Calculate the hydrogen ion concentration in 0.1 M acetic acid solution

1.	Calculate the hydrogen ion concentration in 0.1 M acetic acid solution when the acetic acid is 2% dissociated in the solution.
Answer» The dissociation of acetic acid is represented below : CH₃COOH ⇌ CH₃ – COO^- + H⁺ Given : Dissociation = 2%, C = 0.1 M [H₃O⁺] = ? The concentration of hydrogen ion, [H⁺], is given by the following formula : [H₃O⁺] = αC α = Degree of dissociation = \(\frac{Percent\,dissociation}{100}\) = \(\frac{2}{100}\)= 0.02 [H₃O⁺] = Hydrogen ion concentration = ? C = Molar concentration of acetic acid = 0.1 M = 0.1 mol dm^-3 ∴ [H₃O⁺] = C × α = 0.1 × 0.02 = 0.002 = 2.0 × 10^-3 mol dm^-3 ∴ Hydrogen ion concentration = 2.0 × 10^-3 mol dm^-3

Calculate the hydrogen ion concentration in 0.1 M acetic acid solution when the acetic acid is 2% dissociated in the solution.

Answer»

The dissociation of acetic acid is represented below :

CH₃COOH ⇌ CH₃ – COO^- + H⁺

Given :

Dissociation = 2%, C = 0.1 M

[H₃O⁺] = ?

The concentration of hydrogen ion, [H⁺], is given by the following formula :

[H₃O⁺] = αC

α = Degree of dissociation = \(\frac{Percent\,dissociation}{100}\)

= \(\frac{2}{100}\)= 0.02

[H₃O⁺] = Hydrogen ion concentration = ?

C = Molar concentration of acetic acid

= 0.1 M = 0.1 mol dm^-3

∴ [H₃O⁺] = C × α = 0.1 × 0.02

= 0.002 = 2.0 × 10^-3 mol dm^-3

∴ Hydrogen ion concentration = 2.0 × 10^-3 mol dm^-3