Question

Calculate molar solubilities, concentrations of constituent ions, and solubilities in grams/liter for the following compounds at 25°C. I. copper(I) chloride (Ksp = 1.9 x10?7) II. Barium phosphate (Ksp = 1.3 x10?29) III. lead(II) hydroxide (Ksp = 2.8 x10? 16) IV. silver oxide (Ksp = 2.0 x10?8)

Answer 1

Step-by-step explanation:

1)
`CuCl\rightleftharpoons Cu^++Cl^-`

S S

Solubility product of the copper(I) chloride =
`K_(sp)=1.9* 10^(-7)`

The expression of solubility product is given as :

`K_(sp)=S* S=S^2`

`1.9* 10^(-7)=S^2`

`S=4.3589* 10^(-4) M`

Concentration of copper(I) ions =
`[Cu^+]=4.3589* 10^(-4) M`

Concentration of chloride ions =
`[Cl^-]=4.3589* 10^(-4) M`

Molar solubility of CuCl =
`4.3589* 10^(-4) M`

Solubility of CuCl =
`4.3589* 10^(-4) M* 99 g/mol=0.04315 g/L`

2)
`Ba_3(PO_4)_2\rightleftharpoons 3Ba^(2+)+2PO_4^(3-)`

3S 2S

Solubility product of the Barium phosphate =
`K_(sp)=1.3* 10^(-29)`

The expression of solubility product is given as :

`K_(sp)=(3S)^3* (2S)^2=108S^5`

`1.3* 10^(-29)=108S^5`

`S^5=(1.3* 10^(-29))/(108)`

`S=6.5479* 10^(-7) M`

Concentration of barium ions =
`[Ba^(2+)]=3* 6.5479* 10^(-7) M =1.9643* 10^(-6) M`

Concentration of phosphate ions =
`[PO_4^(3-)]=2* 6.5479* 10^(-7) M =1.3096* 10^(-6) M`

Molar solubility of
`[Ba_3(PO_4)_2]([Ba^(2+)])/(3)`

`(1)/(3)* 1.9643* 10^(-6) M=6.5479* 10^(-7) M`

Solubility of
`Ba_3(PO_4)_2` :

`6.5479* 10^(-7) M* 601 g/mol=0.0003935 g/L`

3)
`Pb(OH)_2\rightleftharpoons Pb^(2+)+2OH^-`

S 2S

Solubility product of the lead(II) oxide =
`K_(sp)=2.8* 10^(-16)`

The expression of solubility product is given as :

`K_(sp)=(S)* (2S)^2=4S^3`

`2.8* 10^(-16)=4S^3`

`S^3=(2.8* 10^(-16))/(4)`

`S=4.1212* 10^(-6) M`

Concentration of lead(II) ions =
`[Pb^(2+)]=1* 4.1212* 10^(-6) M M =4.1212* 10^(-6) M`

Concentration of hydroxide ions =
`[PO_4^(3-)]=2* 4.1212* 10^(-6) M=8.2425* 10^(-6) M`

Molar solubility of
`[Pb(OH)_2]=[Pb^(2+)]=4.1212* 10^(-6) M`

Solubility of
`Pb(OH)_2` :

`4.1212* 10^(-6) M* 241 g/mol=0.0009932 g/L`

4)
`Ag_2O\rightleftharpoons Ag^(+)+O^(2-)`

2S S

Solubility product of the lead(II) oxide =
`K_(sp)=2.0* 10^(-8)`

The expression of solubility product is given as :

`K_(sp)=(2S)^2* (S)=4S^3`

`2.0* 10^(-8)=4S^3`

`S^3=(2.0* 10^(-8))/(4)`

`S=0.001710 M`

Concentration of silver ions =
`[Ag^(+)]=2* 0.001710 M =0.003420 M`

Concentration of hydroxide ions =
`[OH^(-)]=1* 0.001710 M=0.001710 M`

Molar solubility of
`[Ag_2O]=([Ag^(+)])/(2)`

`(1)/(2)* 0.003420 M=0.001710 M`

Solubility of
`Ag_2O` :

`0.001710 M* 232 g/mol=0.3967 g/L`