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The self ionisation constant for pure formic acid,K=[HCOOH_{2}^{+}] [HCOO^{-}] has been estimated as 10-6 at 1.22g/cm3.

The percentage of formic acid molecules in pure formic acid converted to formate ion is

  • Option 1)

    0.002 percentage 

  • Option 2)

    0.004 percentage 

  • Option 3)

    0.006 percentage 

  • Option 4)

    0.008 percentage 

 

Answers (1)

best_answer

As we learned in concept

Ionization constant of weak acids -

Consider 

HX(aq)+H_{2}O(l)\rightleftharpoons H_{3}O^{+}(aq)+\bar{X}(aq)
 

K_{a}=dissociation\:or\:ionization\:constant

- wherein

K_{a}=\frac{C^{2}\:\alpha ^{2}}{C(1-\alpha )}

K_{a}=\frac{C\alpha ^{2}}{1-\alpha }
 

C=initial\:concentration\:of\:undissociated\:acid

\alpha =extent\:upto\:which\:HX\:is\:ionized\:into\:ions.

 

 No. of moles of formic acid in one L of solution =1.22\times 1000\times \frac{1}{46}= 26.5 mol

2HCOOH\rightleftharpoons HCOO^- +HCOOH_2^+

K= 10^{-6}= [HCOO^-][HCOOH_2^+]

[HCOO^-]=[HCOOH_2^+]=10^{-3}M

% dissociation =  \frac{10^{-3}}{26.5}\times 100 =0.004 \%


Option 1)

0.002 percentage 

Incorrect

Option 2)

0.004 percentage 

Correct

Option 3)

0.006 percentage 

Incorrect

Option 4)

0.008 percentage 

Incorrect

Posted by

Aadil

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