A) \[\log \frac{[I{{n}^{-}}]}{[HIn]}=p{{K}_{In}}-pH\]
B) \[\log \frac{[H{{n}^{-}}]}{[I{{n}^{-}}]}=p{{K}_{In}}-pH\]
C) \[\log \frac{[HIn]}{[I{{n}^{-}}]}=pH-p{{K}_{In}}\]
D) \[\log \frac{[I{{n}^{-}}]}{[HIn]}=pH-p{{K}_{In}}\]
Correct Answer: D
Solution :
Acid indicators are generally weak acid. The dissociation of indicator\[HIn\]takes place as follows: \[HIn{{H}^{+}}+I{{n}^{-}}\] \[\therefore \] \[{{K}_{\operatorname{I}n}}=\frac{[{{H}^{+}}][I{{n}^{-}}]}{[HIn]}\] Or \[[{{H}^{+}}]={{K}_{\operatorname{I}n}}\frac{[HIn]}{[I{{n}^{-}}]}\] \[pH=-\log [{{H}^{+}}]\] \[=-\log \left( {{K}_{In}}\frac{[HIn]}{[I{{n}^{-}}]} \right)\] \[=-\log {{K}_{In}}+\log \frac{[I{{n}^{-}}]}{[HIn]}\] \[=p{{K}_{In}}+\log \frac{[I{{n}^{-}}]}{[HIn]}\] Or \[\log \frac{[I{{n}^{-}}]}{[HIn]}=pH-p{{K}_{In}}\]You need to login to perform this action.
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