A) \[7\times {{10}^{-4}}M\]
B) \[7\times {{10}^{-5}}M\]
C) \[9\times {{10}^{-3}}M\]
D) \[9\times {{10}^{-4}}M\]
Correct Answer: D
Solution :
Degree of dissociation, \[\alpha =\frac{{{\lambda }^{o}}_{c}}{{{\lambda }^{o}}_{m}}=\frac{150}{500}=0.3\] Given, \[C=0.007M\] Hydrofluoric acid dissociates in the following manner \[HF\xrightarrow{{}}{{H}^{+}}+{{F}^{-}}\] \[\underset{C(1-\alpha )}{\mathop{\underset{C-C\alpha }{\mathop{C}}\,\,\,\,}}\,\,\,\,\,\,\underset{C\alpha }{\mathop{0}}\,\,\,\,\,\,\,\,\,\underset{C\alpha }{\mathop{0}}\,\,\,\,\underset{\text{At}\,\text{time}\,\text{t}}{\mathop{\text{Initially}}}\,\] Dissociation constant, \[{{K}_{a}}=\frac{[{{H}^{+}}][{{F}^{-}}]}{[HF]}=\frac{C\alpha .C\alpha }{C(1-\alpha )}=\frac{C{{\alpha }^{2}}}{(1-\alpha )}\] On substituting values, we get \[{{K}_{a}}=\frac{0.007\times {{(0.3)}^{2}}}{(1-0.3)}=\frac{63\times {{10}^{-3}}\times {{10}^{-2}}}{0.7}\] \[=9\times {{10}^{-4}}M\]You need to login to perform this action.
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