How do you calculate Molar solubility in grams/100 mL of calcium iodate in water at 25°C? Ksp = 7.1 × 10⁻⁷

To calculate the molar solubility of calcium iodate (Ca(IO3)2) in grams per 100 mL of water at 25°C, follow these steps:

Understanding the Dissociation

Calcium iodate dissociates in water as follows:

• Ca(IO3)2 (s) ⇌ Ca²⁺ (aq) + 2 IO3⁻ (aq)

This means that one mole of calcium iodate produces one mole of calcium ions and two moles of iodate ions.

Setting Up the Expression

The solubility product constant (Ksp) is given by:

• Ksp = [Ca²⁺][IO3⁻]²

Let the molar solubility of calcium iodate be represented by 's' (in moles per liter). Then:

• [Ca²⁺] = s
• [IO3⁻] = 2s

Substituting into the Ksp Expression

Now, substitute these values into the Ksp expression:

• Ksp = s(2s)² = 4s³

Given that Ksp = 7.1 × 10⁻⁷, we can set up the equation:

• 4s³ = 7.1 × 10⁻⁷

Solving for 's'

To find 's', rearrange the equation:

• s³ = (7.1 × 10⁻⁷) / 4
• s³ = 1.775 × 10⁻⁷

Now, take the cube root:

• s ≈ 0.0582 moles/L

Converting to Grams per 100 mL

Next, convert moles to grams. The molar mass of calcium iodate is approximately 391.14 g/mol. Therefore:

• Mass = moles × molar mass
• Mass = 0.0582 moles × 391.14 g/mol ≈ 22.8 grams/L

Since we want the solubility in grams per 100 mL, divide by 10:

• 22.8 g/L ÷ 10 = 2.28 g/100 mL

Final Result

The molar solubility of calcium iodate in water at 25°C is approximately 2.28 grams per 100 mL.