Equilibrium Ques 47

[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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Answer:

Correct Answer: 47.(d)

Solution:

  1. (d) For an acid-base indicator

$HIn \rightleftharpoons H^{+}+In^{-}$

$\therefore K _{In}=\frac{[H^{+}][In^{-}]}{[HIn]}$ 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 [H ] = \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 }} $