Question

.0.2 molar solution of formic acid is 3.2 percent ionized, the ionization constant is

Answer 1

Since

K alpha = {C (alpha)^2}/1-alpha
Here C= 0.2 M
alpha = 3.2% or 0.032
HCOOH = H+ + COOH-

K alpha = [COOH-][H+]/[HCOOH]
=( C alpha × C alpha )/C (1-alpha)
= (0.2×0.032)^2/(0.2×0.968)
= 2.1×10^-4 is right answer

Answer 2

Answer: The ionization constant for the solution of formic acid is $2.11\times 10^{-4}$

Explanation: We are given a solution of formic acid. Ionization equation for this is:

            $HCOOH+H_2O\rightarrow H_3O^++COOH^-$

at t = 0         c                           0            0

at $t=t_{eq}$     $c-c\alpha$                     $c\alpha$          $c\alpha$

Dissociation constant of an acid or ionization constant of the above equation is written as:

$K_a=\frac{c^2\alpha ^2}{c(1-\alpha )}$

We are given:

c = 0.2M

$\alpha =3.2\%=0.032$

Putting the values in above equation, we get:

$K_a=\frac{0.2\times (0.032)^2}{(1-0.032)}$

$K_a=2.11\times 10^{-4}$