MODULE 4

7.13 Analytical Chemistry

7.13.1 Volumetric Analysis

Volumetric analysis relies on methods involving accurate measurement of volumes of liquids (although one or more weighings may be necessary). Volumetric analysis can be carried out in a relatively short time and an experienced worker can be accurate to about 0.2 - 1%.

Titration of iron (II) with acidified manganate (VII).

Potassium manganate (VII) (permanganate), KMnO₄, is a powerful oxidising agent and can be used to estimate reducing agents such as iron(II) salts, ethandioates (oxalates) and hydrogen peroxide solution in redox titration’s.

In acid conditions (produced by adding dilute sulphuric acid) potassium manganate (VII) is reduced.

MnO₄^- + 8H⁺ + 5e^- Mn²⁺ + 4H₂O

Iron(II) solutions can be oxidised.

Fe²⁺ (aq) Fe³⁺ (aq) + e

Combining gives

MnO₄^- + 8H⁺ + 5e^- Mn²⁺ + 4H₂O

5Fe²⁺ (aq) 5Fe³⁺ (aq) + 5e^-

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

Therefore 1 mole of MnO₄^- ions (from KMnO₄) will oxidise 5 moles of Fe²⁺ ions

MnO₄^- Ξ 5Fe²⁺ (aq)

Generally about 0.02 M KMnO₄ is used acidified with a few cm³ Of 1M H₂SO₄.

Indicator and end-point

As the titration proceeds the purple colour of the permanganate is discharged giving in colourless solution. As soon as the permanganate is in excess, the solution turns pink so the potassium permanganate acts as its own indicator. The end-point is the first permanent pink coloration.

Estimation of iron in ammonium iron(II) sulphate Fe(NH₄)₂(SO₄)₂.6H₂O

· Weigh out accurately about 9.8 g of ammonium iron(II) sulphate crystals.

· Dissolve in dilute sulphuric acid which has been boiled to remove air and then cooled (this prevents oxidation of Fe²⁺ to Fe³⁺).

· Make up to 250 cm³ in a volumetric flask with the dilute sulphuric acid.

· Transfer 25 cm³ of the iron(II) solution into a conical flask by pipette.

· Add about 15 cm³ of 1M sulphuric acid.

· Titrate with standard (0.02M) potassium permanganate solution from a burette until the first permanent pink colour is observed.

Example

9.85 g of ammonium iron(II) sulphate crystals were made up to 250 cm³ of solution in cold, boiled out, dilute sulphuric acid. 25.0 cm³ of the solution reacted completely with 23.40 cm³ of 0.02M potassium mangante(VII) solution.

Calculate the concentration of the alum solution and hence the percentage purity of the alum.

[Mr (Fe(NH₄)₂(SO₄)₂.6H₂O) = 392]

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

Amount MnO₄^- = cV = 0.02 x 23.40/1000 = 4.68 x 10^-4 mol

From equation 1 mol MnO₄^- Ξ 5 mol Fe²⁺

Therefore amount Fe²⁺ (in 25 cm³ soln.) = 5 x amount MnO₄^-

= 5 x 4.68 x 10^-4 = 2.34 x 10^-3 mol

Conc. of alum solution = n/V = 2.34 x 10^-3 / 0.0250 = 9.36 x 10^-2 mol dm^-3

Mass of alum crystals in 250 cm³ soln. = c x V x M_r

= 9.36 x 10^-2 mol dm^-3 x 0.250 x 392

= 9.17 g

% purity of crystals = 9.17 x 100 / 9.85 = 93.1%

Simple examples of back titration including determination of the purity of a Group II carbonate and of the solubility of an ammonium salt.

Some substances such CaCO₃ are insoluble in water and would be difficult to titrate directly. These difficulties can be overcome by dissolving a known amount of the substance in an excess of acid, for example, and determining the amount of excess by back titration with alkali. The concentration of the alkali should be less than that of the acid used, as the amount of acid remaining will be considerably less than the amount used up.

Procedure

· Weigh out accurately a suitable amount of the substance.

· Transfer to an excess of standard acid in a beaker.

· Wash any remaining traces of the substance into the beaker.

