Calculating K sp from Solubility You add 10.0 grams of solid copper(II) phosphat

Question

Calculating Ksp from Solubility

You add 10.0 grams of solid copper(II) phosphate, Cu3(PO4)2, 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+(aq) in the water to be 5.01×108 M. What is the Ksp of copper(II) phosphate?

You add 10.0 grams of solid copper(II) phosphate, , to a beaker and then add 100.0  of water to the beaker at  = 298 . The solid does not appear to dissolve. You wait a long time, with occasional stirring and eventually measure the equilibrium concentration of  in the water to be . What is the  of copper(II) phosphate?

Explanation / Answer

Let's write the overall reaction and the ICE chart

r: Cu3(PO4)2(s) <--------> 3Cu2+(aq) + 2PO43-(aq)

i: s 0 0

e: 3s 2s

We already know that the concentration of Cu2+ is 5.01x10-8 M so:

3s = 5.01x10-8 ----> s = 5.01x10-8 / 3 = 1.67x10-8 M

So concentration of PO43-:

[PO43-] = 2*1.67x10-8 = 3.34x10-8 M

So finally the Ksp would be:

Ksp = [Cu2+]3 [PO43-]2

Ksp = (5.01x10-8)3 * (3.34x10-8)2

Ksp = 1.4028x10-37

Hope this helps

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.