. What is the concentration of lead ion when manganese(II) ion begins to precipitate?
PbCO3= 1.5*10^-13
MnCO3= 1.8*10^-11
A solution contains 2.47×10-2 M lead acetate and 2.47×10-2 M manganese(II) nitrate. (NH4)2CO3(s) is added slowly to this mixture?
2016-07-17 9:25 am
回答 (1)
2016-07-17 10:29 am
Initial concentrations :
[Pb²⁺]ₒ = [Mn²⁺]ₒ = 2.47 × 10⁻² M
Ksp(PbCO₃) < Ksp(MnCO₃)
Hence, MnCO₃ will firstly precipitate.
When Mn²⁺ ion begins to precipitate :
[Mn²⁺]ₒ[CO₃²⁻] = 1.8 × 10⁻¹¹
(2.47 × 10⁻²) [CO₃²⁻] = 1.8 × 10⁻¹¹
[CO₃²⁻] = 7.29 × 10⁻¹⁰ M
Also, [Pb²⁺][CO₃²⁻] = 1.5 × 10⁻¹³
[Pb²⁺](7.29 × 10⁻¹⁰) = 1.5 × 10⁻¹³
[Pb²⁺] = 2.06 × 10⁻⁴ M
[Pb²⁺]ₒ = [Mn²⁺]ₒ = 2.47 × 10⁻² M
Ksp(PbCO₃) < Ksp(MnCO₃)
Hence, MnCO₃ will firstly precipitate.
When Mn²⁺ ion begins to precipitate :
[Mn²⁺]ₒ[CO₃²⁻] = 1.8 × 10⁻¹¹
(2.47 × 10⁻²) [CO₃²⁻] = 1.8 × 10⁻¹¹
[CO₃²⁻] = 7.29 × 10⁻¹⁰ M
Also, [Pb²⁺][CO₃²⁻] = 1.5 × 10⁻¹³
[Pb²⁺](7.29 × 10⁻¹⁰) = 1.5 × 10⁻¹³
[Pb²⁺] = 2.06 × 10⁻⁴ M
收錄日期: 2021-04-18 15:17:20
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