Determine the molar solubility of pbso4 in pure water. ksp (pbso4 = 1.82 × 10-8.

PbSO4 -> Pb2+ + SO4 2-

So there are 2 different particles. 1 mol of PbSO4 produces 1 mol of Pb2+, and 1 mol of SO4 2- (one of each).

Kps = [Pb2+] [SO4 2-]

We will call “s” the molarity of PbSO4

So the molarity of Pb2+ is also “s” (same number)

And the molarity of SO4 2- is also “s” (same number)

Kps = s*s = s2

1.82 × 10-8 = s2

So, in order to find s, we have to make the square root of 1.82 × 10-8, which is: 0,00013491 M

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%