Question

The maximum allowable concentration of pb2+ ions in drinking water is 0.05 ppm (i.e., 0.05 g of pb2+ in 1 million grams of water). (a) what is the pb2+ concentration of an underground water supply at equilibrium with the mineral anglesite (pbso4) (ksp = 1.6 × 10−8) ?

Answer 1

PbSO₄ partially dissociates in water. the balanced equation is;

PbSO₄(s) ⇄ Pb²⁺(aq) + SO₄²⁻(aq)
Initial - -
Change -X +X +X
Equilibrium X X

Ksp = [Pb²⁺(aq)] [SO₄²⁻(aq)]
1.6 x 10⁻⁸ = X * X
1.6 x 10⁻⁸ = X²
X = 1.3 x 10⁻⁴ M

Hence the Pb²⁺ concentration in underground water is 1.3 x 10⁻⁴ M.
[Pb²⁺] = 1.3 x 10⁻⁴ M.
= 1.3 x 10⁻⁴ mol / L x 207 g / mol
= 26.91 ppm