An aqueous solution is 0.10 M in both sulfate ion and chromate ion. Powdered lead nitrate(Ⅱ) is slowly dissolved in the well-stirred solution. (Assume no volume change).
(a) What is the concentration of Pb²⁺ when PbSO₄ first start to precipitate?
(b)What is the concentration of Pb²⁺ when PbCrO₄ first start to precipitate?
(c) Which compound precipitate first?
(d)When the second anion just start to precipitate, what is the molar concentration of the anion that precipitate first?
I am working on all these questions. If any member knows correct answers, may reply those answers.
Aqueous solution in both sulfate ion and chromate ion
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Dhamnekar Winod
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Aqueous solution in both sulfate ion and chromate ion
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
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ChenBeier
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Re: Aqueous solution in both sulfate ion and chromate ion
Hint: solubility product.
Re: Aqueous solution in both sulfate ion and chromate ion
These questions are currently occupying my time. Any member who knows the proper answers can reply with them.
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ChenBeier
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Re: Aqueous solution in both sulfate ion and chromate ion
As I wrote check the solubility products of both.
With them you can calculate the lead concentration.
With them you can calculate the lead concentration.
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Dhamnekar Winod
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Re: Aqueous solution in both sulfate ion and chromate ion
Answers
(a)[Pb²⁺] = 1.6 ⨯ 10⁻⁸ ÷ 0.10 =1.6⁻⁷
(b)[Pb²⁺] = 1.7 ⨯ 10⁻¹⁴ ÷ 0.10 = 1.7 ⨯ 10⁻¹³
(c)PbCrO₄ will precipitate first.
(d) CrO₄²⁻ = 1.7 ⨯ 10⁻¹⁴ ÷ 1.6 ⨯ 10⁻⁸ = 1.1 ⨯ 10⁻⁶
(a)[Pb²⁺] = 1.6 ⨯ 10⁻⁸ ÷ 0.10 =1.6⁻⁷
(b)[Pb²⁺] = 1.7 ⨯ 10⁻¹⁴ ÷ 0.10 = 1.7 ⨯ 10⁻¹³
(c)PbCrO₄ will precipitate first.
(d) CrO₄²⁻ = 1.7 ⨯ 10⁻¹⁴ ÷ 1.6 ⨯ 10⁻⁸ = 1.1 ⨯ 10⁻⁶
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]