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

1) The solubility product of the lead(II) chloride is .

2) The solubility of the aluminium hydroxide is .

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride =

                            S     2S

The solubility product of the lead(II) chloride =

The solubility product of the lead(II) chloride is .

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion =

Solubility of aluminium hydroxide in aluminum nitrate solution =

                            S     3S

The solubility product of the aluminium nitrate =

The solubility of the aluminium hydroxide is .

3.

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl =

Volume of the solution = 0.250 L

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

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

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

Solubility of lead(II) chloride =

Ionic product of the lead chloride in solution :

( no precipitation)

The given statement is false.