Equilibrium Question 320
Question: If $ 50,ml $ of $ 0.2,M,KOH $ is added to $ 40,ml $ of $ 0.5,M,HCOOH, $ the $ pH $ of the resulting solution is $ (K_{a}=1.8\times {10^{-4}}) $ [MH CET 2000]
Options:
A) 3.4
B) 7.5
C) 5.6
D) 3.75
Show Answer
Answer:
Correct Answer: A
Solution:
$ pH=-\log K_{a}+log\frac{[\text{ Salt }]}{[\text{ Acid }]} $
$ [\text{ Salt }]=\frac{0.2\times 50}{1000}=0.01 $ ; $ [\text{ Acid }]=\frac{0.5\times 40}{1000}=0.02 $
$ pH=-log(1.8\times {10^{-4}})+log\frac{0.01}{0.02} $
$ pH=4-log(1.8)+log0.5 $
$ pH=4-log(1.8)-0.301 $
$ pH=3.4 $
