A 100.-mL sample of solution contains 10.0 mmol of Ca²⁺ ion.How many mmol of solid Na₂SO₄ must be added in order to cause precipitation of 99.9% of the calcium as CaSO₄? The K_sp of CaSO₄ is 6.1 $\times$ 10^-⁵.Assume the volume remains constant. A)17.4 B)10.0 C)61.0 D)71.0 E)6.1

Question

Quizplus · Accepted Answer

First, calculate the initial concentration of CaS1U1P12+S1S1P0 ion in the solution: 
- 10.0 mmol of CaS1U1P12+S1S1P0 ion in 100 mL of solution gives a concentration of 0.100 M.

Next, use the Ksp value of CaSOS1U1B14S1U1B0 to find the concentration of CaSOS1U1B14S1U1B0 when precipitation begins: 
CaSOS1U1B14S1U1B0 ⇌ CaS1U1P12+S1S1P0 + SO1U1B24S1U1B0 
Ksp = [CaS1U1P12+S1S1P0][SO1U1B24S1U1B0] = 6.1 x 10^-10 
Let x be the concentration of CaSOS1U1B14S1U1B0 at equilibrium. Then: 
6.1 x 10^-10 = (0.100 - x)(x) 
Solve for x: 
x = 6.1 x 10^-6 M

The amount of CaSOS1U1B14S1U1B0 that will precipitate is therefore: 
6.1 x 10^-6 M x 0.100 L = 6.1 x 10^-7 mol

Since we want to precipitate 99.9% of the initial CaS1U1P12+S1S1P0 ion, the amount of CaSOS1U1B14S1U1B0 needed must be slightly greater than this: 
6.1 x 10^-7 mol / 0.999 = 6.101 x 10^-7 mol

Now, calculate the amount of NaS1U1B12S1U1B0SOS1U1B14S1U1B0 needed to supply this amount of SO1U1B24S1U1B0 ion: 
NaSOS1U1B12S1U1B0 ⇌ Na1U1P1+S1S1P0 + SO1U1B24S1U1B0 
1 mol of NaSOS1U1B12S1U1B0 supplies 1 mol of SO1U1B24S1U1B0 ion. Therefore, we need 6.101 x 10^-7 mol of NaS1U1B12S1U1B0SOS1U1B14S1U1B0: 
6.101 x 10^-7 mol x (78.045 g/mol) = 0.0477 g 
Convert to mmol: 
0.0477 g / (78.045 g/mol) x 1000 mmol/mol = 61.0 mmol

Therefore, choice (C) 61.0 is the best answer.

A 100-ML Sample of Solution Contains 10 ×\times× 10-5

A 100-ML Sample of Solution Contains 10 $\times$ 10^-⁵