Question

1)The solubility of silver(I)phosphate at a given temperature is 0.91 g/L. Calculate the K_sp at this temperature. After you get your answer, take the negative log and enter that (so it's like you're taking the pK_sp)

2)At 25 ^oC the solubility of magnesium fluoride is 1.17 x 10^-3 mol/L. Calculate the value of K_sp at this temperature. Give your answer in scientific notation to 2 SIGNIFICANT FIGURES (even though this is strictly incorrect

3)At 25 ^oC the solubility of lead(II) bromide is 2.70 x 10^-2 mol/L. Calculate the value of K_sp at this temperature. Give your answer in scientific notation to 2 SIGNIFICANT FIGURES (even though this is strictly incorrect

Answer 1

1. Solubility product of Silver Phosphate can be written as Ag3(PO₄) K_sp = [Ag⁺] [PO₄^-3]

K_sp = S²

Gicen that Solubility in grams per lit = 0.91 g/L

Molecular weight = 418.58

Solubility in Moles per litre of silver phosphate = 0.91/418.58 = 0.00217 mol/L

So that K_sp = (2.17 x 10^-3)²

= 4.709 x 10^-6 mol/L.

PK_sp = -log [4.709 x 10^-6]

= 6 - 0.673

= 5.327

2. Solubility product of Magnesium fluoride can be written as MgF₂ K_sp = [Mg⁺²] x [F^-]²

K_sp = S x (2S)² = 4S³

MgF₂               Mg⁺² + 2F^-

Given that S = 1.17 x 10^-3 mol/L

K_sp = 4 x (1.17 x 10^-3)³

= 4 x 1.6016 x 10^-9

= 6.406 x 10^-9

= 6.4 x 10^-9 mol/L (2 significant figures)

3. Solubility product of Lead (II) bromide can be written as PbBr₂    K_sp = [Pb⁺²] [Br^-]²

= S x (2S)²

= 4S³

Given that S = 2.70 x 10^-2

K_sp = 4 x (2.70 x 10^-2)³

= 4 x 19.283 x 10^-6

= 78.732 x 10^-6

= 7.9 x 10^-5 mol/L (2 significant figures)