# From the solubility data given for the following compounds- calculate.docx

8 de Feb de 2023
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### From the solubility data given for the following compounds- calculate.docx

• 1. From the solubility data given for the following compounds, calculate their solubility product constants. a. Cu SCN , copper(I) thiocyanate, 4.9 Ã— 10-4 g L Ksp = b. Ag Br, silver bromide, 1.1 x 10-6 g/10 mL c. Ca3 (PO4)2, calcium phosphate, 1.2 Ã— 10-3 g/L d. Ag,CrO4, silver chromate, 0.043 mg/mL K, Solution a) s = 4.9*10^-4 g/L Molar mass of CuSCN, MM = 1*MM(Cu) + 1*MM(S) + 1*MM(C) + 1*MM(N) = 1*63.55 + 1*32.07 + 1*12.01 + 1*14.01 = 121.64 g/mol Molar mass of CuSCN= 121.64 g/mol s = 4.9*10^-4 g/L To covert it to mol/L, divide it by molar mass s = 4.9*10^-4 g/L / 121.64 g/mol s = 4.028*10^-6 g/mol At equilibrium: CuSCN <----> Cu+ + SCN- s s Ksp = [Cu+][SCN-]
• 2. Ksp = (s)*(s) Ksp = 1(s)^2 Ksp = 1(4.028*10^-6)^2 Ksp = 1.623*10^-11 Answer: 1.6*10^-11 b) s = 1.1*10^-6 g/10 mL = 1.1*10^-7 g/mL = 1.1*10^-4 g/L Molar mass of AgBr, MM = 1*MM(Ag) + 1*MM(Br) = 1*107.9 + 1*79.9 = 187.8 g/mol Molar mass of AgBr= 187.8 g/mol s = 1.1*10^-4 g/L To covert it to mol/L, divide it by molar mass s = 1.1*10^-4 g/L / 187.8 g/mol s = 5.857*10^-7 g/mol At equilibrium: AgBr <----> Ag+ + Br- s s Ksp = [Ag+][Br-] Ksp = (s)*(s)
• 3. Ksp = 1(s)^2 Ksp = 1(5.857*10^-7)^2 Ksp = 3.43*10^-13 Answer: 3.4*10^-13 c) s = 1.2*10^-3 g/L Molar mass of Ca3(PO4)2, MM = 3*MM(Ca) + 2*MM(P) + 8*MM(O) = 3*40.08 + 2*30.97 + 8*16.0 = 310.18 g/mol Molar mass of Ca3(PO4)2= 310.18 g/mol s = 1.2*10^-3 g/L To covert it to mol/L, divide it by molar mass s = 1.2*10^-3 g/L / 310.18 g/mol s = 3.869*10^-6 g/mol At equilibrium: Ca3(PO4)2 <----> 3 Ca2+ + 2 PO43- 3s 2s Ksp = [Ca2+]^3[PO43-]^2 Ksp = (3s)^3*(2s)^2 Ksp = 108(s)^5 Ksp = 108(3.869*10^-6)^5 Ksp = 9.36*10^-26
• 4. Answer: 9.36*10^-26 d) s = 0.043 mg/mL s = 0.043 g/L Molar mass of Ag2CrO4, MM = 2*MM(Ag) + 1*MM(Cr) + 4*MM(O) = 2*107.9 + 1*52.0 + 4*16.0 = 331.8 g/mol Molar mass of Ag2CrO4= 331.8 g/mol s = 4.3*10^-2 g/L To covert it to mol/L, divide it by molar mass s = 4.3*10^-2 g/L / 331.8 g/mol s = 1.296*10^-4 g/mol At equilibrium: Ag2CrO4 <----> 2 Ag+ + CrO42- 2s s Ksp = [Ag+]^2[CrO42-] Ksp = (2s)^2*(s) Ksp = 4(s)^3 Ksp = 4(1.296*10^-4)^3 Ksp = 8.706*10^-12 Answer: 8.7*10^-12