A) \[3.5\times {{10}^{-4}}\]
B) \[1.1\times {{10}^{-5}}\]
C) \[1.8\times {{10}^{-5}}\]
D) \[9.0\times {{10}^{-6}}\]
Correct Answer: D
Solution :
Key Idea \[C{{H}_{3}}COOH\](weak acid) and \[C{{H}_{3}}COONa\](conjugated salt) form acidic buffer and for acidic buffer, \[pH=p{{K}_{a}}+\log \frac{[salt]}{acid}\] and \[[{{H}^{+}}]=-anti\log pH\] \[pH=-\log {{K}_{a}}+\log \frac{[salt]}{acid}\] \[[\because p{{K}_{a}}=-\log {{K}_{a}}]\] \[=-\log (1.8\times {{10}^{-5}})+\log \frac{(0.20)}{(0.10)}\] \[=4.74+\log 2\] \[=4.74+0.3010=5.041\] Now, \[[{{H}^{+}}]=anti\log (-5.045)\] \[=9.2\times {{10}^{-6}}mol/L\]You need to login to perform this action.
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