Respuesta :
Explanation:
1) [tex]CuCl\rightleftharpoons Cu^++Cl^-[/tex]
S S
Solubility product of the copper(I) chloride = [tex]K_{sp}=1.9\times 10^{-7}[/tex]
The expression of solubility product is given as :
[tex]K_{sp}=S\times S=S^2[/tex]
[tex]1.9\times 10^{-7}=S^2[/tex]
[tex]S=4.3589\times 10^{-4} M[/tex]
Concentration of copper(I) ions = [tex][Cu^+]=4.3589\times 10^{-4} M[/tex]
Concentration of chloride ions = [tex][Cl^-]=4.3589\times 10^{-4} M[/tex]
Molar solubility of CuCl = [tex]4.3589\times 10^{-4} M[/tex]
Solubility of CuCl = [tex]4.3589\times 10^{-4} M\times 99 g/mol=0.04315 g/L[/tex]
2) [tex]Ba_3(PO_4)_2\rightleftharpoons 3Ba^{2+}+2PO_4^{3-}[/tex]
3S 2S
Solubility product of the Barium phosphate = [tex]K_{sp}=1.3\times 10^{-29}[/tex]
The expression of solubility product is given as :
[tex]K_{sp}=(3S)^3\times (2S)^2=108S^5[/tex]
[tex]1.3\times 10^{-29}=108S^5[/tex]
[tex]S^5=\frac{1.3\times 10^{-29}}{108}[/tex]
[tex]S=6.5479\times 10^{-7} M[/tex]
Concentration of barium ions = [tex][Ba^{2+}]=3\times 6.5479\times 10^{-7} M =1.9643\times 10^{-6} M[/tex]
Concentration of phosphate ions = [tex][PO_4^{3-}]=2\times 6.5479\times 10^{-7} M =1.3096\times 10^{-6} M[/tex]
Molar solubility of [tex][Ba_3(PO_4)_2]\frac{[Ba^{2+}]}{3}[/tex]
[tex]\frac{1}{3}\times 1.9643\times 10^{-6} M=6.5479\times 10^{-7} M[/tex]
Solubility of [tex]Ba_3(PO_4)_2[/tex] :
[tex]6.5479\times 10^{-7} M\times 601 g/mol=0.0003935 g/L[/tex]
3) [tex]Pb(OH)_2\rightleftharpoons Pb^{2+}+2OH^-[/tex]
S 2S
Solubility product of the lead(II) oxide = [tex]K_{sp}=2.8\times 10^{-16}[/tex]
The expression of solubility product is given as :
[tex]K_{sp}=(S)\times (2S)^2=4S^3[/tex]
[tex]2.8\times 10^{-16}=4S^3[/tex]
[tex]S^3=\frac{2.8\times 10^{-16}}{4}[/tex]
[tex]S=4.1212\times 10^{-6} M[/tex]
Concentration of lead(II) ions = [tex][Pb^{2+}]=1\times 4.1212\times 10^{-6} M M =4.1212\times 10^{-6} M[/tex]
Concentration of hydroxide ions = [tex][PO_4^{3-}]=2\times 4.1212\times 10^{-6} M=8.2425\times 10^{-6} M[/tex]
Molar solubility of [tex][Pb(OH)_2]=[Pb^{2+}]=4.1212\times 10^{-6} M[/tex]
Solubility of [tex]Pb(OH)_2[/tex] :
[tex]4.1212\times 10^{-6} M\times 241 g/mol=0.0009932 g/L[/tex]
4) [tex]Ag_2O\rightleftharpoons Ag^{+}+O^{2-}[/tex]
2S S
Solubility product of the lead(II) oxide = [tex]K_{sp}=2.0\times 10^{-8}[/tex]
The expression of solubility product is given as :
[tex]K_{sp}=(2S)^2\times (S)=4S^3[/tex]
[tex]2.0\times 10^{-8}=4S^3[/tex]
[tex]S^3=\frac{2.0\times 10^{-8}}{4}[/tex]
[tex]S=0.001710 M[/tex]
Concentration of silver ions = [tex][Ag^{+}]=2\times 0.001710 M =0.003420 M[/tex]
Concentration of hydroxide ions = [tex][OH^{-}]=1\times 0.001710 M=0.001710 M[/tex]
Molar solubility of [tex][Ag_2O]=\frac{[Ag^{+}]}{2}[/tex]
[tex]\frac{1}{2}\times 0.003420 M=0.001710 M[/tex]
Solubility of [tex]Ag_2O[/tex] :
[tex]0.001710 M\times 232 g/mol=0.3967 g/L[/tex]
