Question

The solubility of silver(I)phosphate at a given temperature is 2.43 g/L. Calculate the Ksp at this temperature. After you calculate the Kspvalue, take the negative log and enter the (pKsp) value with 2 decimal places.

Answer 1

Answer:

Kps = 3.07 x 10⁻⁸

pKsp= 7.51

Explanation:

First, we calculate the molar solubility of  silver(I)phosphate (Ag₃PO₄) from the solubility in g/L by using its molar mass (418.6 g/mol):

2.43 g/L x 1 mol/418.6 g = 5.8 x 10⁻³ mol/L= s

Now, we have to write the ICE chart for the aqueous equilibrium of Ag₃PO₄ as follows:

     Ag₃PO₄(g) ⇄ 3 Ag⁺(aq) + PO₄³⁻

I                               0              0

C                            +3s           +s

E                            3s               s

Ksp = [Ag⁺]³[PO₄³⁻]= (3s)³s= 27s⁴

Since s=5.8 x 10⁻³ mol/L, we calculate Ksp:

Ksp= 27(5.8 x 10⁻³ mol/L)⁴= 3.07 x 10⁻⁸

The pKsp value is:

pKsp= - log Ksp = -log (3.07 x 10⁻⁸) = 7.51