1. Acid/Base Titration Technique used to determine the concentration of an acid or base by comparison with a standard. A neutralization reaction is carried out. Rely on color change to signal end of reaction.

2. Titration: Monoprotic Acid + Monohydroxy Base HA + MOH  HM + H 2 O Use the relationship in the balanced equation to find the unknown concentration. At the end-point, # moles HA = # moles MOH

3. How to calculate # of moles? Remember Molarity! Molarity = # moles solute Rearranging: # moles solute = Liters of solution Molarity X Liters of solution

4. Acid-Base Titration Carefully controlled neutralization reaction. standard solution acid-base indicatorRequires a standard solution and an acid-base indicator. Standard solution has an acid or base of known concentration.

5. Titration Standard solution is slowly added to unknown solution. As the solutions mix, a neutralization reaction occurs. Equivalence pointEventually, enough standard solution is added to neutralize the unknown solution.  Equivalence point. Simple Titration Animation

6. Equivalence point Total number of moles of H + ions donated by acid = total number of moles of H + accepted by base. Total moles H + = total moles OH -

7. Titration End-pointEnd-point = point at which indicator changes color. If the indicator is chosen correctly, the end-point is very close to the equivalence point.

8. Titration of a strong acid with a strong base Volume of 0.100 M NaOH added (ml) pH 0- 14- 7- Equivalence Pt  Phenolphthalein Color change: 8.2 to 10  Between pH of 4 and 10, only drops a few drops of base are added. 0 ml  40ml  20 ml

9. M H+ V H+ = M OH- V OH- M H+ = molarity of H + M OH- = molarity of OH - V H+ = volume of H + V OH- = volume of OH - If you know 3, you can find the 4 th.

10. M a V a = M b V b True for monoprotic acids and monohydroxy bases. True for diprotic acids and dihydroxy bases. Etc.

11. Titration Problem #1 40.0 mL 35.0 mL0.100 M NaOHCalculate the concentration of the nitric acid solution.In a titration of 40.0 mL of a nitric acid solution, the end point is reached when 35.0 mL of 0.100 M NaOH is added. Calculate the concentration of the nitric acid solution.

12. Neutralization Reaction HNO 3 + NaOH  H 2 O + NaNO 3 1 H in acid & 1 OH in base. M a V a = M b V b

13. Variables M a = ? V a = 40.0 mL M b = 0.100 M V b = 35.0 mL

14. Plug and Chug X (40.0 mL) = (0.100 M )(35.0 mL) X =.875 M

15. Titration Problem #2 What is the concentration of a hydrochloric acid solution50.0 mL0.250 M KOH 20.0 mL of the HCl solutionWhat is the concentration of a hydrochloric acid solution if 50.0 mL of a 0.250 M KOH solution are needed to neutralize 20.0 mL of the HCl solution of unknown concentration?

16. Neutralization Reaction KOH + HCl  H 2 O + KCl 1 H in acid & 1 OH in base M a V a = M b V b

17. Variables M a = X V a = 20.0 mL M b = 0.250 M V b = 50.0 mL

18. Plug and Chug X (20.0 mL) = (0.250 M) (50.0 mL) X = 0.625 M

19. Titration Problem #3 What is the concentration of a sulfuric acid solution50. mL of a 0.25 M Ba(OH) 2 20 mL of the H 2 SO 4 solutionWhat is the concentration of a sulfuric acid solution if 50. mL of a 0.25 M Ba(OH) 2 solution are needed to neutralize 20 mL of the H 2 SO 4 solution of unknown concentration?

20. Neutralization Reaction H 2 SO 4 + Ba(OH) 2  2 H 2 O + BaSO 4 H 2 SO 4 is a strong diprotic acid. Ba(OH) 2 is a strong dihydroxy base. # of H’s in acid = # of OH’s in base. M a V a = M b V b

21. Variables M a = X V a = 20. mL M b = 0.25 M V b = 50. mL

22. Plug and Chug X (20.) = (0.25 M) (50. mL) X = 0.625 M = M H+ To calculate the Molarity of the H 2 SO 4, adjust for the fact that the acid is diprotic.

23. [H + ] vs. [H 2 SO 4 ] H 2 SO 4  2H + + SO 4 2- Molarity of H 2 SO 4 = ½ [H + ] ½ (0.625 M) = 0.31 M H 2 SO 4

24. Other kinds of Titrations Titrations based on redox reactions are also used. Same ideas: rely on color change to signal completion of reaction. Use stoichiometric equalities to determine concentration of unknown. Redox titrationAnimation