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 IndicatorsThe 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 + H2O H3O+ + Ph-

• In an acid solution the [H3O+] 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 H3O+ ions in the reaction (H3O+

+ OH- 2H2O). The system shifts to the right to produce more H3O+ and red Ph- ions.

• Some texts use the symbol HIn to signify an indicator. Using bromthymol blue as an example we have:

Equilibrium: HIn + H2O H3O+ + 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 [H3O+] = [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-] > [H3O+].

At the endpoint the concentrations are equal.At the end of the graph [OH-] < [H3O+].

Sample Problems:1. Find the concentration of an H2SO4 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 + H2SO4 Na2SO4 + 2H2O

2. What volume of 3.00 M HCl can be neutralized by a reaction with 13.4 g of sodium carbonate?

Na2CO3 + 2HCl 2NaCl + H2O + CO2

Percent Purity by Volumetric Analysis

Find the percent purity of a sample of Na2CO3 if 250.0 mL of 0.300 M HCl is required to react completely with a 5.0 g sample of the impure salt. Rxn: Na2CO3 + 2HCl 2NaCl + H2O + CO2