The solubility of barium carbonate, BaCO3, is 0.0100 g/L.

The solubility of barium carbonate, BaCO3, is 0.0100 g/L. Its molar mass is 197.3 g/mol. What is the Ksp of barium carbonate? 2. The solubility product constant of calcium sulfate, CaSO4, is 7.10×10^-5. Its molar mass is 136.1 g/mol. How many grams of calcium sulfate can dissolve in 61.0 L of pure water?

The Correct Answer and Explanation is:

Let’s solve both parts of the problem step by step.

1. Solubility of BaCO₃ → Ksp Calculation

Given:

• Solubility of BaCO₃ = 0.0100 g/L
• Molar mass of BaCO₃ = 197.3 g/mol

Step 1: Convert solubility to mol/L (molarity)Molar solubility=0.0100 g/L197.3 g/mol=5.07×10−5 mol/L\text{Molar solubility} = \frac{0.0100 \text{ g/L}}{197.3 \text{ g/mol}} = 5.07 \times 10^{-5} \text{ mol/L}Molar solubility=197.3 g/mol0.0100 g/L​=5.07×10−5 mol/L

Step 2: Write the dissociation equationBaCO₃ (s)⇌Ba2+(aq)+CO₃2−(aq)\text{BaCO₃ (s)} \rightleftharpoons \text{Ba}^{2+} (aq) + \text{CO₃}^{2-} (aq)BaCO₃ (s)⇌Ba2+(aq)+CO₃2−(aq)

Let the molar solubility be x mol/L, then:

• [Ba²⁺] = x
• [CO₃²⁻] = x

Step 3: Write the Ksp expressionKsp=[Ba2+][CO₃2−]=x⋅x=x2K_{sp} = [\text{Ba}^{2+}][\text{CO₃}^{2-}] = x \cdot x = x^2Ksp​=[Ba2+][CO₃2−]=x⋅x=x2Ksp=(5.07×10−5)2=2.57×10−9K_{sp} = (5.07 \times 10^{-5})^2 = 2.57 \times 10^{-9}Ksp​=(5.07×10−5)2=2.57×10−9

✅ Answer for part 1:
Ksp of BaCO₃ = 2.57 × 10⁻⁹

2. Ksp of CaSO₄ → Solubility in grams in 61.0 L

Given:

• Ksp of CaSO₄ = 7.10 × 10⁻⁵
• Molar mass = 136.1 g/mol
• Volume = 61.0 L

Step 1: Write dissociation equationCaSO₄ (s)⇌Ca2+(aq)+SO₄2−(aq)\text{CaSO₄ (s)} \rightleftharpoons \text{Ca}^{2+} (aq) + \text{SO₄}^{2-} (aq)CaSO₄ (s)⇌Ca2+(aq)+SO₄2−(aq)

Let the solubility = s mol/L, then:

• [Ca²⁺] = s
• [SO₄²⁻] = s

Ksp=s2⇒s=7.10×10−5=8.43×10−3 mol/LK_{sp} = s^2 \Rightarrow s = \sqrt{7.10 \times 10^{-5}} = 8.43 \times 10^{-3} \text{ mol/L}Ksp​=s2⇒s=7.10×10−5​=8.43×10−3 mol/L

Step 2: Find moles in 61.0 LMoles=s⋅V=(8.43×10−3)⋅61.0=0.5142 mol\text{Moles} = s \cdot V = (8.43 \times 10^{-3}) \cdot 61.0 = 0.5142 \text{ mol}Moles=s⋅V=(8.43×10−3)⋅61.0=0.5142 mol

Step 3: Convert to gramsMass=moles⋅molar mass=0.5142⋅136.1=69.98 g\text{Mass} = \text{moles} \cdot \text{molar mass} = 0.5142 \cdot 136.1 = 69.98 \text{ g}Mass=moles⋅molar mass=0.5142⋅136.1=69.98 g

✅ Answer for part 2:
69.98 grams of CaSO₄ can dissolve in 61.0 L of water

Summary

1. Ksp of BaCO₃ is 2.57 × 10⁻⁹
2. Mass of CaSO₄ that can dissolve in 61.0 L is 69.98 g

These calculations demonstrate how solubility and the solubility product constant are connected. Using stoichiometry, molar mass, and basic equilibrium principles, we can easily determine both Ksp and the mass of a salt that can dissolve in a specific volume of water.