Chemistry Question

Question

An ionic compound consisting of bromide and an unknown metal M, dissolves in water via the following reaction: MBr2-> M + 2 Br. When 500g of MBr2 is added to 250mL of water, the resulting concentration of bromide ions is 5.5x10^(-3)M. There is plenty of solid MBr2 left behind, implying a saturated solution. Calculate Ksp for MBr2. Is MBr2 more, or less, soluble than AgCl?

Answer 1

MBr2 ==> M²⁺ + 2Br^-

Ksp = [M²⁺][Br^-]²

[Br^-] = 5.5x10^-3

Since there are 2 Br^- for each M²⁺, [M²⁺] = 1/2(5.5x10^-3) = 2.75x10^-3 M

Ksp = (2.75x10^-3)(5.5x10^-3)² = 8.3x10^-8

Using the Ksp for AgCl = 1.8x10^-10, MBr₂ would be MORE soluble than AgCl since the Ksp (8.3x10^-8) is greater.