2004 AIPMT EquilibriumAcid-Base Indicators and Henderson Equation Medium Question 89 of 104 in Equilibrium

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 ratio of the concentrations of the conjugate acid (HIn) and base ($In^-$) forms of the indicator by the expression -

More than one option is correct.

A $\log\dfrac{[HIn]}{[In^-]} = pK_{In} - pH$
B $\log\dfrac{[HIn]}{[In^-]} = pH - pK_{In}$
C $\log\dfrac{[In^-]}{[HIn]} = pH - pK_{In}$
D $\log\dfrac{[In^-]}{[HIn]} = pK_{In} - pH$

The answers are A and C.

Solution

An acid-base indicator is a weak acid: $HIn \rightleftharpoons H^+ + In^-$
$K_{In} = \dfrac{[H^+][In^-]}{[HIn]}$
Taking negative logarithms: $-\log K_{In} = -\log[H^+] - \log\dfrac{[In^-]}{[HIn]}$
$pK_{In} = pH - \log\dfrac{[In^-]}{[HIn]}$
$\Rightarrow \log\dfrac{[In^-]}{[HIn]} = pH - pK_{In}$ (Henderson equation for the indicator)
Multiplying both sides by $-1$ and using $\log\dfrac{[HIn]}{[In^-]} = -\log\dfrac{[In^-]}{[HIn]}$:
$\log\dfrac{[HIn]}{[In^-]} = pK_{In} - pH$
Both forms are the same relation written two ways, so both $\log\dfrac{[In^-]}{[HIn]} = pH - pK_{In}$ and $\log\dfrac{[HIn]}{[In^-]} = pK_{In} - pH$ are correct.
Note: the answer key of the source accepts both of these expressions as correct, so both are marked correct here.

Concept Henderson equation for an indicator

Equilibrium in past papers

104 questions from this chapter have appeared across 27 exam years.