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 $



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