Question

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

2. The Ksp of Al(OH)3 is 1.0 x 10-33. What is the solubility of Al(OH)3 in 0.000010 M Al(NO3)3?
3. A precipitate of lead(II) chloride forms when 3.5 mg of NaCl is dissolved in 0.250 L of 0.12 M lead(II)nitrate.
A. True
B. False

Answer 1

The Ksp of PbCl2 at a given temperature is calculated using the solubility of the compound. For Al(OH)3 in a solution containing Al(NO3)3, the solubility would be influenced by the common ion effect. The potential formation of a PbCl2 precipitate depends on the concentrations of lead(II) and chloride ions.

Step-by-step explanation:

Calculation of Ksp and Solubility

The solubility product (Ksp) represents the equilibrium between a solid and its corresponding ions in a solution. It is a unique value for each soluble compound at a given temperature.

1. Ksp of PbCl2

The dissolution of PbCl2 can be written as:

PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq)

Given the solubility of PbCl2 is 3.1 x 10-2 M, we can deduce the concentration of Pb2+ is 3.1 x 10-2 M and Cl- is 2 x 3.1 x 10-2 M. Thus, the Ksp can be calculated using the formula:

Ksp = [Pb2+][Cl-]2 = (3.1 x 10-2) x (2 x 3.1 x 10-2)2 = 5.88 x 10-5

2. Solubility of Al(OH)3 in Al(NO3)3 Solution

Given the Ksp of Al(OH)3 is 1.0 x 10-33, and the initial concentration of Al3+ from Al(NO3)3 is known, we need to account for the common ion effect which will reduce the solubility of Al(OH)3. This complex calculation would require additional steps to accurately account for the common ion.

3. Formation of PbCl2 Precipitate

NaCl dissociates to give Cl- ions and the Pb2+ from Pb(NO3)2 can form PbCl2, depending on the product of their concentrations relative to the Ksp of PbCl2. This assessment involves comparing the ionic product to the Ksp to determine if a precipitate will form.

Answer 2

1) The solubility product of the lead(II) chloride is
`1.2* 10^(-4)`.

2) The solubility of the aluminium hydroxide is
`1.6* 10^(-10) M`.

3)The given statement is false.

Step-by-step explanation:

1)

Solubility of lead chloride =
`S=3.1* 10^-2M`

`PbCl_2(aq)\rightleftharpoons Pb^(2+)(aq)+2Cl^-(aq)`

S 2S

The solubility product of the lead(II) chloride =
`K_(sp)`

`K_(sp)=[Pb^(2+)][Cl^-]^2`

`K_(sp)=S* (2S)^2=4S^3=4* (3.1* 10^(-2))^3=1.2* 10^(-4)`

The solubility product of the lead(II) chloride is
`1.2* 10^(-4)`.

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion =
`1\timed 0.000010 M=0.000010 M`

Solubility of aluminium hydroxide in aluminum nitrate solution =
`S`

`Al(OH)_3(aq)\rightleftharpoons Al^(3+)(aq)+3OH^-(aq)`

S 3S

The solubility product of the aluminium nitrate =
`K_(sp)=1.0* 10^(-33)`

`K_(sp)=[Al^(3+)][OH^-]^3`

`1.0* 10^(-33)=(0.000010+S)* (3S)^3`

`S=1.6* 10^(-10) M`

The solubility of the aluminium hydroxide is
`1.6* 10^(-10) M`.

3.

`Molarity=(Moles)/(Volume (L))`

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl =
`(0.0035 g)/(58.5 g/mol)=6.0* 10^(-5) mol`

Volume of the solution = 0.250 L

`[NaCl]=(6.0* 10^(-5) mol)/(0.250 L)=0.00024 M`

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

`[Cl^-]=[NaCl]=0.00024 M`

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

`n=0.12 M]* 0.250 L=0.030 mol`

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

`[Pb^(2+)]=[Pb(NO_2)_3]=0.030 M`

`PbCl_2(aq)\rightleftharpoons Pb^(2+)(aq)+2Cl^-(aq)`

Solubility of lead(II) chloride =
`K_(sp)=1.2* 10^(-4)`

Ionic product of the lead chloride in solution :

`Q_i=[Pb^(2+)][Cl^-]^2=0.030 M* (0.00024 M)^2=1.7* 10^(-9)`

`Q_i<K_(sp)` ( no precipitation)

The given statement is false.