A) it carries negative charge
B) it is a pseudo halide
C) it can accept electrons from metal species
D) it forms high spin complexes with metal species
Correct Answer: B
Solution :
\[\text{Hin}\rightleftharpoons {{\text{H}}^{\text{+}}}\text{+I}{{\text{n}}^{\text{-}}}\] \[\therefore \]\[{{K}_{\text{ln}}}=\frac{[{{H}^{+}}][I{{n}^{-1}}]}{[HIn]}\] or\[[{{H}^{+}}]={{K}_{\text{ln}}}.\frac{[HIn]}{[I{{n}^{-1}}]}\] \[pH=-\log \,[{{H}^{+}}]\] \[=-\log \left( {{K}_{\text{ln}}}\cdot \frac{[HIn]}{[I{{n}^{-}}]} \right)\] \[=-\log \,{{K}_{\ln }}+\log \frac{[I{{n}^{-}}]}{[HIn]}\] \[=p{{K}_{\ln }}+\log \frac{[I{{n}^{-}}]}{[HIn]}\] or\[\log \frac{[I{{n}^{-}}]}{[HIn]}=pH-p{{K}_{\ln }}\]
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