Question

Ksp for PbBr2 is equal to: 4.67 x 10^-6. How many grams of PbBr2 will dissolve in 250.0 mL of solution? (Molar mass of PbBr2 = 367.01 g/mol)

Answer 1

PbBr₂ <=> Pb⁺² (aq) + 2Br^- (aq)

Ksp = [Pb⁺²] [Br^-]²………………………………………eq(1)

There is a 1:1 molar ratio between PbBr₂ and Pb²⁺, but there is a 1:2 molar ratio between PbBr₂ and Br¯. This means that, when s moles per liter of PbBr₂ dissolves, it produces s moles per liter of Pb²⁺ and it produces 2s moles per liter of Br¯ in solution.

Therfore eq (1) will be

Ksp = [s] [2s]²

Ksp = 4.67 x 10^-6

So, 4.67 x 10^-6 = 2s³

(4.67 x 10^-6 )/2 = s³

(2.335 x 10^-6) = s³………………taking cube root

s = 0.01327

therefore concentration of Pb = 0.01327 moles/L and Br = (2x 0.01327)² = 0.00070 moles/L

The molar solubility of PbBr2 solution is 0.01327 moles/L………………….(2)

Volume of solution = 250 ml

Molecular weight of PbBr2 = 367.01g/mol

Molarity = (no. of moles of PbBr2 x 1000)/volume of solution

(Molarity x volume of solution)/1000 = n

( 0.01327 x 250 )/ 1000 = n

0.00332 moles = n

Mass of PbBr2 = n x Molar mass

Mass of PbBr2 = 0.00332 x 367.01 = 1.2175 g