Question

Formic acid is 4.5% dissociated in 0.1 N solution at 20 C.Ionization constant of formic acid

Answer 1

Answer ⇒ Ka = 2.1 × 10⁻⁴ mole/L.

Explanation ⇒ Since, Acetic acid is an Mono basic acid, thus it released only one H positive ion when dissociated.

Let the degree of dissociation of acetic acid be α.

Then,

HCOOH ⇒ H⁺ + HCOO⁻

Conc. of HCOOH = c(1 - α)

Conc. of H⁺ = cα and Conc. of COOH⁻ = cα

Thus, Using the concept of Ionic Equilibrium,

Ka = c²α²/c(1 - α)

Ka = cα²/(1 - α)

This formula was also the statement of the Ostwald dilution law.

Now, α = 4.5 %

and c = molarity = Normality (since acid is mono basic). = 0.1 M.

∴ Ka = [0.1 × 4.5²/100²] ÷ [1 - 4.5/100]

∴ Ka = 0.0002025 ÷ 0.955

∴ Ka = 0.00021 = 2.1 × 10⁻⁴ mole/L.

Unit has been written from C as there is no other term containing any unit. Thu, unit of Ka will be the unit of Concentration only.

This is the required value of Ionization constant of acid.

Hope it helps.