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