JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Sample Paper Topic Test - Ionic Equilibrium

The ionization constant of \[HCOOH\] is \[1.8\times {{10}^{-4}}\]. What is the percent ionization of a 0.001 M solution?

A) 66%

B) 42%

C) 34%

D) 58%

Correct Answer: C

Solution :

[c] \[\underset{0.001-x}{\mathop{HCOOH}}\,\,{{\underset{x}{\mathop{HCOO}}\,}^{-}}+{{\underset{x}{\mathop{H}}\,}^{+}}\]

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

\[=\frac{\text{Ionised}\,HCOOH}{\text{Total}\,HCOOH}=\frac{3.4\times {{10}^{-4}}}{0.001}\times 100=34%\]