Question

A 0.50 M solution of formic acid, HCOOH, has a pH of 2.02. Calculate the percent ionization of HCOOH

Answer 1

Answer : 1.91 %

Explanation : The steps to solve this problem are explained below;

1. HCOOH ⇄
`HCOO^(-) + H^(+)`

Here Ka =(
`[HCOO^(-)]_(eq) X  [H^(+)]_(eq)` )/
`[HCOOH]_(eq)`

As the equilibrium concentration of
`H^(+)` will be the pH of the solution.

∴
`[H^(+)]_(eq)` =
`10^((-2.02)) = 9.55 x [tex] 10^(-3)` M

2. The initial concentration of HCOOH. When it loses x moles from it as the acid undergoes dissociation to form
`HCOO^(-)` and
`H^(+)`.

3. The moles present will be as

[HCOOH] (M)
`[H^(+)]`(M)
`[HCOO^(-)]`(M)

Initial 0.50 0.00 0.00

After Change -x +x +x

Equilibrium ( 0.50 -x) x x

∴ Ka = (x) x (x) / (0.50 - x)

4. Assuming that all of the
`H^(+)` comes from the acid, and none from water.

As
`[H^(+)]_(eq)` = 9.55 x
`10^(-3)` which is much higher than the 1.0 x
`10^(-7 )` M [tex[H^{+}[/tex] from water.

Also, the concentration of HCOOH will change very little, from 0.50 to 0.50 - 9.55 x
`10^(-3)`.

The change in concentration can be ignored if it is less than 5% of the original concentration.

∴ 0.50 M x 5% = 0.025, so the change in [HCOOH] in this problem can be ignored.

Now, Ka = (x)(x)/0.50 = (9.55 x
`10^(-3))^(2)` /0.50= 1.82 x
`10^(-4)`

Now, calculating the percent ionization for this problem.

which will represent the relative number of acid molecules which dissociate. It is calculated as :

`[H^(+)]_(eq)` x 100 /
`[HCOOH]_(i)`

∴ percent ionization = {(9.55 x
`10^(-3))`/ (0.50)}x 100 = 1.91 %

This value of 1.91 % indicates that very little of this acid dissociates (ionizes) under these conditions.

For strong acids and bases, the percent ionization is 100%.

Answer 2

Answer is: the percent ionizationof formic acid is 1,82%.
Chemical reaction: HCOOH(aq) ⇄ H⁺(aq) + HCOO⁻(aq).
pKa(HCOOH) = 3,77.

Ka(HCOOH) = 1,7·10⁻⁴.

c(HCOOH) = 0,5 M.

[H⁺] = [HCOO⁻] = x; equilibrium concentration.
[HA] = 0,1 M - x.
Ka = [H⁺] · [HCOO⁻] / [HCOOH].
0,00017 = x² / 0,5 M - x.
Solve quadratic equation: x = 0,0091 M.
α = 0,0091 M ÷ 0,5 M · 100% = 1,82%.