0.2 moler solution of formic acid is 3.2% ionized. Its ionization constant is 

  • Option 1)

    9.6x10-3

  • Option 2)

    2.1x10-4

  • Option 3)

    1.25x10-6

  • Option 4)

    4.8x10-5

 

Answers (1)

As we discussed 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.

 

 C=0.2M\\\alpha =3.2\% \: or \: 0.032

\underset{c(1-\alpha)}{HCOOH} \rightleftharpoons \underset{c\alpha }{H^+} + \underset{c\alpha }{COOH^-}

 

Ionization constant K_{a}=\:\:\:\:\:\:\:\frac{[COOH^{-})][H^{+}]}{[HCOOH]}

                              K_{a}\:=\:\frac{c\alpha \times c\alpha }{c(1-\alpha )}

                           =   \frac{(0.2\times 0.032)^{2}}{0.2\times 0.968}

                           = 2.1\times 10^{-4}

 


Option 1)

9.6x10-3

This option is incorrect.

Option 2)

2.1x10-4

This option is correct.

Option 3)

1.25x10-6

This option is incorrect.

Option 4)

4.8x10-5

This option is incorrect.

Preparation Products

Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout BITSAT-JEE Main 2021

An exhaustive E-learning program for the complete preparation of JEE Main and Bitsat.

₹ 27999/- ₹ 16999/-
Buy Now
Exams
Articles
Questions