A) \[3.4\times {{10}^{-4}}\]
B) \[3.4\times {{10}^{-3}}\]
C) \[6.8\times {{10}^{-4}}\]
D) \[6.8\times {{10}^{-3}}\]
Correct Answer: D
Solution :
\[C{{H}_{3}}COOH\,\rightleftharpoons \,C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}\] \[{{K}_{a}}=\frac{[C{{H}_{3}}CO{{O}^{-}}]\,[{{H}^{+}}]}{[C{{H}_{3}}COOH]}\] Given that \[[C{{H}_{3}}CO{{O}^{-}}]=[{{H}^{+}}]\,=3.4\times {{10}^{-4}}\,M\] \[{{K}_{a}}\,for\,C{{H}_{3}}COOH=1.7\times {{10}^{-5}}\] \[C{{H}_{3}}COOH\] is weak acid, so in it \[(C{{H}_{3}}COOH)\] is equal to initial concentration. Hence, \[1.7\times {{10}^{-5}}=\frac{(3.4\times {{10}^{-4}})\,(3.4\times {{10}^{-4}})}{[C{{H}_{3}}COOH]}\] \[[C{{H}_{3}}COOH]=\frac{3.4\times {{10}^{-4}}\times 3.4\times {{10}^{-4}}}{1.7\times {{10}^{-5}}}\,\] \[=6.8\times {{10}^{-3}}\,M\]
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