· When effervescence ceases transfer to a volumetric flask, including washings, and make up to the mark with distilled water.

· Shake well.

· Withdraw 25 cm³ portions of the solution and titrate with standard alkali, using methyl orange or methyl red as indicator.

Example

1.500 g of a sample of limestone was dissolved in 50 cm³ of 1M hydrochloric acid. The resulting solution was mode up to 250 cm³ with distilled water. 25.0 cm³ of this solution required 21.05 cm³ of 0.1M sodium hydroxide for neutralisation. Assuming all the basic material in the rock to be calcium carbonate, calculate the percentage of CaC0₃ present.

[M_r(CaCO₃) = 100]

HCl + NaOH NaCl + H2O

Soln 1 HCl M₁ = ? V₁ = 25.00 cm³ n₁ = 1

Soln 2 NaOH M₂ = 0.1 V₂ = 21.05 cm³ n₂ = 1

M₁V₁ = n₁

M₂V₂ n₂

25.00 x M₁ = 1

0.1 x 21.05 1

M₁ = 0.1 x 21.05/ 25.00 = 8.42 x 10^-2 mol dm^-3

Amount HCl (in 250 cm³ soln) = c x V = 8.42 x 10^-2 x 0.250 = 2.110 x 10^-2 mol

Original amount HCl = c x V = 1 x 0.0500 = 5.000 x 10^-2 mol

Therefore amount HCl reacted = (5.000 – 2.110) x 10^-2 = 2.890 x 10^-2 mol

CaCO₃ + 2HCl CaCl₂ + H₂O + CO₂

From equation 1 mol CaCO3 Ξ 2 mol HCl

Amount CaCO₃ reacted = ½ x amount HCl reacted = 2.890 x 10^-2 / 2 = 1.450 x 10^-2 mol

Mass CaCO₃ reacted = 1.450 x 10^-2 x 100 = 1.450 g

% CaCO₃ in limestone = 1.450 x 100 / 1.500 = 96.7 %
Exercise 1

1 A 1.00 g sample of limestone is allowed to react with 100 cm³ of 0.200 mol dm^-3 hydrochloric acid. The excess acid required 24.8 cm³ of 0.100 mol dm^-3 sodium hydroxide solution. Calculate the percentage of calcium carbonate in the limestone.

[ 87.6% ]

2 An impure sample of barium hydroxide of mass 1.6524 g was allowed to react with 100 cm³ of hydrochloric acid of concentration 0.200 mol dm^-3. When the excess of acid was titrated against sodium hydroxide 10.9 cm³ of sodium hydroxide were required. 25.0 cm³ of the sodium hydroxide required 28.5 cm³ of the hydrochloric acid in a separate titration. Calculate the percentage purity of the sample of barium hydroxide.

[ 90.6% ]

3 Calculate the relative molecular mass of the carbonate of a divalent metal X, and hence the relative atomic mass of X from the data: 1.000 g of the anhydrous normal carbonate of X was added to 50 cm³ of 1M HCl. The excess acid required 30.0 cm³ of 1M NaOH for neutralisation.

[ 100 ; 40 ]

4 1.600 g of a metallic oxide of type MO were dissolved in 100 cm³ of 1M hydrochloric acid. The resulting liquid was made up to 500 cm³ with distilled water. 25.0 cm³ of the solution then required 21.02 cm³ of 0.1020M sodium hydroxide for neutralisation. Calculate the mass of the oxide reacting with 1 mole of hydrochloric acid and hence the molar mass of the oxide and the relative atomic mass of the metal.

[ 28.01 g ; M_r = 56.02 ; A_r = 40.02 ]

5 0.800 g of a metallic oxide of type MO were dissolved in 50 cm³ of 1M hydrochloric acid. The resulting liquid was made up to 250 cm³ with distilled water and 25.0 cm³ of this solution required 21.4 cm³ of 0.1M NaOH for neutralisation.

Calculate the mass of the oxide reacting with 1 mole of HCl and hence the relative molecular mass of the oxide and the relative atomic mass of M.

