VOLUMETRIC

ANALYSYIS

1.

This is a method of quantitative chemical analysis, involving the accurate measurement of volumes of 2 solutions which will react together completely.

2.

In VOLUMETRIC ANALYSIS, the concentration of a solution is found by measuring the volume of solution that will react with a known volume of a standard solution.

3.

A STANDARD SOLUTION is a solution whose concentration is accurately known and used to determine the concentration of the other solution.

4.

TITRATION is the process of adding a standard solution (titrant) to a fixed volume of the other (unknown) solution until the reaction between the two solutions is complete, i.e. equivalent amounts of reactants are present.

5.

2 main types of reaction ¾ ACID-BASE reactions and REDOX reactions.

6.

STOICHIOMETRIC POINT (equivalence point, theoretical end-point) of a titration is the theoretical volume of acid (or base) required for complete reaction, i.e. the exact point in the titration at which precisely equivalent amounts of the reactants are present together in the reaction vessel.

7.

The END-POINT of a titration

a) is the visible point at which the reaction is complete

b) is usually marked by a colour change of the solution

c) would vary using different indicators

8.

STANDARDISATION is the process by which the concentration of a solution is determined by using a standard solution.

9.

INDICATORS are substances which will show a different colour in acid and alkaline medium, i.e. when the pH of its environment is altered.

Occasionally the solutions themselves can act as their own indicators. e.g. No indicator is required when using KmnO4 solution in redox titrations. The purple permanganate is usually run in from the burette into the second reactant (usually a colourless or pale green solution). The equivalence point is reached when the first permanent pink tinge is seen. Permanganate solutions become colourless on reduction. Hence persistence of the pink colour indicates that the reducing agent has been used up.

10.

ACID-BASE TITRATION : Reactions between acids and alkalis are particularly suitable for use in titrimetric analysis as they are rapid, essentially complete and the point at which the reaction is complete is easily detected.

11.

Consider the equation of a reaction (a moles of A react with b moles of B):

a A + b B ® products

In a titration, if VA cm3 of an acid solution of concentration MA mol dm-3 react with VB cm3 of a base solution of concentration MB mol dm-3,

then MA VA/ MB VB = a / b

12.

For Dilution Problems:

No. of moles before dilution = No. of moles are dilution

M1V1 = M2V2

M1 ¾ original conentration (mol dm-3)

V1 ¾ volume taken (dm3)

M2 ¾ concentration of the diluted solution (mol dm-3)

V2 ¾ final volume (dm3)

13.

Acid-Base Titration Problems:

example: 25.0 cm3 of a sodium hydroxide solution were required to neutralise 22.50 cm3 of 0.100 mol dm-3 H2SO4. Calculate the concentration of the sodium hydroxide solution in mol dm-3.

method 1

Write the equation for the reaction

2 NaOH + H2SO4 ® Na2SO4 + 2 H2O

Calculate the no. of moles of H2SO4 required.

22.50/1000 x 0.1 = 2.25 x 10-3 mol contained in 22.50 dm3

Calculate the no. of moles of NaOH neutralised.

H2SO4 º 2 NaOH

2 x 2.25 x 10-3 = 4.50 x 10-3 mol contained in 25 dm3

 Calculate the no. of moles of NaOH in 1 dm3 solution

(4.50 x 10-3 x 100) / 25 = 0.18 mol

\ Conc. of NaOH = 0.18 mol dm-3

Method 2

Write the equation for the reaction

2 NaOH + H2SO4 ® Na2SO4 + 2 H2O

Apply the mole ratio formula

MA VA/ MB VB = 1 / 2

(0.1 x 22.5) / (MNaOH x 25.0) = 1 / 2

MNaOH = 0.18 mol dm-3

 14.

 In back titrations, a known amount of excess reactant A is added to reactant B and the amount of excess A remaining unreacted after the reaction with B is found by titration. By subtracting the excess A from the initial amount of A added, the amount of A that actually reacted with B can be determined.

example:

FA1 is a mixture of NaCl and NaHCO3 and FA2 is a solution containing 30.0 g dm-3 of FA1. 40.0 cm3 of a 0.200 mol dm-3 HCl is added to 25.0 cm3 of FA2 in a conical flask and the excess acid remaining in the flask requires 21.60 cm3 of 0.150 mol dm-3 NaOH for complete neutralisation. Calculate the % by mass of NaHCO3 in FA1.

no. of mol of NaOH required for neutralisation = 0.15 x (21.6/1000) = 3.24 x 10-3

no. of mol of excess HCl remaining in flask = 3.24 x 10-3 (NaOH º HCl)

Amt of HCl added = 40/1000 x 0.2 = 8 x 10-3 mol

Amt of HCl reacted with NaHCO3 = 8 x 10-3 - 3.24 x 10-3 = 4.76 x 10-3 mol

no. of mol of NaHCO3 in 25.0 cm3 of FA2 = 4.76 x 10-3 (HCO3- º H+)

mass of NaHCO3 in 1 dm3 of FA2 = 4.76 x 10-3 x 1000/25 x 84 = 15.99 g

% by mass of NaHCO3 in FA1 = 15.99/30 x 100 = 53.3%

K Solve this #1: A sample containing ammonium sulphate was warmed with 250 cm3 of 0.800 mol dm-3 sodium hydroxide solution. After the evolution of ammonia had ceased, the excess sodium hydroxide solution was neutralised by 85.00 cm3 of 0.500 mol dm-3 hydrochloric acid solution. What mass of ammonium sulphate did the sample contain?

The two reactions taking place are :

(1) Rxn betw. the ammonium salt and the alkali

(NH4)2SO4 (s) + 2 NaOH (aq) ® 2 NH3 (g) + Na2SO4 (aq) + 2 H2O (l)

(2) Rxn betw. the excess alkali and the HCl

NaOH (aq) + HCl (aq) ® NaCl (aq) + H2O (l)

[10.4 g]


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