicator is ionised, $\%\left[I{n}^{-}\right]=50\%$
and unionised $\%\left[HIn\right]=50\%$
If the initial concentration of indicator is $C$
Then, ratio $\left(\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}\right)=\left(\frac{0.5C}{0.5C}\right)=1$
Option C is correct.
$pH=p{K}_{in}+\log \left(\frac{\left[I{n}^{-}\right]}{\left[HIn\right]}\right)pH=p{K}_{in}+\log \left(1\right)pH=p{K}_{in}-log\left[H^{+}\right]=-log\left[K_{in}\right]\left[H^{+}\right]=K_{in}$