1. To Precipitate or not
6-6

2. Steps 1. Balanced chemical equation.
2. Find the molarity of the reactants
CaVa = CbVb
Va is the total volume ( add the volume of the reactants).
Ca = (CbVb) /Va
3. Dissociate each reactant, find the concentration of each ion.
4. Write the Ksp expression of the solid.
5. Find the trial Ksp or Q.
6. Is Q > Ksp, if so a precipitate forms.

3. Will a precipitate of silver chloride form if one drop of (0
Will a precipitate of silver chloride form if one drop of (0.050 ml) of 5.0 M silver nitrate solution is added to 1.5 L of 0.20 M sodium chloride,
if the Ksp is 1.8 x ?
Balance the equation
AgNO3 (aq) + NaCl (aq)→ AgCl (s) +NaNO3 (aq)

4. 2. New concentration
AgNO3 NaCl
0.05 ml of 5.0 M 1.5 l of 0.20 M
Ca = (CbVb) /Va Ca = (CbVb) /Va
=( l)(5.0 mol/l) =(1.5 l)(0.20mol/l)
1.5 l l
( add the two volumes together, the l is insignificant to the 1.5 l)

5. Molarity of reactants AgNO3 NaCl 0.00025 mol 0.30 mol 1.5 L 1.5 L
= 1.67 x 10-4 mol/l = 0.20 mol/l

6. 3. dissociation →[Ag+] + [NO3-] 1.67 x 10-4 M 1.67 x 10-4 M
→ [Na+] [Cl-]
0.20 M M

7. 4. Write Ksp expression for silver chloride.
AgCl (s) > Ag+ (aq) + Cl- (aq)
5. Solve for Q
Ksp = [Ag + (aq) ] [Cl - (aq)]
=( 1.67 x 10-4)(0.20)
= 3.34 x 10-5

8. 6. Compare Q to Ksp If Q > ksp 3.34 x 10-5 > 1.8 x 10-8
A ppt will form.

9. Determine if a precipitate of Ca(OH)2 will form is 20 ml of 0
Determine if a precipitate of Ca(OH)2 will form is 20 ml of 0.30 mol/l CaCl2 solution is mixed with 10 ml of 0.90 mol/l NaOH, if the Ksp for Ca(OH)2 is 5.1 X 10-6.

10. solution CaCl2 (aq) + 2 NaOH (aq) 20 ml of 0.30M 10 ml of 0.90M
→ Ca(OH)2 (s) + 2 NaCl (aq)

11. CaCl2 (aq) NaOH (aq)
Ca = (CbVb) /Va Ca = (CbVb) /Va
=(20 ml)( mol/l) =(10 ml)(0.90mol/l)
30 ml ml
= 0.2 M = 0.30 M
→Ca Cl- → 2 Na Cl-

12. Ksp expression for Ca(OH)2 is:
Ca(OH)2 ----> Ca 2+ (aq) + 2 OH- (aq)
Ksp = [Ca 2+ ] [OH-] 2
Q = ( 0.20)(0.30)2
Q =
Ksp = 5.1 x 10 -6
Q > Ksp therefore a ppt forms