INTRODUCTION TO TITRIMETRY

Most common types of titrations :

• acid-base titrations

• oxidation-reduction titrations

• complex formation

• precipitation reactions

In a titration, increments of titrant are added to the analyte until their reaction is complete.

From the quantity of titrant required, the quantity of analyte that was present can be calculated.

TITRATIONS IN PRACTICE

Accurately add of specific volume of sample solution to a conical flask using a pipette

Known: volume of sample

Unknown: concentration of analyte in sample

1

Slowly add standard solution from a burette to the sample solution

Known: concentration of the titrant

2

Add until just enough titrant is added to react with all the analyte

The end point is signaled by some physical change or detected by an instrument

Known: volume of the titrant

Note the volume of titrant used

3

If we have:

HA + BOH BA + H2O

Then from the balanced equation we know:

1 mol HA reacts with 1 mol BOH

analyte titrant

We also know:

CBOH, VBOH and VHA and2

22

1

11

n

vc

n

vc

1

vc

1

vc BOHBOHHAHA HA

BOHBOHHA v

vcc

STANDARD SOLUTIONS

Standard solution:

Reagent of known concentration

Primary standard:

highly purified compound that serves as a reference material in a titration.

Determine concentration by dissolving an accurately weighed amount in a suitable solvent of known volume.

Secondary standard:

compound that does not have a high purity

Determine concentration by standardisation.

Titrate standard using another standard.

Standard solutions should:

• Be stable

• React rapidly with the analyte

• React completely with the analyte

• React selectively with the analyte

EQUIVALENCE

POINTThe amount of added titrant is the exact amount necessary for stoichiometric reaction with the analyte in the sample.

An estimate of the equivalence point that is observed by some physical change associated with conditions of the equivalence point.

Aim to get the difference between the equivalence point and the end point as small as possible.

Titration error: Et = Veq – Vep

Estimated with a blank titration

END POINTVS

Indicators

used to observe the end point (at/near the equivalence point)

Thymol blue indicator

Instruments can also be used to detect end points.

Respond to certain properties of the solution that change in a characteristic way.

E.g.: voltmeters, ammeters, ohmmeters, colorimeters, temperature recorders, refractometers etc.

Add excess titrant and then determine the excess amount unreacted by back titration with a second titrant.

BACK TITRATION

Used when:

• end point of back titration is clearer than end point of direct titration

• an excess of the first titrant is required to complete reaction with the analyte

If we have:

HA + BOH BA + H2O

Then from the balanced equation we know:1 mol HA reacts with 1 mol BOH

analyte titrant

If I add excess titrant and then react the excess with a second titrant as follows:

HX + BOH BA + H2Otitrant 2 excess

Then from the balanced equation we know:

1 mol HX reacts with 1 mol BOH

We also know:

CBOH, CHX and VHX and 2

22

1

11

n

vc

n

vc

1

vc

1

vc HXHXBOHBOH (excess)

BOH

HXHXBOH c

vcv (excess)

We also know vBOH(total)

vBOH(reacted) = vBOH(total) – vBOH(excess)

1

vc

1

vc (reacted)BOHBOHHAHA

HA

BOHBOHHA v

vcc

(reacted)

From our initial titration:

HA + BOH BA + H2O analyte titrant

we then know:

CBOH, VBOH(reacted) and VHA

and we want to find CHA!

In summary:

HA + BOH BA + H2Oanalyte titrant reacted

VHA CBOH

HX + BOH BA + H2Otitrant 2 titrant excess

VHX

CHX CBOH

BOH

HXHXBOH c

vcv (excess)

vBOH(reacted) = vBOH(total) – vBOH(excess)

2

22

1

11

n

vc

n

vc

CHA?

HA

BOHBOHHA v

vcc

(reacted)

Example:

50.00 ml of HCl was titrated with 0.01963M Ba(OH)2. The end point was reached (using bromocresol green as indicator) after 29.71 ml Ba(OH)2 was added.

What is the concentration of the HCl? 2HCl + Ba(OH)2 BaCl2 + 2H2O

50.00 ml 29.71 ml

0.01963M

2

22

1

11

n

vc

n

vc

(1)

L)mM)(29.71(0.01963

(2)

Lm0050c1 ).(

C1 = 0.02333 M = [HCl]

Example:

A 0.8040 g sample of iron ore is dissolve in acid. The iron is reduced to Fe2+ and titrated with 0.02242 M KMnO4. 47.22 ml of titrant was added to reach the end point. Calculate the % Fe in the sample.

MnO4- + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O

0.02242 M

47.22 ml

n = cv

n = (0.02242 M)(0.04722 L)

n = 1.059x10-3 mol

BUT

nFe2+ =

MnO4- + 5Fe2+ …..

0.02242 M

47.22 ml

1.059x10-3 mol

5.293x10-3 mol

Mfe = 55.847 g/mol

m = (5.293x10-3 mol)(55.847 g/mol)

m = 0.2956 g Fe in sample

100sampleofmassanalyteofmass

%Fe

Example:

The CO in a 20.3 L sample of gas was converted to CO2 by passing the gas over iodine pentoxide heated to 150oC:

I2O5(s) + 5CO(g) 5CO2(g) + I2(g)

The iodine distilled at this temperature was collected in an absorber containing 8.25 mL of 0.01101 M Na2S2O3:

I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6

2-(aq)

The excess Na2S2O3 was back titrated with 2.16 mL of 0.00947 M I2 solution.

Calculate the mg CO per liter of sample.

I2O5(s) + 5CO(g) 5CO2(g) + I2(g)

I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6

2-(aq)

8.25 mL

0.01101 Madded

I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6

2-(aq)

2.16 mL

0.00947 M

n = 2.05x10-5 mol

nadded = 9.08x10-5 mol

nreacted = nadded - nexcess = 4.99x10-5 mol

nexcess = 2(2.05x10-5) mol

nexcess = 4.09x10-5 mol

AD

DE

DE

XC

ES

S

REACTED

I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6

2-(aq)

nreacted = 4.99x10-5 mol

2nI2 produced = nreacted

nI2 = 2.50x10-5 mol

I2O5(s) + 5CO(g) 5CO2(g) + I2(g)

I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6

2-(aq)

I2O5(s) + 5CO(g) 5CO2(g) + I2(g)

nCO = 5nI2

nCO = 1.25x10-4 mol

nI2 produced = 2.50x10-5 mol

nCO = 1.25x10-4 mol

Calculate the mg CO per liter of sample.

MCO = 28.01g/mol

mCO = 3.49x10-3 g

Vsample = 20.3 L

3.49 mg / 20.3 L = 0.172 mg/L CO in the sample