Question

The Ksp of barium sulfate (BaSO4) is 1.1 × 10–10. What is the solubility concentration of sulfate ions in a saturated solution at 25°C?
a.1.0 x 10^-5m
b. 1.5 x10^-5m
c. 5.5 x 10^-11m
d.7.4 x 10^-6m

Answer 1

Each mole of barium sulfate releases equal moles of barium ions and sulfate ions. Let the concentration of one substance be x. Thus: 
Ksp = [Ba⁺²][SO₄⁻²]
1.1 x 10⁻¹⁰ = (x)(x)
x = 1.05 x 10⁻⁵

The answer is A.

Answer 2

Given:

Ksp of BaSO4 = 1.1 * 10⁻¹⁰

To determine:

The solubility concentration of SO4²⁻ ions

Explanation:

BaSO4 ↔ Ba²⁺(aq) + SO4²⁻(aq)

Ksp = [Ba²⁺][SO4²⁻]

if 's' is the solubility of the ions then we have

Ksp = s²

s = √ksp = √1.1*10⁻¹⁰ = 1.05*10⁻⁵M

Ans: solubility of sulfate ions is 1.05*10⁻⁵ M