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One method in volumetric analysis is titrationIn titration:

- substance to be analysed is known as the analyte

- the solution added to the analyte is known as the titrant

- titrant is usually delivered from a buret

Volumetric analysis refers to any procedure in which the volume of reagent needed to react with the analyte is being measured.

- the point when the quantity of added titrant is the exact amount necessary for stoichiometric reaction with the analyte is called the equivalence point

Example :

5HO-C-C-OH + 2MnO4- + 6H+ 10CO2 + 2Mn2+ + 8H2O

If the unknown contains 5.0 mmol of oxalic acid, the equivalence point is reached when 2.0 mmol of MnO4

- has been added.

Equivalence point is the ideal theoretical result we seek in a titration

What we actually measure is the end point, which is marked by a sudden change in the physical property of the solution

TitrationTitration

Methods for determining the end point are:- detecting a sudden change in the voltage or current

between a pair of electrodes- observing an indicator colour change.

An indicator is a compound which changes colour abruptly near the equivalence point. This change is caused by the disappearance of the analyte or appearance of excess titrant

- monitoring the absorption of light

Estimation of titration error – blank titration :

5HO-C-C-OH + 2MnO4- + 6H+ 10CO2 + 2Mn2+ + 8H2O

-solution containing no oxalic acid was titrated with MnO4- to deter

mined how much is needed to form an observable purple colour

-this volume of MnO4- is then subtracted from the volume observed

in the analytical titration

What is the difference between equivalence point and end point?

Titration error – by choosing a physical property whose change is easily observed one can minimise the titration error so that the end point is very close to the equivalence point

The concentration of titrant used can be derived if :

- the titrant was prepared by dissolving a weighed amount of pure reagent in a known volume of solution

- such a reagent is known as a primary standard because the reagent is pure enough (should be 99.9% pure)to be weighed and used directly. See Table 28-8; p836

What if a primary standard is not available ?

- titrate reagent (to be used as titrant in analysis) against a weighed primary standard in order to determine the concentration of titrant – process is known as standardization

- after standardization, the reagent (to be used as titrant in analysis) is known as the standard solution

7-2 Titration Calculations7-2 Titration CalculationsExample : pp151The calcuim content of urine can be determined by:1. precipitating Ca2+ as calcium oxalate in basic solution:

Ca2+ + C2O42- Ca(C2O4).H2O(s)

2. wash the precipitate and dissolve it in acid to give Ca2+ and

H2C2O4 in solution

3. heat the dissolved oxalic acid to 60oC and titrate with standardized KMnO4 until the purple end point is observed

-

Standardization :Suppose 0.3562 g of Na2C2O4 is dissolved in a 250.0 ml volumetric flask. Concentration of the oxalate solution is:(0.3562g Na2C2O4)(134.00 g/mol Na2C2O4)

0.250 ml

= 0.01063M

Moles of C2O42- in 10.0ml = (0.01063)(0.010l)

= 1.063 x 10-4 mol

Since:

5 NaO-C-C-ONa + 2MnO4- + 6 H+ 10CO2 + 2Mn2+ + 8H2O

2 mol MnO4- requires 5 mol oxalate

moles of MnO4- =

= 0.04253 mmol

molarity of MnO4- =

= 8.794 x 10-4 M

Analysis of Unknown :

Ca in a 5.0 ml urine sample was precipitated, redissolved and required 16.17ml of standard MnO4

- solution. Find the concentration of Ca2+ in the urine.

(2mol MnO4-)

(5mol C2O42-) (mol C2O4

2-)

0.04253 mmol48.36 ml

In 16.17ml of standard MnO4- solution, there are:

(0.01617l)(8.794 x 10-4 M) = 1.422 x 10-4 mol MnO4

-

1.422 x 10-4 mol MnO4- will react with:

mol C2O42- =

= 0.03555 mmolFrom :

Ca2+ + C2O42- Ca(C2O4).H2O(s)

there is one oxalate ion for each Ca2+ in Ca(C2O4) .H2O(s)

there must be 0.03555 mmol Ca2+ in 5.00ml urine[Ca2+] = (0.03555 mmol )/5ml

= 0.00711M

(5mol C2O42-)

(2mol MnO4-)

(mol C2O42-)

When titrant is added to the analyte until the reaction is complete direct titration

Some reactions require an excess of the titrant for complete reaction with the analyte. In such cases, the first titration will be followed by a titration with a second standard reagent. The second standard reagent is used to titrate the excess of the first reagent back titration

Back titration is also applicable if the end point of the back titration is clearer than the end point of the direct titration. Or Excess reagent is required for complete reaction.

