1. Acid/Base Titration

2. Acid–Base Titration The concentration of a weak acid or a weak base in water is difficult – if not impossible – to measure directly. But we can calculate the concentration from the results of titration. The concentration of a weak acid or a weak base in water is difficult – if not impossible – to measure directly. But we can calculate the concentration from the results of titration.

3. Acid–Base Titration A titration is a carefully controlled neutralization rxn. To run a titration on a soln of unknown conc of an acid or a base, you need a 2 nd soln called a standard soln. A standard soln contains acid or base in a known conc A titration is a carefully controlled neutralization rxn. To run a titration on a soln of unknown conc of an acid or a base, you need a 2 nd soln called a standard soln. A standard soln contains acid or base in a known conc

4. Acid–Base Titration If the unknown soln is an acid, the standard soln is a base –And vice a versa Another substance involved in a titration is an indicator. –which is a substance that changes color at a certain pH If the unknown soln is an acid, the standard soln is a base –And vice a versa Another substance involved in a titration is an indicator. –which is a substance that changes color at a certain pH

5. Acid–Base Titration Most of the time we use the common indicator Phenolphthalein. –Phenolphthalein is clear in acid –Light pink in neutral –Dark pink in base Most of the time we use the common indicator Phenolphthalein. –Phenolphthalein is clear in acid –Light pink in neutral –Dark pink in base

6. Procedure for Titration the standard is slowly added to the unknown soln the 2 solns mix, the acid in one neutralizes the base in the other Eventually, enough standard soln is added to neutralize all of the acid or base in the unknown soln. the standard is slowly added to the unknown soln the 2 solns mix, the acid in one neutralizes the base in the other Eventually, enough standard soln is added to neutralize all of the acid or base in the unknown soln.

8. Procedure for Titration The point at which this occurs is called the equivalence point. –[H 3 0+] = [OH-] The point at which the indicator changes color is called the end point of the titration. The point at which this occurs is called the equivalence point. –[H 3 0+] = [OH-] The point at which the indicator changes color is called the end point of the titration.

9. Procedure for Titration If the indicator is chosen correctly the end point is very close to the equivalence point. Therefore, at approximately the end pt of a titration the total mols of H+ donated by the acid is equal to the total mols of H+ accepted by the base. If the indicator is chosen correctly the end point is very close to the equivalence point. Therefore, at approximately the end pt of a titration the total mols of H+ donated by the acid is equal to the total mols of H+ accepted by the base.

10. Procedure for Titration For Arrhenius Acids and Bases we can say: Total moles of H + from the acid Total moles of OH - from the base = = This is key to solving titration calcs

22. Weak Acid/Strong Base Titration A solution that is 0.10 M CH 3 COOH is titrated with 0.10 M NaOH Endpoint is above pH 7

23. Strong Acid/Strong Base Titration A solution that is 0.10 M HCl is titrated with 0.10 M NaOH Endpoint is at pH 7

24. Strong Acid/Strong Base Titration A solution that is 0.10 M NaOH is titrated with 0.10 M HCl Endpoint is at pH 7 It is important to recognize that titration curves are not always increasing from left to right.

25. Strong Acid/Weak Base Titration A solution that is 0.10 M HCl is titrated with 0.10 M NH 3 Endpoint is below pH 7

26. Sample Problem Solns of NaOH are used to unclog drains. A 43 ml volume of NaOH was titrated with 32 ml of.100M HCl. What is the molarity of the NaOH soln?

27. Analyze: At the equivalence pt, the added acid has completely neutralized the unknown base. It took 32 ml of acid to neutralize 43 ml of base We know that mols of base = mols of acid, at the equiv pt At the equivalence pt, the added acid has completely neutralized the unknown base. It took 32 ml of acid to neutralize 43 ml of base We know that mols of base = mols of acid, at the equiv pt

28. Solve: moles of acid used: –(.100mols/L)(.032L)=.0032mol of acid mols of acid = mols of base –.0032 mol base used in rxn. Molarity of NaOH –.0032mol/.043L =.074M moles of acid used: –(.100mols/L)(.032L)=.0032mol of acid mols of acid = mols of base –.0032 mol base used in rxn. Molarity of NaOH –.0032mol/.043L =.074M