Question

Calculate the molar solubility of Mg(OH) 2 ksp = 1.8 * 10 ^ - 11

a) in water

b) in 0.25 M NaOH (aq)

Answer 1

Calculating solubility in pure water molar solubility=S solubility Product Constant of Mg(OH)2 = Ksp reaction Mg(OH)2 + Initial(M) S Change (M) -S Equilibrium(M) 0 = 1.8x10-11 1 Mg2+ 0 + 2 OH- O +25 25 Ksp = [Mg2+ ] [ OH- } 1.8x10-11 = [ S ] [ 25 ? 1.8x10-11 = 453 S = 1.65x104 M Calculating solubility in NaOH molarity of NaOH = 0.25 M reaction Mg(OH)2 + Mg2+(aq) + Initial(M) S 0.000 Change (M) -S +S Equilibrium(M) 0 Due to low Ksp value 0.25+25 = 0.250 2 OH-(aq) 0.2500 +2S 0.25+25 Ksp = 1.8x10-11 = S = [Mg2+ ] [ OH- )? [ S ] [ 0.250 2.88x10-10 M ?