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 :
[b] \[\text{Hin}\rightleftharpoons {{\text{H}}^{\text{+}}}\text{+I}{{\text{n}}^{\text{-}}}\] |
\[\therefore \]\[{{K}_{\text{ln}}}=\frac{[{{H}^{+}}][I{{n}^{-1}}]}{[HIn]}\] |
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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