A) \[1.8\times {{10}^{-3}}M\]
B) \[1.6\times {{10}^{-5}}M\]
C) \[CaS{{O}_{4}}.{{H}_{2}}O\]
D) \[MgS{{O}_{4}}.7{{H}_{2}}O\]
Correct Answer: D
Solution :
\[C{{H}_{3}}COOH\]ionize as follows: \[\underset{C(1-\alpha )}{\mathop{C{{H}_{3}}COOH}}\,\underset{C\alpha }{\mathop{C{{H}_{3}}CO{{O}^{-}}}}\,+\underset{C\alpha }{\mathop{{{H}_{3}}{{O}^{+}}}}\,\] \[[{{H}_{3}}{{O}^{+}}]=C\alpha \] And \[\alpha =\sqrt{\frac{{{K}_{a}}}{C}}\](for any weak electrolyte) \[[{{H}_{3}}{{O}^{+}}]=\sqrt{K\alpha C}\] \[=\sqrt{1.8\times {{10}^{-5}}\times 0.1}\] \[=1.34\times {{10}^{-3}}M.\] Hence, the correct option is [d].
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