Question

What are the answers to these 2 chemistry problems?
1. At a certain temperature, the solubility product constant* of copper (II) iodate, Cu(IO3)2, is 7.88x10-8 M3. Calculate the solubility of this compound for this temperature.

*Unless stated otherwise all solubility product constants are for water.

2. For the reaction: PbCl2(s) ↔ Pb2+(aq)+2Cl1-(aq), what is Q* when 2.0 mL of 0.023 M lead nitrate is added to 16 mL of 0.025 M sodium chloride?
Ksp of lead chloride is 1.6 x 10-5 M3.

Answer 1

....................Cu(IO3)2 ==> Cu^2+ + 2IO3^-
I.........................solid..................0............0
C........................solid-x...............x............2x
E........................ solid..................x............2x
so Ksp = 7.88E-8 = (Cu^2+)(IO3^-)^2
Substitute The E line into Ksp expression and solve for x = solubility of Cu(IO3)2.

2. A little different. You want Qsp.
..............PbCl2 ==> Pb^2+ + 2Cl^-
Q = (Pb^2+)(Cl^-)^2
Just calculate [Pb(NO3)2] and (NaCl) and plug those concentration into the Q expression and solve for Q.
You're adding 2 mL of the Pb salt to 16 mL of the NaCl so the total volume is 18 mL.
(Pb^) = 0.023 M x (2/18) = ?
(Cl^-) = 0.025 M x (16/18) = ?
Don't lose site of what this problem does for you. If Qsp is > Ksp you know a ppt of PbCl2 will occur. If Qsp < Ksp you know a ppt of PbCl2 will not occur. If Qsp = Ksp you know you have a saturated solution of PbCl2.
Post your work if you get stuck on either problem or don't understand something.






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