Answer to Question #187275 in Inorganic Chemistry for mue

Question

You add 10.0 grams of solid copper(II) phosphate, Cu31PO422,

to a beaker and then add 100.0 mL of water to the beaker at

T = 298 K. The solid does not appear to dissolve. You wait

a long time, with occasional stirring and eventually measure

the equilibrium concentration of Cu2 + 1aq2 in the water to be

5.01 * 10-8 M. What is the Ksp of copper(II) phosphate?

(a) 5.01 * 10-8 (b) 2.50 * 10-15

 (c) 4.20 * 10-15

(d) 3.16 * 10-37

 (e) 1.40 * 10-37

Answer 1

$$Cu_{3}(PO_4)_{2} = 3[Cu^{2+}] + 2[PO_4^{3-}]$$

$$[PO_4]^{3-} = 3.34*10^{-8}$$

$$K_{sp} = [Cu^{2+}]^3[PO_4^{3-}]^2$$

$$K_{sp} = (5.01*10^{-8})^3*(3.34*10^{-8})^2 = 1.40*10^{-37}$$

ANSWER: (e)