Kjeldahl Nitrogen AnalysisKjeldahl Nitrogen Analysis

- developed in 1883- one of the most accurate and widely used method for determining nitrogen in substances

Procedure :

- substance is decomposed and dissolved (digested) in boiling H2SO4 to which K2SO4 has been added. Selenium-coated boiling chips are used to catalyze the reaction

organic C H N NH4+ + CO2 + H2O

- after digestion, the solution containing NH4+ is made basic

NH4+ + OH- NH3 + H2O

- the liberated NH3 is distilled into a known amount of HClNH3 + H+ NH4

+

- Excess unreacted HCl is titrated with standard NaOH to determine how much HCl has been consumed

H+ + OH- H2O

boiling H2SO4

7-4. Precipitation Titration Curve7-4. Precipitation Titration CurveA titration curve is a graph showing how the concentration of one of the reactants varies as titrant is addedTitration curve tells us how concentrations of analyte and titrant vary during a titration. From it we can:- understand the chemistry that occurs during a titration- learn how experimental control can be exerted to influence the quality of an analytical titration- since concentration varies over many orders of magnitude, the graph plots the p function { pX = -log[X]} against the volume of XExample :

Consider the titration of 25.00 ml of 0.1000M I - with 0.05000M Ag+, given:

AgI(s) I- + Ag+ Ksp = [Ag+][I-]

= 8.3 x 10-17

Each aliquot of Ag+ reacts completely with I- . At the equivalence point, there will be a sudden increase in the Ag+ concentration because all the I- has been consumed and now unreacted Ag+ exists in the solutionWhat volume of Ag+ is needed to reach the equivalence point ? Let Ve denote the volume of Ag+ is needed to reach the equivalence point

Since 1 mol of Ag+ reacts with 1 mol of I- , thus(0.02500l)(0.1000 mol I- /l) = Ve(0.050000 mol I- /l)

Ve = 0.05000 l = 50.00 ml

Since the equilibrium constant for the titration reaction is large, this means the equilibrium lies far to the right. What does this mean experimentally?

This implies: I- + Ag+ AgI(s) K = 1/ Ksp =1.2 x 1016

Before the equivalence point :

Consider the point when 10.00ml Ag+ has been added to the I- solution:- solution has more moles of I- than Ag+

- moles of I- =original moles of I- - moles of Ag+

added =(0.02500 l)(0.100 mol/l) – (0.01000 l)(0.05000 mol/l) =0.002000 mol I-

Since the volume of solution is now 35.00 ml,[I- ] = = 0.05714 M

[Ag+ ] = = 1.4 x 10-15 M

pAg+ = -log[Ag+ ] = 14.84

0.002000 mol I-0.03500 l

Ksp[I- ]

At the equivalence point :

All the AgI precipitates and some may redissolve to give equal concentrations of Ag+ and I- . Hence let [Ag+] = [I- ] = x

Ksp = [Ag+][I- ] = x2 = 8.3 x 10-17

x = 9.1 x 10-9

pAg+ = -log [Ag+]= -log(9.1 x 10-9) = 8.04After the equivalence point :

When V > Ve, the concentration of Ag+ is determined by the excess Ag+ added from the buretSuppose that V = 52.00ml

mol Ag+ =(0.00200l)(0.05000 mol Ag+/l) = 0.000100 mol

Ag+

Since the total volume = 77.00ml,

[Ag+] = (0.000100mol)(0.07700l)= 1.30 x 10-3 M

pAg+ = -log (1.30 x 10-3 ) = 2.89 - equivalence point is the steepest point of the curve- concentration of reactant affects the titration- for stoichiometries other than 1:1, the equivalence point is not at the centre of the steepest section of the curve- in practice, conditions are chosen such that the titration curves are steep

Ksp also affects the steepness of the titration curve

At the equivalence point, the titration curve is steepest for the least soluble precipitate

The larger the equilibrium constant for any titration reaction, the more pronounced will be the change in concentration near the equivalence point

Ag+

7-7 End Point Detection

Ag+

Ag+