BaSO4 Ksp=1.1*10^-10
PbSO4 Ksp=6.3*10^-7
Lead or Barium?
A solution contains [Ba+2]=0.075M and [Pb+2]=0.015M. Which one will precipitate first if 1M Na2SO4 is added to the solution?
2018-04-09 11:44 pm
回答 (1)
2018-04-09 11:55 pm
Minimum [SO₄²⁻] to precipitate Ba²⁺ ion
= Ksp(BaSO₄) / [Ba²⁺]
= (1.1 × 10⁻¹⁰) / (0.075) M
= 1.5 × 10⁻⁹ M
Minimum [SO₄²⁻] to precipitate Pb²⁺ ion
= Ksp(BaSO₄) / [Pb²⁺]
= (6.3 × 10⁻⁷) / (0.015)
= 4.2 × 10⁻⁵ M
Ba²⁺ ions need a lower concentration to precipitate than Pb²⁺ ions. Hence, BaSO₄ will precipitate first.
= Ksp(BaSO₄) / [Ba²⁺]
= (1.1 × 10⁻¹⁰) / (0.075) M
= 1.5 × 10⁻⁹ M
Minimum [SO₄²⁻] to precipitate Pb²⁺ ion
= Ksp(BaSO₄) / [Pb²⁺]
= (6.3 × 10⁻⁷) / (0.015)
= 4.2 × 10⁻⁵ M
Ba²⁺ ions need a lower concentration to precipitate than Pb²⁺ ions. Hence, BaSO₄ will precipitate first.
收錄日期: 2021-05-01 14:49:36
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20180409154424AA5ZHmn