[ 28.01 g ; M_r = 56 ; A_r = 40 ]

Estimation of Ammonium Salts

The ammonium salt (as solid or solution) is boiled with excess sodium hydroxide.

NH₄⁺ (aq) + OH^- (aq) NH₃ (g) + H₂O (l)

The excess of sodium hydroxide is determined by back-titration with acid.

Procedure

· Place the sample of ammonium salt in a conical flask.

· Run a fixed volume of standard alkali from a burette into the conical flask.

· Insert a rubber stopper fitted with a reflux condenser into the flask to prevent loss of solution (alternatively a glass funnel con be used, with care).

· Boil the flask contents for about 15 minutes to expel all the ammonia gas.

· Wash the condenser (or glass funnel) with a little water to wash any alkali solution back into the flask.

· Pour the contents of the flask, when cold, into a volumetric flask, adding several washings, and make up to the mark. Shake well.

· Titrate portions of the solution against standard hydrochloric acid using methyl orange as indicator.

Example

25 cm³ of a solution of ammonium chloride was boiled with 100 cm³ of 2M NaOH until no more ammonia was evolved. The residual solution was made up to 250 cm³ with water and 25 cm³ of this required 22.40 cm³ of 0.5M HCl for neutralisation.

Calculate the concentration of the original ammonium chloride solution in mol dm^-3 and hence the solubility of ammonium chloride in g dm^-3.

[ c = 3.52 mol dm^-3 ; solubility = 188 g dm^-3 ]

Exercise2

1 0.500 g of impure ammonium chloride is warmed with an excess of sodium hydroxide solution. the ammonia liberated is absorbed in 25.0 cm³ of 0.200 mol dm^-3 sulphuric acid. The excess of sulphuric acid requires 5.64 cm³ of 0.200 mol dm^-3 sodium hydroxide solution for titration.

Calculate the percentage of ammonium chloride in the original sample.

[ 95.0% ]

2 5.000g of ammonium chloride contaminated with sodium chloride were boiled with 100 cm³ of 2M NaOH until no more ammonia was evolved. The residual solution was made up to 250 cm³ with water and 25 cm³ of this required 22.4 cm³ of 0.5M HCl for neutralisation.

What was the mass of sodium chloride in the ammonium chloride ?

[ 0.292 g ]

3 Calculate the percentage of ammonium sulphate in a sample of this compound contaminated with sodium sulphate from the following data: 1.65 g of the ammonium sulphate was made up to 250 cm³ of aqueous solution. 25 cm³ of this were boiled with 50 cm³ O. 1M NaOH and the excess alkali neutralised by 25.4 cm³ of 0. 1 M HCl.

[ 98.37% ]

Use of double indicator method to determine the composition of a mixture such as a sodium hydrogencarbonate/sodium carbonate mixture.

Estimation of NaOH / Na₂CO₃ mixtures

The reaction between NaOH and HCl occurs in a single step.

NaOH + HCl NaCl + H₂O eqn 1

The reaction between sodium carbonate and hydrochloric acid occurs in two steps;

Na₂CO₃ + HCl NaHCO₃ + NaCl eqn2

NaHCO₃ + HCl NaCl + H₂O + CO₂ eqn3

If phenolphthalein is used as the indicator, the end-point (pink colourless) occurs when reactions 1 and 2 are complete.

If a separate sample is used with methyl orange as indicator then the end-point (yellow pink) occurs when reactions 1, 2 and 3 are complete.

The amount of acid required for reactions 2 and 3 are the same.

The difference between the titrations with methyl orange and phenolphthalein is the volume of acid required for reaction 3.

Therefore

Volume of acid to titrate Na₂CO₃ = 2 x difference between methyl orange

and phenolphthalein titres.

Volume of acid to titrate NaOH = methyl orange - volume of acid to

titre titrate Na₂CO₃

Example

25.0 cm³ of a solution containing NaOH and Na₂CO₃ required 21.50 cm³ of 1M HCl with phenolphthalein as indicator. A separate 25.0 cm³ of the same solution required 27.85 cm³ of the same acid with methyl orange as indicator. Calculate the concentration of each compound in

g dm^-3 of the solution.

