1. 1 Chemistry 111 Chapter 16.10 – 16.12

2. 2 Concentration - Computations Dilution –Dilution Equation –Setting Up Problems Titration –Solid Titrant (like lab) –Liquid Titrant Stoichiometry

3. 3 Dilution Dilution means “watering down” a solution. Steps: 1.Start with known solution 2.Pipette a known amount into another container 3.Add Water to bring fluid level up. M 1 V 1 = M 2 V 2 Note: –Concentration always goes DOWN! M1M1 M 2, V 2 V1V1

4. 4 Titration Titration is a mole-based experiment using the buret. We usually put acid in the bottom (with phenolphthalein) and base in the top. Base: –Usually NaOH –Known Concentration, Measure Volume on Buret Acid: –Can be Solid (Type I) or Liquid (Type II) Math: n acid = n base –Either M 1 V 1 =M 2 V 2 or M 1 V 1 = Mass / MM M 1 _____ V 1 _____ Type I Mass _____ MM _____ Type II M 2 _____ V 2 _____

5. 5 Titration Example 25.00 mL of an unknown acid solution requires 38.32 mL of standardized 0.632 M NaOH to titrate. What is the concentration of the acid solution? This problem is Type II. M 1 _____ V 1 _____ Type II M 2 _____ V 2 _____ 25.00 mL ???? 38.32 mL 0.632 M

6. 6 Solution Stoichiometry Concentration: To use concentration in stoichiometry, get the Liters to cancel & you’ve got MOLES!

7. 7 Moles of B Moles of A  Molar Mass Gas Laws  Avogadro’s Number  Molarity  Molar Mass Gas Laws  Avogadro’s Number  Molarity Mass of B Volume of Gas B Number of particles of B Volume of Solution B Mass of A Volume of Gas A Number of particles of A Volume of Solution A Moles B Moles A Stoichiometry Map - Solutions

8. 8 Stoichiometry Example Titrations are perfect stoichiometery examples – but we have to add in the balanced equation (making them tougher). Kim tests the base solutions from lab by titrating sulfuric acid with them. If 44.23 mL of NaOH solution is required to neutralize 25.00 mL of 0.450 M H 2 SO 4, what is the concentration of the base?

9. 9 Solving the Stoichiometry Balanced chemical Equation –H 2 SO 4 + NaOH  H 2 O + Na 2 SO 4 Concentration: –M 1 = ????V 1 = 44.23 mL –M 2 = 0.450 MV 2 = 25.00 mL Mole Ratio =0.509 M