1. Precipitation Reactions
The cation from one reactant combines with the anion from another reactant to form an insoluble compound, called a precipitate.

2. Activity: Precipitation Reactions
When a solution of lead(II) nitrate is mixed with a solution of potassium chromate, a yellow precipitate forms according to the equation:
Pb(NO3)2(aq) + K2CrO4(aq)  2 KNO3(aq) + PbCrO4(s)
What volume of M lead(II) nitrate is required to react with mL of M potassium chromate? What mass of PbCrO4 solid forms?

3. Activity Solution: Precipitation Reactions
Pb(NO3)2(aq) + K2CrO4(aq)  2 KNO3(aq) + PbCrO4(s)
What volume of M lead(II) nitrate is required to react with mL of M potassium chromate? What mass of PbCrO4 solid forms?
To solve for volume of Pb(NO3)2:
M × V = moles

4. Activity Solution: Precipitation Reactions
Pb(NO3)2(aq) + K2CrO4(aq)  2 KNO3(aq) + PbCrO4(s)
What volume of M lead(II) nitrate is required to react with mL of M potassium chromate? What mass of PbCrO4 solid forms?
To solve for mass of PbCrO4:

5. Acid-Base Titrations Titration Equivalence point End point
The process of determining the concentration of one substance in solution by reacting it with a solution of another substance that has a known concentration
Equivalence point
The point in a titration at which the moles of the acid equal the moles of the base
End point
The point in a titration where the indicator changes color
Slightly different than the equivalence point, but a good estimation

6. Acid-Base Titrations
Figure 11.25

7. Activity: Acid-Base Titrations
We need mL of M HCl solution to completely titrate mL of Ba(OH)2 solution of unknown concentration. What is the molarity of the barium hydroxide solution?

8. Activity Solution: Acid-Base Titrations
We need mL of M HCl solution to completely titrate mL of Ba(OH)2 solution of unknown concentration. What is the molarity of the barium hydroxide solution?
First, write the balanced chemical equation:
2 HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2 H2O(l)
Next, find the moles of HCl:

9. Activity Solution: Acid-Base Titrations
HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + H2O(l)
Next, we need to find the volume (in L) of Ba(OH)2:
Finally, divide the moles of Ba(OH)2 by the volume of Ba(OH)2:

10. Acid-Base Titrations
Figure from p. 454
Figure 11.25