Question

The generic metal hydroxide M(OH)2 has Ksp = 5.45×10−18. (NOTE: In this particular problem, because of the magnitude of the Ksp and the stoichiometry of the compound, the contribution of OH− from water can be ignored. However, this may not always be the case.)

Part A

What is the solubility of M(OH)2 in pure water?

Part B

What is the solubility of M(OH)2 in a 0.202 M solution of M(NO3)2?

Answer 1

Solubility product of a salt is equal to product of concentration of the dissolved ion raised to the power of the stoichiometric coefficients of the ions. Given that, K, for M(OH), =5.45x10-18 Consider the reaction, M(OH), (aq) → M*- (aq) + 20H (aq) Let Solubility $ $ 28 The solubility product expression for the above reaction will be, K, [M][OH] Substitute the value of K to calculate the solubility of ions, 5.45*10-18 =sx (28) 5.45x10-18 = 45 s=1.10x10 M Hence, the molar solubility of M(OH), in water is 1.10X10ⓇM. Given that, Concentration of M(NO3), = 0.202 M K for M(OH), =5.45x10-18

Consider the reaction, M(OH), (aq)=M?- (aq)+20H- (aq) The solubility constant product for M(OH), is K,=[mº-ICOH) 5.45x10-18=[M?][CH] .... (1) Metal nitrate is a strong electrolyte and due to the common ion effect, it will suppress the dissociation of metal hydroxide. Suppose the molar solubility of M(OH), in M(NO3), is 'S'. Then, [Mºd]=(S+0.202)M [OH-]=2SM Substitute the above concentrations in equation (1), 5.45*10*8 =[M2+][OH-] 5.45*10=(S+0.202) (25) S=2.59x10'M Hence, the molar solubility of M(OH), in a 0.202 M MNO), solution is 2.59x10°M.