Question

A 5.0 gram sample of Ag2SO4, will dissolve in 1.0 L of water. Calculate the solubility product

constant for this salt.​

Answer 1

Ksp = 1.64 ×10⁻⁵

Step-by-step explanation:

Given data:

Mass of Ag₂SO₄ = 5.0 g

Volume of water = 1.0 L

Solubility product = Ksp = ?

Solution:

First of all we will calculate the molarity.

Molarity = Number of moles of solute / Volume of solution in L

Number of moles = mass/ molar mass

Number of moles = 5.0 g/311.74 g/mol

Number of moles = 0.016 mol

Molarity = 0.016 mol / 1 L

Molarity = 0.016 M

Dissociation equation:

Ag₂SO₄ ⇔ 2Ag⁺ + SO₄²⁻

Ksp = [Ag⁺]² [SO₄²⁻]

Ksp = [0.032]²[0.016]

Ksp = 1.64 ×10⁻⁵