Equilibrium Question 886
Question: The ionization constant of $ HCOOH $ is $ 1.8\times {10^{-4}} $ . What is the percent ionization of a 0.001 M solution'
Options:
A) 66%
B) 42%
C) 34%
D) 58%
Show Answer
Answer:
Correct Answer: C
Solution:
$ \underset{0.001-x}{\mathop{HCOOH}}{{\underset{x}{\mathop{HCOO}},}^{-}}+{{\underset{x}{\mathop{H}},}^{+}} $
As $ \alpha =\sqrt{\frac{K_{a}}{C}}=\sqrt{\frac{1.8\times {10^{-4}}}{0.001}}=0.42 $
As $ \alpha >0.1 $ so can’t be neglected, $ 1.8\times {10^{-4}}=\frac{x^{2}}{0.001-x} $ and $ x=3.4\times {10^{-4}} $
Now, % ionization $ =\frac{Ionised,HCOOH}{Total,HCOOH}=\frac{3.4\times {10^{-4}}}{0.001}\times 100=34 $%
