Question

Design experiments in the virtual lab to answer the follow questions, 1). Use the virtual lab to determine the solubility product (Ksp) for the following solids: (a). AgCl (b). SrSO4 (c). Ag2CO3 (d). Sr(IO3)2 2 1.5x100 S = 1.27x10-4 a) Kop c) Ksp = (29) ² x $=45° = 8-1x102 s=71.8x10o = 1.34*10-smolle (8.1x10-12 b) Ksp = 5x (25)² =493 = 1.14x10? v3.5*100 = 5.92x104 mol, =3.05*10-3 mol/L) samoly d) kse s? = 3.5810) S (1.1481o- 4 2). What is the solubility of the solids listed in problem 1, in grams liter" grams per liter)? 3). For any pair of solids, does the one with the largest Ksp always have the larger solubility? Why or why not?

Complete Table 1: K_sp data

Salt	[cation] (M)	[anion] (M)	molar solubility (M)	Ksp
AgCl
SrSO4
Ag2CO3
Sr(IO3)2

1) Rank the salts in order of increasing molar solubility.

2) Rank the salts in order of increasing K_sp (remember 10^–10 < 10^–5)

3) If these rankings are not in the same order, why might K_sp not always scale directly with molar solubility?

Answer 1

3) C - 13ч х? 2 (a) Ag ce f Ag ches = Ag fo (S = solubility) At t=0 с S At teg C-s При ртора О C - , 3х10 Molar man molle 4 : 143.32 glaol xt43. 32 ( (glued) xuraffe -Г 192.ОчB 8х10 Egle

Srsoy x 10-4 mall solubility = s=5.92* (from problem ) Molar was of Ersoy = 183.68 glund s = 5.924104 x 183.68 lglund) x (mol/L) * 1087.3856 x 10-4 gle © Aga CO₂ prob Solubility s= 1.278104 molle (from len ①) Molar man of AgaCO3 cO₂ = 275.745 /mnd S = 1.274 104 * 27507451glued) (mallu s = 350. 196 350.196 g/l

= 3.058 103 (d) So 1334.07 X 10^ Sr(10₃), udlu Molar mass sr (103)2 = 437.4 glund S = 3.05X108 x 437.4 Gluco) (mol/L) gle 3

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