Volumetric Analysis – Titrations Volumetric Analysis a process of analysis accomplished by adding together particular volumes of 2 solutions the objective may be to determine the molar concentration of a solution of unknown strength, the percent purity of a sample, etc.

Titration that part of a volumetric analysis which is carried out by carefully adding a controlled volume of one substance to another sample by use of a burette Endpoint that point reached in a titration when enough of one reagent has been added to just completely react with the other reagent and produce a colour change in the indicator being used E.g. A + 2B  C + D An endpoint would be reached when enough A (.30 moles) has been added to just use up the amount of B that was present (.60 moles). Endpoints are usually signaled by the colour change of an indicator.

Behavior of Indicators The way indicators like phenolphthalein, litmus, & bromthymol blue change colour is readily explained on the basis of equilibrium shifts. Acid-base indicators are really just very weak acids or bases which are slightly dissociated and set up equilibrium between the molecular and ionic forms, which are different colours.

Example:phenolphthalein (HPh) It is a weak acid which in water sets up this equilibrium: HPh + H 2 O  H 3 O + + Ph - In an acid solution the [H 3 O + ] is large, the equilibrium is shifted far to the left to form colorless HPh molecules In a basic solution the OH - ions use up the H 3 O + ions in the reaction (H 3 O + + OH -  2H 2 O). The system shifts to the right to produce more H 3 O + and red Ph - ions. Some texts use the symbol HIn to signify an indicator. Using bromthymol blue as an example we have: Equilibrium:HIn + H 2 O  H 3 O + + In -

Titration Curve This is a graph which shows how the pH changes during a titration. GENERAL SHAPES: 1) Acid titrated with a base. At the equivilence or endpoint [H 3 O + ] = [OH - ] This is where the indicator colour changes & pH changes rapidly.

2)Base titrated with an acid. At the top of the graph the [OH - ] > [H 3 O + ]. At the endpoint the concentrations are equal. At the end of the graph [OH - ] < [H 3 O + ].

Sample Problems: 1. Find the concentration of an H 2 SO 4 solution if 15.0 mL of it is needed to neutralize 40.0 mL of 0.60 M NaOH solution, according to the equation: 2NaOH + H 2 SO 4  Na 2 SO 4 + 2H 2 O

2.What volume of 3.00 M HCl can be neutralized by a reaction with 13.4 g of sodium carbonate? Na 2 CO 3 + 2HCl  2NaCl + H 2 O + CO 2

Percent Purity by Volumetric Analysis Find the percent purity of a sample of Na 2 CO 3 if mL of M HCl is required to react completely with a 5.0 g sample of the impure salt. Rxn: Na 2 CO 3 + 2HCl  2NaCl + H 2 O + CO 2