Step 1: Write the equation for the dissociation of copper(II) iodate.

Cu(IO 3 )2(s)?Cu(aq)2++ 2IO3(aq)–

Step 2: Write down the concentrations during the process of dissociation.

3 3 10..###--^3 M Cu IO^ 3 h 2 "3 3 10^33 M Cu^2 +_aqi+2 3 3 10_ .-M IOi 3 - _aqi

.

.

Cu M

IO M

33 10

66 10

3

3

3

(^2) #

=

+

7

7

A

A

Step 3: Write the equilibrium expression.

Ksp= [Cu2+][IO 3 – ]^2

Step 4:Substitute the equilibrium concentration into the Kspexpression.

Ksp= (3.3 × 10 –3)(6.6 × 10 –3)^2 = 1.4 × 10 –7

9. Lead iodide has a Kspvalue of 1.08 × 10 –7at 20°C. Calculate its solubility at 20°C.

A. 5.00 × 10 –8

B. 3.00 × 10 –6

C. 1.00 × 10 –4

D. 6.00 × 10 –3

E. 3.00 × 10 –3

Answer: E

Step 1: Write the equilibrium equation for the dissociation of lead iodide.

PbI 2 (s) ↔Pb2+(aq) + 2 I–(aq)

Step 2: Write the equilibrium expression.

Ksp= [Pb2+][I–]^2 = 1.08 × 10 –7

Step 3: Set up a chart that expresses initial and final concentrations (at equilibrium) of the
Pb2+(aq) and I–(aq).

Final Concentration

Species Initial Concentration at Equilibrium

Pb2+ 0 M x

I– 0 M 2 x

Part II: Specific Topics