Question

The rapid change of pH near the stoichiometric point of an acid-base titration is the basis of indicator detection. pH of the solution is related to the ratio of the concentrations of the conjugate acid $\left(HIn\right)$ and base $\left(In\right)$ forms of the indicator by the expression:

• A. $\log \frac{\left[HIn\right]}{\left[I{n}^{-}\right]}=p{K}_{In}-pH$
• B. $\log \frac{\left[HIn\right]}{\left[I{n}^{-}\right]}=pH-p{K}_{In}$
• C. $\log \frac{\left[I{n}^{-}\right]}{\left[HIn\right]}=pH-p{K}_{In}$
• D. $\log \frac{\left[I{n}^{-}\right]}{\left[HIn\right]}=p{K}_{In}$ - pH

Answer 1

Answer: (C)

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

$Hln\rightleftharpoons H^{+}+{ln}^{-}$

$K_{ln}=\frac{\left[H^{+}\right]\left[{ln}^{-}\right]}{\left[Hln\right]}$

$\text{Taking logarithm on both sides we get}$

$-\log K_{ln}=-\log \left[H^{+}\right]-log\frac{\left[{ln}^{-}\right]}{\left[Hln\right]}$

$\Rightarrow p{K}_{ln}=p{K}_a-\log \frac{\left[{ln}^{-}\right]}{[Hln]}$

$\Rightarrow \log \frac{{[ln}^{-}]}{\left[Hln\right]}=pH-p{K}_{ln}$