1. Do Now: Look at the examples and fill in the blanks  Neutralization reactions always make ______, almost always make________, and sometimes make ________. HCl + NaOH  H 2 O + NaCl HCl + NH 3  NH 4 Cl HCl + NaHCO 3  CO 2 + H 2 O + NaCl

2. Titration Calculations

3. Titrations (don’t write this)  Titration is defined as the controlled addition and measurement of the amount of a solution of known concentration required to react completely with a measured amount of a solution of unknown concentration.  You must have a balanced equation to do the calculations.

4. Given information  The names or formula of the acid and base.  The volume and molarity of one reactant (reactant A).  The volume of the second reactant (reactant B).

5. Indicator is selected on a case by case basis according to where the color change needs to happen. Phenothaline is selected for strong base/weak acid titrations.

6. You need to find  1. A balanced equation for the titration.  2. The number of moles from the volume and molarity of “reactant A”.  3. The number of moles consumed of “reactant B”.  4. The molarity of reactant B.  (once you understand the process you can do all the conversions in one step if it just asks you for the concentration)

7. Example  A student is given 20 mL of HCl with an unknown molarity. She titrates this solution with 30 mL of 0.5 M NaOH. What is the concentration of the HCl?

8. You need to find  1. A balanced equation for the titration.  2. The number of moles from the volume and molarity of “reactant A”.  3. The number of moles consumed of “reactant B”.  4. The molarity of reactant B. HCl + NaOH  NaCl + H 2 O 30 mL 1 L 0.5 mol = 0.015 mol NaOH 1000 mL 1 L 0.015 mol NaOH 1 mol HCl = 0.015 mol HCl 1 mol NaOH M = mol = 0.015 mol HCl = 0.75 M HCl L 0.020 L

9. Another Example  If 29.96 mL of a solution of Ba(OH) 2 requires 16.08 mL of a 2.303 M solution of HNO 3 for complete titration, what is the concentration of the Ba(OH) 2 solution? Answer: 0.6180 M = 0.6180 mol dm -3

10. Another Example  By titration, 17.6 mL of aqueous H 2 SO 4 neutralized 27.4 mL of 0.0165 M LiOH solution. What was the molarity of the aqueous acid solution? Answer: 0.0128 M H 2 SO 4 = : 0.0128 mol dm -3 H 2 SO 4

11. Stop Here

12. What if you know the molarity?  What volume in milliliters of 1.250 M HCl is required to neutralize 50.0 mL of a 0.300 M NaOH solution? Answer: 12 mL of HCl