K sp = [M x+ ] [N x– ] Solubility Equilibria -- involve the dissolution or precipitation of “insoluble” salts Consider a saturated solution of a typical salt: For these cases, the solubility-product constant is equal to (respectively): MN(s) M x+ (aq) + N x– (aq) MN 2 (s) M 2x+ (aq) + 2 N x– (aq) etc. K sp = [M 2x+ ] [N x– ] 2 K sp expressions NEVER have denominators because… the “reactant” is always a solid, and we never write solids in equil. eqs.

small K sp … -- K sp is the equilibrium constant between undissolved and dissolved ionic solute in a saturated aqueous solution. Write the solubility-product constant expression for lead(II) chloride. PbCl 2 (s) Pb 2+ (aq) + 2 Cl – (aq) K sp = [Pb 2+ ] [Cl – ] 2 large K sp … Not much dissolves. “Quite a bit” dissolves. MN(s) M x+ (aq) N x– (aq) M x+ (aq) N x– (aq) M x+ (aq) N x– (aq)

Copper(II) azide has K sp = 6.3 x 10 –10. Find the solubility of Cu(N 3 ) 2 in water, in g/L. Cu(N 3 ) 2 (s) Cu 2+ (aq) + 2 N 3 – (aq) K sp = 6.3 x 10 –10 = [Cu 2+ ] [N 3 – ] 2 ** In “plain-old” Cu(N 3 ) 2,: [N 3 – ] = 2 [Cu 2+ ] So…6.3 x 10 –10 = x (2x) 2 = 4x 3 x = 5.40 x 10 –4 M = [Cu 2+ ] From eq. at top, 5.40 x 10 –4 M is also the [ ] of Cu(N 3 ) 2 that dissolves x 10 –4 mol Cu(N 3 ) 2 L g 1 mol () g L = 8.0 x 10 –2 (Let [Cu 2+ ] = x)

Find the solubility of zinc hydroxide (K sp = 3.0 x 10 –16 ) in a soln buffered at pH = Zn(OH) 2 (s) Zn 2+ (aq) + 2 OH – (aq) K sp = 3.0 x 10 –16 = [Zn 2+ ] [OH – ] x 10 –16 = [Zn 2+ ] (2.69 x 10 –3 ) 2 from given pH… [Zn 2+ ] = 4.1 x 10 –11 M From eq. at top, 4.1 x 10 –11 M is also the [ ] of Zn(OH) 2 that dissolves.