the solubility product The solubility product, K of PbBr_2 is 8.9 times 10^-6. C
ID: 507445 • Letter: T
Question
the solubility product The solubility product, K of PbBr_2 is 8.9 times 10^-6. Calculate the molar solubility in: i. Pure water. ii. 0.20 M KBr. Calculate the OH^- and H_3O^+ concentrations, and pH and pOH, of a 0.200 M solution of the weak base methylamine, CH_3NH_2 (K = 4.8 times 10^-4) The equilibrium constant for the reaction: 2Fe^3+(aq) + Hg_2^2+(aq) 2Fe^2+(aq) + 2Hg^2+(aq) is K = 9.1 times 10^-6 at 298 K. Calculate Delta G when [Fe^3+] = 0.20 M, [Hg_2^2+] = 0.010 M, [Fe^2+] = 0.010 M, and [Hg^2+] = 0.025 M (i.e. the system is not at equilibrium). Based on your answer, in which direction will the reaction proceed?Explanation / Answer
Ksp of PbBr2 is 8.9 x 10-6
i) In Pure water
Let us consider solubility of PbBr2 be 'S' mole / litre
PbBr2 ------ Pb+2 + 2 Br-1
Ksp = [Pb+2] + [Br-1]2 = S x S2 = S3
8.9 x 10-6 = S3
S = 2.073 x 10-2 mole / litre
ii) In 0.2 M KBr
PbBr2 ------ Pb+2 + 2 Br-1
Ksp = [Pb+2] + [Br-1]2
8.9 x 10-6 = S x (0.2)2
By solving
S = 3.56 x 10 -7 mole / litre
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