Titrations

Acid Base Titrations

Indicators

Calculations

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.

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.

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.

Titration Setup

• The long glass tubes are burets

• Generally a flask is used to avoid splashing.

• The buret is very precise (± 0.01mL)

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.

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) H2O(l) + NaCl(aq)

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.

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– Molarityunknown = moles acid/volume acid (L)

– Munknown = 0.0212 moles/0.0200L = 1.06M HCl

Titration Example

• 15.00mL of Sr(OH)2 were neutralized via titration with 13.5mL of 2.00M HNO3. What is the concentration of Sr(OH)2

• Sr(OH)2 + 2HNO3 Sr(NO3)2 + 2H2O• 13.5mL = 0.0135L; 0.0135L x 2.00M = 0.0270

moles HNO3

• 0.0270 moles HNO3 x 1Sr(OH)2/2HNO3 = 0.0135 mole Sr(OH)2

• M = 0.0135 mole/0.015L = 0.90M Sr(OH)2

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.