Question

A solution of 0.5M aluminum chloride and 1M calcium chloride, and WE add sodium phosphate to it. The total volume is 2L. How much % is left of the 1st fabric that precipitates when the 2nd fabric just starts to precipitate.
How much mass of sodium phosphate should you add so that the 2nd substance is just beginning to precipitate

Answer 1

Concentration of AlCl3= 0.5M & CaCl2= 1M

Reactions will be as follows:

AlCl3+ Na3PO4​ --> AlPO4+ 3NaCl

3CaCl2+ 2Na3PO4 -->Ca3(PO4)2+ 6NaCl

From the solubility product,

Aluminium Phosphate[AlPO4] Ksp= 6.3×10−19

Ksp = [Al3+][PO43-]

6.3×10−19=0.5 x [PO43-]

[PO43-]= 1.26×10−18

Calcium Phosphate [Ca3(PO4)2 ] Ksp= 2.07 × 10−33

Ksp = [Ca2+]3[PO43-]2

2.07 × 10−33= 1 x [PO43-]2

[PO43-]= 4.54 × 10−17

Now, since the phosphate concentration of Calcium Phosphate is lesser than Aluminium Phosphate. Hence Ca(PO4)3 will precipitate first.

In a total volume of 2L,

Number of moles of AlCl3= 0.5M x 2= 1 mole

Number of moles of CaCl2​​​​​​​= 1M x 2= 2 moles

• From the reaction, 3CaCl2​​​​​​​+ 2Na3PO4 -->Ca3(PO4)2+ 6NaCl,

we can see, 3 moles of CaCl2 will react with 2 moles of Na3PO4 to form 1 mole of Ca3(PO4)2

That means 1 mole of Calcium Chloride will react with 2/3 moles of Sodium Phosphate.

Since we have only 2 moles of Calcium Chloride, therefore, it will react with 4/3 moles of Sodium Phosphate to form 2/3 moles of Calcium Phosphate.

Hence, when 2nd fabric that is AlPO4 starts to precipitate, 1st fabric that is Ca3(PO4)2 will completely precipitate.

Part 2 : Since, 1 mole of Calcium Chloride will react with 2/3 moles of Sodium Phosphate, and we have 2 moles of Calcium Chloride in the solution,

therefore we need 2x2/3=4/3 moles of Sodium Phosphate so that 2nd substance (AlPO4) is just beginning to precipitate.

Mass of sodium phosphate required so that 2nd substance (AlPO4) is just beginning to precipitate= Molar mass of Na3PO4 x number of moles of Sodium Phosphate

Molar mass of Na3PO4=163.94gm

Mass of sodium phosphate required so that 2nd substance (AlPO4) is just beginning to precipitate= 163.94 x 4/3 gm

= 218.58gm

Hence, we will add 218.58 gm of sodium phosphate required so that 2nd substance (AlPO4) is just beginning to precipitate.