A) \[\log \,\frac{[HIn]}{[I{{n}^{-}}]}=p{{K}_{In}}+pH\]
B) \[\log \,\frac{[HIn]}{[I{{n}^{-}}]}=pH-p{{K}_{In}}\]
C) \[\log \,\frac{[I{{n}^{--}}]}{[HIn]}=pH-p{{K}_{In}}\]
D) \[\log \,\frac{[I{{n}^{--}}]}{[HIn]}=p{{K}_{In}}-pH\]
Correct Answer: C
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