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The molar solubility of silver phosphate, Ag3PO4, is 4.68E-6 mol/L. (1) Express the solubility in units of grams per liter. (

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Answer #1

1)

Molar mass of Ag3PO4,

MM = 3*MM(Ag) + 1*MM(P) + 4*MM(O)

= 3*107.9 + 1*30.97 + 4*16.0

= 418.67 g/mol

Given:

S = 4.68*10^-6 mol/L

= 4.68*10^-6 mol/L * 418.67 g/mol

= 1.96*10^-3 g/L

Answer: 1.96*10^-3 g/L

2)

Ag3PO4 dissociates as:

Ag3PO4 <-> 3Ag+ + PO43-

So,

[Ag+] = 3s

= 3*4.68*10^-6 mol/L

= 1.40*10^-5 mol/L

Answer: 1.40*10^-5 mol/L

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