1. C. Johannesson III. Titration (p. 493 - 503) Ch. 15 & 16 - Acids & Bases

2. LPChem:Wz A. Neutralization  Chemical reaction between an acid and a base.

3. LPChem:Wz A. Neutralization  Chemical reaction between an acid and a base.  Products are a salt (ionic compound) and water.

4. C. Johannesson A. Neutralization ACID + BASE  SALT + WATER HCl + NaOH  NaCl + H 2 O Salts can be neutral, acidic, or basic! Neutralization does not always mean pH = 7. strong acid strong base neutral salt

5. C. Johannesson A. Neutralization ACID + BASE  SALT + WATER HCl + CuOH  CuCl + H 2 O HC 2 H 3 O 2 + NaOH  NaC 2 H 3 O 2 + H 2 O weak acid strong acid weak base strong base acidic salt basic salt

6. LPChem:Wz A. Neutralization ACID + BASE  SALT + WATER Strong + Strong  Neutral + water Acid Base Salt. Strong + Weak  Acidic + water Acid Base Salt. Weak + Strong  Basic + water Acid Base Salt.

7. C. Johannesson B. Titration  Titration Analytical method in which a standard solution is used to determine the concentration of an unknown solution. standard solution unknown solution

8. LPChem:Wz  Equivalence point  Point at which equal amounts of H 3 O + and OH - have been added.  (moles H 3 O + = moles OH - ) Mathematically determined by… Rapid pH change B. Titration

9. C. Johannesson  End Point  Visual end of titration indicator color change Should be very similar to equivalence point. B. Titration

10.  Strong Acid & Strong Base =  Titration equivalence is at pH = 7 LPChem:Wz Neutral Salt

11. B. Titration  Strong Acid & Weak Base =  Titration equivalence at pH < 7 LPChem:Wz Acidic Salt On your homework today, do #3 like this.

12. B. Titration  Weak Acid & Strong Base =  Titration equivalence at pH > 7 LPChem:Wz Basic Salt

13. C. Johannesson B. Titration moles H 3 O + = moles OH - M  V  n = M  V  n M:Molarity V:volume n:# of H + ions in the acid or OH - ions in the base

14. LPChem:Wz B. Titration  10.0 mL of 2.3M LiOH are required to neutralize 20.0 mL of HNO 2. Find the molarity of HNO 2. H3O+H3O+ M = ? V = 20.0 mL n = 1 OH - M = 2.3M V = 10.0 mL n = 1 MV# = MV# (2.3M)(10.0mL)(1) = M(20.0mL)(1) M = 1.15M HNO 2

15. C. Johannesson B. Titration  42.5 mL of 1.3M KOH are required to neutralize 50.0 mL of H 2 SO 4. Find the molarity of H 2 SO 4. H3O+H3O+ M = ? V = 50.0 mL n = 2 OH - M = 1.3M V = 42.5 mL n = 1 MV# = MV# (1.3M)(42.5mL)(1) =M(50.0mL)(2) M = 0.55M H 2 SO 4