Question

025 M Sr(NOh M Sr(NO,) 3)2- Calculate the molar solubility of Sr,(PO)h in 0.025 M S 2. -3 mol/L at a certain temperature. 19. The molar solubility of Zn(OH)2 is 5.7 x 10 Calculate the value of Ksp for Zn(OH)2 at this temperature

Answer 1

18. In order to answer this question, we need to know the molar solubility of Sr₃(PO₄)₂ which is 1.0*10^-31. We need to write down the dissociation equation and set up the ICE chart.

Sr₃(PO₄)₂ ç=====è 3 Sr²⁺ + 2 PO₄

Let x mole/L be the molar solubility of Sr₃(PO₄)₂. Then the solubility of Sr²⁺ and PO₄^3- will be 3x and 2x as per the dissociation equation. Also, we have 0.025 M Sr(NO₃)₂ which furnishes 0.025 mole/L Sr²⁺. Therefore,

K_sp = [Sr²⁺]³[PO₄^3-]² = (0.025 + 3x)³(2x)² where K_sp is the solubility product of Sr₃(PO₄)₂. We need to make an assumption here: x << 0.025 and hence [Sr²⁺] = 0.025. Therefore,

1.0*10^-31 = (0.025)³(2x)² = (1.5625*10^-5).(4x²) = 6.25*10^-5.x²

=è x² = 1.6*10²⁷

=è x = 4.0*10^-14

The molar solubility of Sr₃(PO₄)₂ in 0.025 M Sr(NO₃)₂ is 4.0*10^-14 M (ans).

19. The molar solubility of Zn(OH)₂ is 5.7*10^-3 mol/L at a certain temperature. The dissociation of Zn(OH)₂ occurs as

Zn(OH)₂ ç==è Zn²⁺ + 2 OH^-

[Zn²⁺] = 5.7*10^-3 mol/L

[OH^-] = 2*5.7*10^-3 mol/L; this is as per the dissociation equation. Therefore,

K_sp = [Zn²⁺][OH^-]² = (5.7*10^-3).(2*5.7*10^-3)² = 7.407*10^-7 ≈ 7.41*10^-7 (ans).

The molar solubility of CuCl in water is 1.0*10^-3 mol/L. The dissociation of CuCl takes place as

CuCl ç===è Cu⁺ + Cl^-

[Cu⁺] = 1.0*10^-3 mol/L

[Cl^-] = 1.0*10^-3 mol/L; this is as per the dissociation equation. Therefore,

K_sp = [Cu⁺][Cl^-] = (1.0*10^-3).(1.0*10^-3) = 1.0*10^-6 (ans).