A) 2
B) 0.5
C) 4
D) 0.25
Correct Answer: D
Solution :
As pH of HCOOH sol. = pH of\[C{{H}_{3}}COOH\]sol.\[[{{H}^{+}}]\]in HCOOH sol.\[=[{{H}^{+}}]\]in\[C{{H}_{3}}COOH\]sol. \[{{K}_{1}}=\frac{[HCO{{O}^{-}}][{{H}^{+}}]}{[HCOOH]}=\frac{{{[{{H}^{+}}]}^{2}}}{[HCOOH]}\] \[{{K}_{2}}=\frac{[C{{H}_{3}}CO{{O}^{-}}][{{H}^{+}}]}{[C{{H}_{ & 3}}COOH]}=\frac{{{[{{H}^{+}}]}^{2}}}{[C{{H}_{3}}COOH]}\] \[\Rightarrow \]\[{{K}_{1}}[HCOOH]={{K}_{2}}[C{{H}_{3}}COOH]\] \[\therefore \]\[\frac{[HCOOH]}{[C{{H}_{3}}COOH]}=\frac{{{K}_{2}}}{{{K}_{1}}}=\frac{1}{4}=0.25\]
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