A) \[\log \frac{\left[ I{{n}^{-}} \right]}{\left[ HIn \right]}=p{{K}_{In}}-pH\]
B) \[\log \frac{\left[ HIn \right]}{\left[ I{{n}^{-}} \right]}=p{{K}_{In}}-pH\]
C) \[\log \frac{\left[ HIn \right]}{\left[ I{{n}^{-}} \right]}=pH-p{{K}_{In}}\]
D) \[\log \frac{\left[ I{{n}^{-}} \right]}{\left[ HIn \right]}=pH-p{{K}_{In}}\]
Correct Answer: D
Solution :
[d] For an acid-base indicator \[HIn{{H}^{+}}+I{{n}^{-}}\] \[\therefore \]\[{{K}_{In}}=\frac{[{{H}^{+}}][I{{n}^{-}}]}{[HIn]}\] or \[[{{H}^{+}}]={{K}_{In}}\times \frac{[HIn]}{[I{{n}^{-}}]}\] or \[log\,{{H}^{+}}=\log {{K}_{In}}+\log \frac{[HIn]}{[I{{n}^{-}}]}\] Taking negative on both sides \[-log[{{H}^{+}}]=-\log {{K}_{In}}-\log \frac{[HIn]}{[I{{n}^{-}}]}\] or we can write \[pH=p{{K}_{In}}+\log \frac{[I{{n}^{-}}]}{[HIn]}\] or \[\log \frac{[I{{n}^{-}}]}{[HIn]}=pH-p{{K}_{In}}\]
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