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The solubility of barium chloride (BaCl₂, 208.23 g/mol) at 20°C is 35.8 g in 100 mL of water. What is the Ksp at this temperature?

Answer :

The Ksp of barium chloride (BaCl₂) at 20⁰C is 1.17 x 10⁻³.

Barium chloride (BaCl₂) dissociates into Ba²⁺ and 2Cl⁻ ions in water, according to the equation: BaCl₂ ⇌ Ba²⁺ + 2Cl⁻. Given that 35.8 g of BaCl₂ dissolves in 100 ml of water at 20⁰C, we can calculate the molar concentration of BaCl₂:

Molarity (M) = moles of solute / volume of solution (in liters)

First, we find the moles of BaCl₂:

Moles = mass / molar mass = 35.8 g / 208.23 g/mol = 0.172 mol.

Now, we find the molarity of BaCl₂:

Molarity = 0.172 mol / (100 ml / 1000) = 1.72 M.

Since BaCl₂ dissociates into 1 Ba²⁺ and 2 Cl⁻ ions, the concentration of Ba²⁺ ions is equal to the molarity of BaCl₂, and the concentration of Cl⁻ ions is twice that. Thus, [Ba²⁺] = 1.72 M and [Cl⁻] = 2 * 1.72 M = 3.44 M.

Now, we use these concentrations to calculate the Ksp using the formula:

Ksp = [Ba²⁺] * [Cl⁻]².

Plugging in the values:

Ksp = (1.72 M) * (3.44 M)² = 1.17 x 10⁻³.

Therefore, the Ksp of BaCl₂ at 20⁰C is 1.17 x 10⁻³, indicating its solubility product under these conditions.

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