Title: Lesson 13 Titration Learning Objectives: Understand how we can use titration to work out an unknown concentration

If we had an unlabelled bottle of HCl, how can we find out it’s concentration using neutralisation? We can react an acid with a standard solution of alkali such as NaOH, and determine the exact volumes reacting together. If we have the volumes of both and the concentration of one we can work out the unknown concentration.

Titrations Can be used to find: Concentration of a solution Molar Mass (Mm) Formula

Carrying out a titration Using a pipette add a measured volume of one solution to a conical flask. The other (unknown) solution is placed in a burette. Solution on the burette is added to the solution in the conical flask until the reaction has just completed. This is the end-point of the titration. The volume of the solution added from the burette is measured.

The Equivalence-Point Can be identified using an indicator. The indicator must be a different colour in the acidic solution than in the neutral or basic solution.

Technique/Procedure

Example results table

Calculation 25cm3 of 0.01M sulfuric acid was added to neutralize 50cm3 of sodium hydroxide. Calculate the concentration of the sodium hydroxide in g dm-3.

Step 1 Write a balanced equation:

Step 2 Calculate the number of moles of acid added from the burette: Dividing by 1000 converts cm3 to dm3

Step 3 Use the ratio to work out the number of moles in the sample of alkali:

Step 4 Calculate the concentration. (convert to g dm3 if necessary). Concentration (Mol dm3) = Moles Volume (dm3) Concentration (Mol dm3) = 50 x 10-3 (50 cm/1000) Concentration (Mol dm3) = 0.01Mol dm3 Dividing by 1000 converts cm3 to dm3

Step 5 Convert to g dm3 if required (usually for solubility) Mass = 0.01 x 40 Concentration = 0.4 g dm3

Example: It is found by titration that 25.0 cm3 of an unknown solution of sulfuric acid is just neutralised by adding 11.3 cm3 of1.00 mol dm-3 sodium hydroxide. What is the concentration of sulfuric acid in the sample. H2SO4 + 2 NaOH  Na2SO4 + 2 H2O Use: n2C1V1 = n1C2V2 2 x C1 x 25.0 = 1 x 1.00 x 11.3 C1 = (1 x 1.00 x 11.3) / (2 x 25.0) = 0.226 mol dm-3 Where: n = coefficient C = concentration V = volume ‘1’ refers to H2SO4 ‘2’ refers to NaOH

Questions You have 75.0 cm3 of a 0.150 mol dm-3 solution of zinc sulphate (ZnSO4). What mass of zinc sulphate crystals will be left behind on evaporation of the water? What mass of copper (II) chloride (CuCl2) should be added to 240 cm3 water to form a 0.100 mol dm-3 solution? A 10.0 cm3 sample is removed from a vessel containing 1.50 dm3 of a reaction mixture. By titration, the sample is found to contain 0.0053 mol H+. What is the concentration of H+ in the main reaction vessel? The titration of 50.0 cm3 of an unknown solution of barium hydroxide was fully neutralised by the addition of 12 cm3 of 0.200 mol dm-3 hydrochloric acid solution. What concentration is the barium hydroxide solution? Ba(OH)2 + 2 HCl  BaCl2 + 2 H2O

More questions… Complete the test yourself questions on page 45. Check your answers on page 559.

More Titration Problems…

Solutions

Done in reverse by returning to the end point after it is passed. Used when end point is hard to identify or when one of the reactants is impure. A known excess of one reactant is added. The unreacted excess is then determined by titration against a standard solution. Subtract the unreached amount from the original amount.

Questions…

Questions…

Questions…

Problem Time Write three stoichiometry problems using any of the ideas from the unit so far (make sure you know the answer). In ten minutes time you will need to give your problems to a classmate to solve.

Key Points

Alcohol is not the answer but it is a solution*! Remember! Alcohol is not the answer but it is a solution*! *Of ethanol, water and various other bits and pieces