Question

Calculate the $pH$ at which an acid indicator $\left(HIn\right)$ having $K_{In}={10}^{-5}$ changes its colour.
Given, the $HIn$ concentration is ${10}^{-3}M$.

• A. $2$
• B. $3$
• C. $5$
• D. $8$

Answer 1

The correct option is C $5$

The colour of an indicator always changes at midpoint , i.e. when it is $50\%$ ionized.

$HIn\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+I{n}^{-}\left(aq\right)$
$K_{In}=\frac{\left[H^{+}\right]\left[I{n}^{-}\right]}{\left[HIn\right]}$
When colour changes,
$\left[I{n}^{-}\right]=\left[HIn\right]$
Putting in the above equation we get,
$K_{In}=\left[H^{+}\right]-\log \left(K_{In}\right)=-log\left[H^{+}\right]Since,pH=-\log \left[H^{+}\right]\therefore pH=-\log \left(K_{In}\right)pH=-\log \left({10}^{-5}\right)pH=5$
Note: Range of color change of the indicator is $p{K}_{In}-1$ to $p{K}_{In}+1$