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There is FOUR parts to this question. (A, B, C, & D) ANSWER ALL PARTS. SHOW YOUR

ID: 1000054 • Letter: T

Question

There is FOUR parts to this question. (A, B, C, & D) ANSWER ALL PARTS. SHOW YOUR WORK, CLEARLY!

Will a precipitate of Callo3)2 form when 400. mL of 0.015 M CaCl2 M of are added to 300. mil. of a precipitate of CaflO3)2 form when 4 0.15 M NalO3? .5 The Ksp of Ca(l03)2 is 6.17 x10 a) What is the reaction for the solubility of Ca(lO3)2(s)? ) What is the reaction for the solubility of Calloso b) Given that the final volume is 700 mL, dletermine the concentrations of [Ca2 1 and [10l. of[C.2+1 and 110), c) Calculate Qto determine if Ca(IO3)/2 precipitates. (s) in pure water?

Explanation / Answer

CaCl2 + 2NaIO3 ----> Ca(IO3)2 + 2NaCl the kps of  Ca(IO3)2 is 6.47x10-5

nCaCl2 = C1 x V1 = 0.4l x 0.015mol/l = 6x10-4mol CaCl2

nNaIO3 = C2 x V2 = 0.3l x 0.15mol/l = 4.5x10-2 NaIO3

the limit reagent is the CaCl2 because he reacction whit 1.2x10-3mol of NaIO3 and i have 4.5x10-2 NaIO3

the mol of Ca(IO3)2 formed are = to mol of CaCl2 =) nCa(IO3)2 = 1.2x10-3mol

a) Ca+2 + 2IO3-   -----> Ca(IO3)2(s)       [Ca+2][IO3]2 = Kps

b) to begining before the reacction nCa+2 = nCaCl2 = 6x10-4mol nIO3- = nNaIO3 = 4.5x10-2mol  

[Ca+2] = 6x10-4mol /0.7l = 8.57x10-4M and the [IO3] =4.5x10-2mol / 0.7l = 6.48x10-2M

C)

late of reaction   [Ca+2] = [Ca+2]o -X and [IO3] = [IO3]o - X as the Kps control the concentrations in solutions

[Ca+2][IO3]2 = Kps = [Ca+2]o -X ([IO3]o - X)2 = Kps as the Kps is little Kps = 6.47x10-5 and the [IO3] > [Ca+2] we cant say than the concentration of Ca is verry little and for so X is negligible so get

[Ca+2][IO3]2 = Kps =)   [Ca+2] = Kps/[IO3]2 = 6.47x10-5 mol3/(6.48x10-2M)2 =) [Ca+2] = 1.5x10-2M and i have a fewer than in solution therefore all specs are present in solution dont presipite

D) in water pure [Ca+2][IO3]2 = Kps we say than S = to solubility of sustances and replacing get

[S][S]2 = Kps =) Kps =  [S]3 so the S = (Kps)1/3 = ( 6.47x10-5 mol3)1/3 = 4x10-2M

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