更新1:
ksp*
A solution contains [Ba+2]=0.75 M and [Pb+2]=0.015 M. What is the [SO4] needed to selectively precipitate the barium?
回答 (1)
2018-04-10 4:22 pm
BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq) …… Ksp = [Ba²⁺] [SO₄²⁻] = 1.1 × 10⁻¹⁰
[SO₄²⁻] needed to selectively precipitate the barium
= Ksp / [Ba²⁺]
= (1.1 × 10⁻¹⁰) / 0.75 M
= 1.5 × 10⁻¹⁰ M
[SO₄²⁻] needed to selectively precipitate the barium
= Ksp / [Ba²⁺]
= (1.1 × 10⁻¹⁰) / 0.75 M
= 1.5 × 10⁻¹⁰ M
收錄日期: 2021-04-24 01:01:31
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20180410002716AA8OT1l