Question

A $0.2$ molar solution of formic acid is $3.2\%$ ionised, its ionisation constant is:

• A. $9.6\times {10}^{-3}$
• B. $2.1\times {10}^{-4}$
• C. $1.25\times {10}^{-6}$
• D. $2.1\times {10}^{-8}$

Answer 1

Answer: (B)

$2.1\times {10}^{-4}$

$\underset{\text{At equillibrium}:}{\underset{\text{Initial amount}:}{}}\underset{(0.2-x)}{\underset{0.2}{HCOOH}}\leftrightharpoons \underset{x}{\underset{0}{{HCOO}^{-}}}+\underset{\left(x\right)}{\underset{0}{H^{+}}}$

$K_a=\frac{\left[{HCOO}^{-}\right]\left[H^{+}\right]}{\left[HCOOH\right]}=\frac{x\times x}{(0.2-x)}$

Given that $3.2\%$ of 0.2 M formic acid is ionised.

$\therefore x=\frac{3.2}{100}\times 0.2=0.0064$

Therefore, at equillibrium,

$\left[HCOOH\right]=0.2-0.0064=0.1936$

$\left[{HCOO}^{-}\right]=0.0064$

$\left[H^{+}\right]=0.0064$

$\therefore K_a=\frac{0.0064\times 0.0064}{0.1936}=\frac{0.4\times {10}^{-4}}{0.1936}=2.1\times {10}^{-4}$

Hence, option B is correct.