In 100 mL of distilled water, 1.0 g of Ba(BrO3)2 is added. What will be the excess
mass of solid? In other words, how much of the 1.0 g of Ba(BrO3)2 doesn’t dissolve? The Ksp
of Ba(BrO3)2 is 7.8 x 10-6
chemistry?
2016-06-16 3:28 pm
回答 (1)
2016-06-16 4:46 pm
Let s mol/L be the solubility of Ba(BrO₃)₂ in water.
Ba(BrO₃)₂(s) ⇌ Ba²⁺(aq) + 2BrO₃⁻(aq)
At equilibrium : [Ba²⁺] = s mol/L, [BrO₃⁻] = 2s mol/L
Ksp = [Ba²⁺] [BrO₃⁻]²
7.8 × 10⁻⁶ = s × (2s)²
4s³ = 7.8 × 10⁻⁶
s = ∛[(7.8 × 10⁻⁶)/4]
s = 0.0125
Solubility of Ba(BrO₃)₂ = 0.0125 mol/L
Molar mass of Ba(BrO₃)₂ = 137.3 + 79.9×2 + 16.0×6 = 393.1 g/mol
Mass of Ba(BrO₃)₂ dissolved = (0.0125 mol/1000 mL) × (393.1 g/mol) × (100 mL) = 0.49 g
Excess mass of Ba(BrO₃)₂ solid = (1.0 - 0.49) g = 0.51 g
Ba(BrO₃)₂(s) ⇌ Ba²⁺(aq) + 2BrO₃⁻(aq)
At equilibrium : [Ba²⁺] = s mol/L, [BrO₃⁻] = 2s mol/L
Ksp = [Ba²⁺] [BrO₃⁻]²
7.8 × 10⁻⁶ = s × (2s)²
4s³ = 7.8 × 10⁻⁶
s = ∛[(7.8 × 10⁻⁶)/4]
s = 0.0125
Solubility of Ba(BrO₃)₂ = 0.0125 mol/L
Molar mass of Ba(BrO₃)₂ = 137.3 + 79.9×2 + 16.0×6 = 393.1 g/mol
Mass of Ba(BrO₃)₂ dissolved = (0.0125 mol/1000 mL) × (393.1 g/mol) × (100 mL) = 0.49 g
Excess mass of Ba(BrO₃)₂ solid = (1.0 - 0.49) g = 0.51 g
收錄日期: 2021-04-18 15:08:45
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160616072858AAQzVoH