Equilibrium - Result Question 49
####51. 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
[2004]
(a) $\log \frac{[In^{-}]}{[HIn]}=pK _{In}-pH$
(b) $\log \frac{[HIn]}{[In^{-}]}=pK _{In}-pH$
(c) $\log \frac{[HIn]}{[In^{-}]}=pH-pK _{\text{In }}$
(d) $\log \frac{[In^{-}]}{[HIn]}=pH-pK _{\text{In }}$
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Solution:
- (d) For an acid-base indicator
$HIn \rightarrow H^{+}+In^{-}$
$\therefore K _{In}=\frac{[H^{+}][In^{-}]}{[HIn]}$ or $[H^{+}]=K _{In} \times \frac{[HIn]}{[In^{-}]}$
or $\log H^{+}=\log K _{\text{In }}+\log \frac{[HIn]}{[In^{-}]}$
Taking negative on both sides
$ \log \begin{bmatrix} H & ] \end{matrix} \log K _{In} \log \frac{[HIn]}{[In]}. $
or we can write $pH=pK _{\text{In }}+\log \frac{[In^{-}]}{[HIn]}$
$ \text{ or } \log \frac{[In^{-}]}{[HIn]}=pH-pK _{\text{In }} $
