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The self-ionization constant for pure formic acid \mathrm{K=\left[\mathrm{HCOOH}_{2}^{+}\right]\left[\mathrm{HCOO}^{-}\right]} has been estimated as 10^{-7} room temperature. What percentage of formic acid molecules in pure formic acid is converted to formate ions? The density of formic acid is \mathrm{1.25\ g/cm^{3}}

Option: 1


Option: 2


Option: 3


Option: 4


Answers (1)


The density of HCOOH  is given as \mathrm{1.25\ g/cm^{3}}

\therefore Mass of HCOOH in 1 litre  solution \mathrm{= 1.25 \times 10^{3}\ g} 

The concentration of HCOOH , 

\mathrm{c = \frac{1.25\times 10^{3}}{46} = 27.17\: M}

As it is given the HCOOH goes to auto-ionization, so \left[\mathrm{HCOOH}_{2}^{+}\right]=\left[\mathrm{HCOO}^{-}\right]

Also, \left[\mathrm{HCOOH}_{2}^{+}\right]\left[\mathrm{HCOO}^{-}\right]= 10^{-7}

\therefore \left[\mathrm{HCOO}^{-}\right]= \sqrt{10^{-7}} =3.16\times 10^{-4}

 \text{\% dissociation of HCOOH = }\frac{\left[\mathrm{HCOO}^{-}\right] \times 100}{[\mathrm{HCOOH}]}

\Rightarrow \alpha = \frac{3.16\times 10^{-4}\times 100}{27.17} = 0.001\%

Hence, the correct answer is Option (2)

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