Question

The rapid change in $pH$ near the equivalence point of an acid-base titration is the basis of indicator detection. $pH$of the solution is related to ratio of the concentrations of the acidic form ($HIn$) and basic form ($I{n}^{-})$ of the indicator by the expression :

• A. $\log \frac{\left[I{n}^{-}\right]}{\left[HIn\right]}=pH-p{K}_{In}$
• B. $\log \frac{\left[I{n}^{-}\right]}{\left[HIn\right]}=p{K}_{In}-pH$
• C. $\log \frac{\left[HIn\right]}{\left[I{n}^{-}\right]}=pH-p{K}_{In}$
• D. None of the above

Answer 1

The correct option is A $\log \frac{\left[I{n}^{-}\right]}{\left[HIn\right]}=pH-p{K}_{In}$

Let’s represent an indicator by formula $HIn$.
$HIn\left(aq\right)\stackrel{K_{In}}{\rightleftharpoons }{H}^{+}\left(aq\right)+I{n}^{-}\left(aq\right)$
$AcidicFormBasicForm$
The equilibrium constant is written as :
$K_{In}=\frac{\left[H^{+}\right]\left[I{n}^{-}\right]}{\left[HIn\right]}$
$\left[H^{+}\right]=\frac{K_{In}\left[HIn\right]}{\left[I{n}^{-}\right]}-\log \left[H^{+}\right]=-\log \left[K_{In}\right]-\log \frac{\left[HIn\right]}{\left[I{n}^{-}\right]}$
$pH=p{K}_{In}+\log \frac{\left[I{n}^{-}\right]}{\left[HIn\right]}$
$\log \frac{\left[I{n}^{-}\right]}{\left[HIn\right]}=pH-p{K}_{In}$