Given the information that follows, calculate the concentrations of all molecular and ionic species, and the pH in aqueous solutions

Question

Given the information that follows, calculate the concentrations of all molecular and ionic

species, and the pH in aqueous solution.

A) 0.01M acetic acid

B) 0.25M ammonium chloride

I don't even know where to start so any help would be great.

Answer 1

A) Acetic acid is a weak acid, so to start, you need to know the Ka. Looking it up, I find it to be 1.8x1^-5

Know we use that to find the [H+] when acetic acid, CH3COOH ionizes:

CH3COOH ==> CH3COO^- + H⁺ Ka = 1.8x10^-5

Ka = [CH3COO^-][H⁺] / [CH3COOH]

1.8x10^-5 = (x)(x) / 0.01 - x and if we assume x is small relative to 0.01 we can ignore it in denominator

1.8x10^-5 = x² / 0.01

x² = 1.8x10^-7

x = [H⁺] = 4.2x10^-4 (which is only ~4% of 0.01 so our assumption was valid, i.e. <5%)

pH = -log [H⁺] = -log 4.2x10^-4

pH = 3.38

Species present = 4.2x10^-4 M CH3COOH (molecular), 9.6x10^-3 M CH3COO^- (ionic) and 4.2x10^-4 M H+ (ionic). Also 1x10^-7 M OH^- will be present from the water that is present.

B). Ammonium chloride = NH4Cl. Being a soluble salt, it will dissociate completely in water as follows:

NH₄Cl(s) ==> NH₄⁺(aq) + Cl^-(aq)

Species present = NH₄⁺ (ionic) and Cl^- (ionic). Also present will be H⁺ and OH^-. (see below for concentrations)

To find the pH and concentrations of species present, we need to look at the hydrolysis of NH₄⁺ acting as as acid. The final pH should be <7 since NH₄Cl is the salt of a strong acid (HCl) and a weak base (NH₃).

NH₄⁺ + H₂O ==> NH₃ + H₃O⁺ ... hydrolysis of NH₄⁺ (Ka NH₄⁺ = 5.6x10-¹⁰, I looked it up)

Ka = [NH3][H3O⁺] / [NH₄⁺]

5.6x10^-10 = (x)(x) / 0.25 - x and assume x is small and neglect it in denominator

x² = 1.4x10-¹⁰

x = [H3O⁺] = 1.18x10^-5 M

pH = -log [H3O⁺] = -log 1.18x10^-5

pH = 4.93

Species present = 1.18x10^-5 M NH₃, 0.25 M NH₄⁺, 0.25 M Cl^-, 1.18x10^-5 M H⁺ and 8.47x10^-10 M OH^-