A) \[\log \left[ \frac{I{{n}^{-}}}{HIn} \right]=p{{K}_{a}}+pH\]
B) \[\log \left[ \frac{HIn}{I{{n}^{-}}} \right]={{K}_{a}}+pH\]
C) \[\log \left[ \frac{HIn}{I{{n}^{-}}} \right]=pH+p{{K}_{a}}\]
D) \[\log \left[ \frac{I{{n}^{-}}}{HIn} \right]=pH-p{{K}_{a}}\]
Correct Answer: D
Solution :
For \[HIn\rightleftharpoons {{H}^{+}}+I{{n}^{-}}\] \[{{K}_{a}}=\frac{[{{H}^{+}}][I{{n}^{-}}]}{[HIn]}\] \[\therefore \]\[pH=p{{K}_{a}}+\log \frac{[I{{n}^{-}}]}{[HIn]}\]You need to login to perform this action.
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