Question

1. How many grams of PbCl2 must be weighed to prepare 500 ml of a saturated solution at room temperature if its Kps is 10^-4.79.
2.The Kps of fluorite at 25ºC is 4 x 10^-11. What is its solubility in g / L of solution at that temperature?
3.Calculate the solubility of calcium oxalate in g / L
4. Bismuth Iodide has a solubility of 7.76 mg / L. What is its Kps?

Answer 1

Ans 1

Tge dissociation reaction

PbCl2 = Pb2+ + 2Cl-

Let the solubility of PbCl2 = s

Equilibrium constant expression of the reaction

Ksp = [Pb2+] [2Cl-]²

10^-4.79 = s(2s)²

1.62*10^-5 = 4s³

s = 1.59*10^-2

Moles of PbCl2 = 1.59*10^-2 mol/L x 500 mL x 1L/1000mL

= 7.97*10^-3 mol

Mass of PbCl2 = moles x molecular weight

= 7.97*10^-3 mol x 278.1 g/mol

= 2.22 g