1. Solubility Product Determining Precipitation Conditions

2. Problem Solving Strategies  Write the net ionic equation.  Calculate the concentrations of each ion species;  volume x Molarity / total volume  Calculate the ion product, Q.  If Q<K sp then no ppt will form  If Q>K sp then a ppt will form  Run the ppt reaction to completion; RICE chart to determine LR and excess reagents

3. Strategies….continued  Calculate the [ ] of the excess reagent  Establish a BRA and RICE charts for the dissolving of the solid by using K sp value for the solid

4. Example #1 A 100.0ml sample of 0.0500M Pb(NO 3 ) 2 is mixed with 200.0ml of a 0.100M NaI solution. Determine the Pb 2+ and I - concentrations at equilibrium.  Calculate the [Pb 2+ ] and [I - ] [Pb 2+ ] =.100L (0.0500M) /.300L = 1.67 x 10 -2 M [I - ] =.200L (0.100M) /.300L = 6.67 x 10 -2 M  Calculate Q = [Pb 2+ ][I - ] 2 = 1.67 x 10 -2 M (6.67 x 10 -2 M) 2 = 7.43 x 10 -5  Since Q > K sp, the ppt PbI 2 will form

5. Example #1 Continued  Create a BRA chart to determine which is LR  Pb 2+ + 2 I - PbI 2  B.005mol.020mol  R -.005 -2 x.005  A 0.010mol  This means that [I - ] =.010mol/.300L =.0333M  Some of the PbI 2 will dissolve at equilibrium  R PbI 2 Pb 2+ + 2 I -  I 0.0333M  C +x +2x  E x.0333 + 2x

6. Example #1 Continued  K sp = x (.0333 + 2x) 2  1.4 x 10 -8 = x (.0333) 2  1.26 x 10 -5 M= x  [Pb 2+ ] = 1.26 x 10 -5 M