Molar solubility help?
2020-04-07 1:11 am
The Ksp of strontium carbonate, SrCO3, is 5.60×10^−10M^2. Calculate the molar solubility, S, of this compound
回答 (2)
2020-04-07 1:19 am
SrCO₃(s) ⇌ Sr²⁺(aq) + CO₃²⁻(aq) ___ Ksp = 5.60 × 10⁻¹⁰ M²
Initial: _____ 0 M _____ 0 M
Change: __ +s M ____ +s M
Eqm: _____ s M ______ s M
At equilibrium:
Ksp = [Sr²⁺] [CO₃²⁻]
5.60 × 10⁻¹⁰ = s²
s = √(5.60 × 10⁻¹⁰) = 2.37 × 10⁻⁵
Molar solubility of SrCO₃ = 2.37 × 10⁻⁵ M
Initial: _____ 0 M _____ 0 M
Change: __ +s M ____ +s M
Eqm: _____ s M ______ s M
At equilibrium:
Ksp = [Sr²⁺] [CO₃²⁻]
5.60 × 10⁻¹⁰ = s²
s = √(5.60 × 10⁻¹⁰) = 2.37 × 10⁻⁵
Molar solubility of SrCO₃ = 2.37 × 10⁻⁵ M
2020-04-07 1:15 am
SrCO3(s) <--> Sr2+(aq) + CO32-(aq)
Ksp = [Sr2+][CO32-] = 5.60X10^-10
Let S = [Sr2+] = CO32-] = molar solubility. Then,
S^2 = 5.60X10^-10
S = 2.37X10^-5 M
Ksp = [Sr2+][CO32-] = 5.60X10^-10
Let S = [Sr2+] = CO32-] = molar solubility. Then,
S^2 = 5.60X10^-10
S = 2.37X10^-5 M
收錄日期: 2021-05-01 22:34:40
原文連結 [永久失效]:
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