A) 1.35
B) 0.5
C) 3.7
D) 1.85
Correct Answer: C
Solution :
\[{{[{{H}^{+}}]}_{HCOOH}}=\sqrt{{{K}_{a}}.C}\] \[=\sqrt{2\times {{10}^{-4}}\times 0.0002}\] \[pH=-\log [{{H}^{+}}]=-\log (2\times {{10}^{-4}})\] \[=3.69\approx 3.7\]
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