1. Titrations Acid Base Titrations Indicators Calculations

2. Titration “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.” Titration is an analytical process designed to determine an unknowns concentration The most common titrations are acid/base.

3. Titration Process The idea behind titration is that at one particular point in a neutralization reaction the moles of H + and OH - are equal (equivalence point). If we can use some chemical (indicator) to signify to use when this occurs we can use the amounts added to that point to find the unknown.

4. Indicators Indicators change color in a specific pH range. You choose the indicator by looking at the pH at equivalence point. The most common indicator is phenolphthalein.

5. Titration Setup The long glass tubes are burets Generally a flask is used to avoid splashing. The buret is very precise (± 0.01mL)

6. Calculations We need to be able to determine the moles in a certain volume. Molarity is moles/L so: –moles = molarity x volume(L) When the titration is done then –moles of H + = moles of OH - We should be able to find the unknown by find the unknown moles and dividing my the volume of unknown used.

7. Example Lets consider the titration of a strong acid by a strong base. We have 20.0mL of a HCl with unknown concentration. We titrate it with a 0.500M NaOH solution. At equivalence point we used 42.4mL of NaOH solution. NaOH(aq) + HCl(aq)  H 2 O(l) + NaCl(aq)

8. Example (cont.) First find the moles of known substance. –moles NaOH = moles OH - = 0.50M x 0.0424L –moles OH - = 0.0212 –moles OH - = moles H + = 0.0212 We now use the balanced chemical equation to calculate the moles of acid. –0.0212 moles H + x 1HCl/1H + = 0.0212 mole HCl –**Note: if the acid is diprotic then this calculation would change appropriately.

9. Example (cont.) We now have the moles of acid that were present at equivalence point. The original amount of acid added is 20.0mL Using this and the moles of acid we can find concentration –Molarity unknown = moles acid/volume acid (L) –M unknown = 0.0212 moles/0.0200L = 1.06M HCl

10. Titration Example 15.00mL of Sr(OH) 2 were neutralized via titration with 13.5mL of 2.00M HNO 3. What is the concentration of Sr(OH) 2 Sr(OH) 2 + 2HNO 3  Sr(NO 3 ) 2 + 2H 2 O 13.5mL = 0.0135L; 0.0135L x 2.00M = 0.0270 moles HNO 3 0.0270 moles HNO 3 x 1Sr(OH) 2 /2HNO 3 = 0.0135 mole Sr(OH) 2 M = 0.0135 mole/0.015L = 0.90M Sr(OH) 2

11. Finally The previous example was with an unknown acid, of course, we could also do the same thing with an unknown base. The most common mistake is in not writing the balanced neutralization reaction and using it to change between moles of acid and moles of base, or vice versa.