OH^- + H⁺ H₂O 1

CO₃^2- + H⁺ HCO₃^- 2 21.50 cm³ 27.85 cm³

HCO₃^- + H⁺ H₂O + CO₂ 3

Difference in titres = volume of acid for reaction 3 = 27.85 – 21.50 = 6.35 cm³

Volume of acid for CO₃^2- titration = 2 x 6.35 = 12.70 cm³

Volume of acid for OH^– titration = 27.85 – 12.70 = 15.15 cm³

Concentration of CO₃^2-

CO₃^2- + 2H⁺ H₂O + CO₂

Soln. 1 OH^-/CO₃^2- M₁ = ? V₁ = 25.0 cm³ n₁ = 1

Soln. 2 HCl M₂ = 1 V₂ = 12.70 cm³ n₂ = 2

M₁V₁ = n₁

M₂V₂ n₂

M₁x 25.0 = 1

1 x 12.70 2

M₁ = 12.70/ 2 x 25.0 = 2.54 x 10^-1 mol dm^-3

M_r(Na₂CO₃) = 106

Conc. of Na₂CO₃ = M₁ x M_r(Na₂CO₃) = 2.54 x 10^-1 x 106 = 26.9 g dm^-3

Conc. of OH^-

OH^- + H⁺ H₂O

Soln. 1 OH^-/CO₃^2- M₁ = ? V₁ = 25.0 cm³ n₁ = 1

Soln. 2 HCl M₂ = 1 V₂ = 15.15 cm³ n₂ = 1

M₁V₁ = n₁

M₂V₂ n₂

M₁x 25.0 = 1

1 x 15.15 1

M₁ = 15.15 / 25.0 = 6.06 x 10^-1 mol dm^-3

M_r(NaOH) = 40

Conc. of NaOH = M₁ x M_r(NaOH) = 6.06 x 10^-1 x 40 = 24.2 g dm^-3

OR

CO₃^2- + 2H⁺ H₂O + CO₂

2000 cm³ 1M HCl reacts with 106 g of Na₂CO₃

12.70 cm³ 1M HCl react with 106 x 12.70 / 2000 g Na₂CO₃ = 6.73 x 10^-1 g

This is the mass of Na₂CO₃ in 25.0 cm³ solution

Mass of Na₂CO₃ in 1 dm³ of solution = 6.73 x 10^-1 x 1000 / 25.0 = 26.9 g

Concentration of Na₂CO₃ = 26.9 g dm^-3

OH^- + H⁺ H₂O

1000 cm³ 1M HCl reacts with 40 g of NaOH

15.15 cm³ of 1M HCl reacts with 40 x 15.15 / 1000 g NaOH = 6.06 x 10^-1 g

This is the mass of NaOH in 25.0 cm³ solution

Mass of NaOH in 1 dm³ of solution = 6.06 x 10^-1 x 1000 / 25.0 = 24.2 g

Concentration of NaOH = 24.2 g dm^-3

Exercise 3

1 20.0 cm³ of a solution containing NaOH and Na₂CO₃ required 19.20 cm³ of 0.5M HCl with phenolphthalein as indicator. With methyl orange, a further 5.10 cm³ of the acid were needed.

What is the concentration of each compound in the original solution, in g dm^-3 ?

[ 14.10 g ; 13.52 g ]

2 25 cm³ of a solution NaOH and Na₂CO₃ required 28.0 cm³ of 0.1 M HCl for neutralisation with methyl orange as indicator. The carbonate ions were removed from 25.0 cm³ of the solution by the addition of excess barium chloride solution. The resulting mixture was titrated slowly with the same acid, and with phenolphthalein as indicator 18.0 cm³ of the acid were needed. Calculate the mass of anhydrous NaOH and Na₂CO₃ per dm³ of solution.

[ 2.88 g ; 2.12 g ]

Estimation of NaHCO₃ / Na₂CO₃ mixtures

The same method can be applied to the estimation of sodium carbonate and sodium hydrogencarbonate together.

With phenolphthalein as indicator, the end-point is given when all of the carbonate is converted to hydrogencarbonate.

