The Ksp of manganese(II) carbonate, MnCO3, is 2.42×10−11. Calculate the solubility of this compound in g/L.
g/L
chem help?
2020-04-03 11:59 pm
回答 (1)
2020-04-04 12:13 am
✔ 最佳答案
Molar mass of MnCO₃ = 54.9 + 12.0 + 16.0×3 = 114.9 g/mol__MnCO₃(s) ⇌ Mn²⁺(aq) + CO₃²⁻(aq) ___ Ksp = 2.42 × 10⁻¹¹
Equilibrium: ___ s M ______ s M
At quilibrium:
Ksp = [Mn²⁺] [CO₃²⁻]
2.42 × 10⁻¹¹ = s²
s = √(2.42 × 10⁻¹¹) = 4.92 × 10⁻⁶
Solubility = (4.92 × 10⁻⁶ mol/L) × (114.9 g/mol) = 5.65 × 10⁻⁴ g/L
收錄日期: 2021-04-23 23:06:21
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