Question

A chemist attempts to separate barium ions from lead ions by using the sulfate ion as a precipitating agent. the ksp values of baso4 and pbso4 are 1.1 × 10−10 and 1.6 × 10−8 , respectively. what two sulfate ion concentrations are required for the precipitation of baso4 and pbso4 from a solution containing 0.099 m ba2+(aq) and 0.099 m pb2+(aq)? 1. 2.47501 × 10−9 m, 3.60002 × 10−7 m 2. 1.02154 × 10−9 m, 1.48587 × 10−7 m 3. 2.17763 × 10−9 m, 3.16746 × 10−7 m 4. 1.11111 × 10−9 m, 1.61616 × 10−7 m 5. 3.68485 × 10−9 m, 5.35977 × 10−7 m

Answer 1

4. 1.11111 x 10⁻⁹ M, 1.61616 x 10⁻⁷ M.

Step-by-step explanation:

• The main idea here to solve this problem is that to precipitate a substance, the ionic products of the ions that form this substance should be ≥ its solubility product (Ksp).

• To precipitate BaSO₄, The ionic product of BaSO₄ ([Ba²⁺][SO₄²⁻]) should be ≥ Ksp of BaSO₄ (Ksp = 1.1 x 10⁻¹⁰).

BaSO₄ ↔ Ba²⁺ + SO₄²⁻

Ksp = [Ba²⁺][SO₄²⁻]

• [SO₄²⁻] = Ksp / [Ba²⁺] = (1.1 x 10⁻¹⁰) / (0.099) = 1.11111 x 10⁻⁹ M.

By the same way; the other precipitate:

• To precipitate PbSO₄, The ionic product of PbSO₄ ([Pb²⁺][SO₄²⁻]) should be ≥ Ksp of PbSO₄ (Ksp = 1.6 x 10⁻⁸).

PbSO₄ ↔ Pb²⁺ + SO₄²⁻

Ksp = [Pb²⁺][SO₄²⁻]

• [SO₄²⁻] = Ksp / [Pb²⁺] = (1.6 x 10⁻⁸) / (0.099) = 1.61616 x 10⁻⁷ M.

So, the right answer is 4. 1.11111 x 10⁻⁹ M, 1.61616 x 10⁻⁷ M.