CO₃^2- + H⁺ HCO₃^- eqn. 1

With methyl orange as indicator the end-point appears when the original hydrogencarbonate and that produced in eqn. 1 are both converted to CO₂ and H₂O.

HCO₃^- + H⁺ H₂O + CO₂ eqn. 2

(from CO₃^2-)

HCO₃^- + H⁺ H₂O + CO₂ eqn. 3

(from original solution)

Since reactions 1 and 2 require the same amount of acid;

Volume of acid for CO₃^2- titration = 2 x phenolphthalein titre

Volume of acid for HCO₃^- titration = methyl orange titre - 2 x phenolphthalein titre

Example

Anhydrous Na₂CO₃ contaminated with NaHCO₃ was made up to 250 cm³ of solution. 25 cm³ of this solution required 11.2 cm³ of 1M HCl with phenolphthalein as indicator and 24.5 cm³ of the same acid with methyl orange as indicator.

Calculate the percentage by mass of NaHCO₃ in the mixture.

Volume of acid for CO₃^2- titration = 2 x phenolphthalein titre = 2 x 11.2 = 22.4 cm³

Volume of acid for HCO₃^- titration = methyl orange titre - 2 x phenolphthalein titre

= 24.5 – 22.4 = 2.1 cm³

CO₃^2- + 2H⁺ H₂O + CO₂

From equation 1 mol CO₃^2- Ξ 2 mol H⁺

Amount CO₃^2- in 25 cm³ soln. = ½ x amount H⁺ = ½ (c x V) = ½ (1 x 22.4 /1000) = 1.12 x 10^-2 mol

Mass Na₂CO₃ in 25 cm3 soln. = 1.12 x 10^-2 x Mr(Na₂CO₃) = 1.12 x 10^-2 x 106 = 1.187g

HCO₃^- + H⁺ H₂O + CO₂

From equation 1 mol HCO₃^- Ξ 1 mol H⁺

Amount HCO₃^- in 25 cm³ soln. = amount H⁺ = c x V = 1 x 2.1 / 1000 = 2.1 x 10^-3 mol

Mass NaHCO₃ in 25 cm³ soln. = 2.1 x 10^-3 x M_r(NaHCO₃) = 2.1 x 10^-3 x 84 = 1.76 x 10^-1 g

Total mass = 1.187 + 0.176 = 1.363 g

% by mass of NaHCO3 = 0.176 x 100/ 1.363 = 12.91%

Use of edta to determine (separately) magnesium and copper ions in solution. Names and colour changes of indicators.

The complexes formed by metal ions with bis[di(carboxymethyl)amino]ethane (HO₂CCH₂)₂NCH₂CH₂N(CH₂CO₂H)₂ are very stable, and can be used for the estimation of metal ions by titration.

In practice the disodium salt is used to increase the solubility of the reagent. The actual ligand is

* *

-OOCCH₂ * * CH₂COO-

N CH₂ CH₂ N

-OOCCH₂ CH₂COO-

* *

ethylene-diaminetetra acetate (edta)

This is a hexadentate ligand (* denotes the positions at which the ligand attaches itself to the central metal ion), and the complexes formed are known as chelates (Greek ‘chelos’ – a crab’ claw), because the polydentate ligand forms a claw-like grip on the central metal ion.

Mⁿ⁺ + edta^4- (4-n)^-

Erichrome Black T is used as indicator, which is itself a weak complexing agent.

The metal-indicator colour of red is seen at the start of the titration. As the edta is added, the metal ions are extracted from the indicator complex and complex with the edta.

metal-indicator complex + edta metal-edta complex + indicator

(RED) (BLUE)

At the end-point the colour changes from red to the blue of the free indicator.

The solution is buffered to give the optimum pH for the metal-edta complex formation.

Mg²⁺ pH 10

Cu²⁺ pH 3

Exercise 4

Calculate the concentration of a solution of magnesium sulphate from the following data. 25.0 cm³ of the solution when added to an alkaline buffer and Erichrome Black T indicator, required 22.3 cm³ of a 1.05 x 10^-2 mol dm^-3 solution of edta for titration.

Mg²⁺ (aq) + edta^4- (aq) [ Mg(edta)]^2- (aq)