Respuesta :

Dejanras
Reaction of dissociation: Ag₂SO₄ → 2Ag⁺ + SO₄²⁻.
m(Ag₂SO₄) = 4 g.
V(Ag₂SO₄) = 1 l.
n(Ag₂SO₄) = m(Ag₂SO₄) ÷ M(Ag₂SO₄).
n(Ag₂SO₄) = 4 g ÷ 311,8 g/mol.
n(Ag₂SO₄) = 0,0128 mol.
n(Ag⁺) = 2 · 0,0128 mol = 0,0256 mol.
n(Ag₂SO₄) = n(SO₄²⁻) = 0,0128 mol.
c(Ag⁺) = n ÷ V = 0,0256 mol ÷ 1 l = 0,0256 mol/l.
Ksp = c(Ag⁺)² · c(SO₄²⁻).
Ksp = (0,0256 mol/l)² · 0,0128 mol/l.
Ksp = 8,3·10⁻⁶.

Eduard22sly

The solubility product (Ksp) for silver(i) sulphate Ag₂SO₄ in the solution is 8.39×10¯⁶

How to determine the mole of Ag₂SO₄

  • Mass of Ag₂SO₄ = 4 g
  • Molar mass of Ag₂SO₄ = (108×2) + 32 + (16×4) = 312 g/mol
  • Mole of Ag₂SO₄ =?

Mole = mass / molar mass

Mole of Ag₂SO₄ = 4 / 312

Mole of Ag₂SO₄ = 0.0128 mole

How to determine the molarity of Ag₂SO₄

  • Mole of Ag₂SO₄ = 0.0128 mole
  • Volume = 1 L
  • Molarity of Ag₂SO₄ =?

Molarity = mole / Volume

Molarity of Ag₂SO₄ = 0.0128 / 1

Molarity of Ag₂SO₄ = 0.0128 M

How to determine the solubility product

Dissociation equation

Ag₂SO₄ → 2Ag⁺ + SO₄²⁻

  • Molarity of Ag₂SO₄ = 0.0128 M
  • Molarity of Ag⁺ = 2 × 0.0128 = 0.0256 M
  • Molarity of SO₄²⁻ = 0.0128 M
  • Solubility product (Ksp) =?

Ksp = [Ag⁺]² × [SO₄²⁻]

Ksp = (0.0256)² × 0.0128

Ksp = 8.39×10¯⁶

Learn more about stoichiometry:

https://brainly.com/question/14735801

Otras preguntas