Question

9. Calculate the solubility product of each of the following ions from the solubility given:

(a) AgBr, 5.7 x 10 ^{- 7} mole/L

(b) PbF₂ , 2.1 x 10 ^{- 3} mole/L

(d) Ag ₂CrO ₄ , 4.3 x 10 ^{- 2} g//L

Answer 1

Determine the K_sp of silver bromide, given that its molar solubility is 5.71 x 10¯⁷ moles per liter.

When AgBr dissolves, it dissociates like this:

AgBr (s) ⇌ Ag⁺ (aq) + Br¯ (aq)

The K_sp expression is:

K_sp = [Ag⁺] [Br¯]

There is a 1:1 molar ratio between the AgBr that dissolves and Ag⁺ that is in solution. In like manner, there is a 1:1 molar ratio between dissolved AgBr and Br¯ in solution. This means that, when 5.71 x 10¯⁷ mole per liter of AgBr dissolves, it produces 5.71 x 10¯⁷ mole per liter of Ag⁺ and 5.71 x 10¯⁷ mole per liter of Br¯ in solution.

Putting the values into the K_sp expression, we obtain:

K_sp = (5.71 x 10¯⁷) (5.71 x 10¯⁷) = 3.26 x 10¯¹³

K_sp = (5.71 x 10¯⁷) (5.71 x 10¯⁷) = 3.26 x 10¯¹³

(b) PbF2 , 2.1 x 10 - 3 mole/L

PbF₂ (s) ⇌ Pb²⁺ (aq) + 2F¯ (aq)

The K_sp expression is:

K_sp = [ Pb²⁺] [F¯]²

      = (2.1 x 10^-3) (4.41 x 10¯³)² =

Ksp = 4.0841 x 10^-8

(d) Ag 2CrO 4 , 4.3 x 10 - 2 g//L

Ag 2CrO 4 (s) ⇌ 2Ag⁺ (aq) + CrO₄¯ (aq)

K_sp = [ Ag⁺]² [CrO₄¯]

       = [8.6 x 10^-2]² [4.3 x 10^{- 2}]

Ksp= 3.18 x 10^-5