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Doc Scientia - IEB Senior Chemistry Textbook and Workbook Book 2

8. Consider the reactants and predict the product(s) that will form.Mg + H2SO4 →

A. MgHSO4 B. MgSO4 + H2 C. MgH2 + MgSO4 D. MgH2 + SO4

9 Calculate the hydroxide ion concentration of each of the following solutions:9.1 [HNO3] = 2,51 × 10-3 mol⋅dm-3 9.2 [H2SO4] = 0,5 × 10-6 mol⋅dm-3

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13 Titrations13.1 What is a titration?

When an acid is gradually added to a base, the concentration of the H3O+ ions in the solution increases

gradually. The pH of the solution decreases gradually from a high value (therefore basic) to a low value (therefore acidic). The more acid is added to the base, the lower the pH becomes. When the addition of the acid is stopped, the moment when the acid neutralises the base, the equivalence point of the titration is reached. The inverse is also true. If a base is added gradually to an acid, the concentration of the H3O

+ ions in the solution, gradually decrease and the pH of the solution gradually increases.

A titration is used to experimentally determine the unknown concentration of an acid or a base.

TITRATION

A titration is therefore the process during which an acid is added to a base until they neutralise one another completely, and the equivalence point is reached. A suitable indicator can be used to indicate the end point of the titration by changing colour. Care must be taken to ensure that the end point is as close as possible to the equivalence point.

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A titration where a weak acid and a weak base are used produces very unreliable results. The rate at which the two weak species react is so slow that the indicator’s colour change occurs too slowly. The end point cannot be indicated accurately.

When the results of a titration are used to determine the concentration of a solution, we refer to the process as volumetric analysis.

Titration curvesConsider the pH of the titration mixture before, during, and after the neutralisation.Firstly look at a titration of a strong acid with a strong base. Diagram A shows the graph for the titration of a 25,0 mℓ sample of 0,1 mol⋅dm-3 hydrochloric acid with 0,1 mol⋅dm-3 sodium hydroxide. The graph is called a titration curve. The pH increases slowly at first, increases rapidly in the middle portion of the curve, and then increases slowly again. The midpoint of the vertical part of the curve is the equivalence point for the titration as indicated. It indicates when equivalent quantities of acid and base are present. For the titration of a strong acid with a strong base, the equivalence point occurs at a pH of 7.

The titration of a weak acid with a strong base (or of a weak base with a strong acid) is more complicated, but it follows the same general principles. Study the titration of 25,0 mℓ of 0,1 mol⋅dm-3 acetic acid (a weak acid) with 0,1 mol⋅dm-3 sodium hydroxide and compare the titration curve with that of the strong acid. Diagram B shows the titration curve.

Although the initial volume and concentration of the acids are the same, there are important differences between the two titration curves. The titration curve for the weak acid begins at a higher value (less acidic) and maintains higher pH values up to the equivalence point. This is because acetic acid is a weak acid, which is only partially ionised.

Titration of strong acid

equivalence point pH, 7,00pH

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12

11

10

9

8

7

6

5

4

3

2

10

5045403530252015105

volume of 0,100 M NaOH added (mℓ)(A)

pH13

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9

8

7

6

5

4

3

2

10

5045403530252015105

volume of 0,100 M NaOH added (mℓ)(B)

Titration of weak acid

equivalence point pH, 8,72

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The pH at the equivalence point is also higher (8,72 rather than 7,00) due to the hydrolysis of acetate, that raises the pH.The higher pH of the equivalence point indicates that the acetate salt is basic.

After the equivalence point, the two curves are identical because the pH is dependent on the excess of hydroxide ion in both cases.

As indicated on the diagram, there are three ranges where the equivalent point may be situated. If the equivalence point is at a pH between:• 3–4,5, it is an acidic salt and methyl orange should be used as indicator for that titration;• 5–8, it is an neutral salt and bromothymol blue should be used as indicator for that titration;• 8,5–10, it is a basic salt and phenolphthalein should be used as indicator for that titration.

13.2 How is a titration conducted?

• Rinse out a burette with a little of the acid solution.• Fill the burette carefully with some of the acid – use a funnel. Take the initial reading on the burette.• Rinse a conical (Erlenmeyer) flask with distilled water.• Rinse the pipette with the base solution.• A measured quantity of the base solution is transferred with a pipette to the flask.• Add approximately two drops of suitable indicator to the base in the flask.• Place a white tile or paper under the flask so the colour change of the indicator will be clearly observable.• During the addition of the acid, the flask must be continually and carefully swirled to ensure that the solutions mix well. • Take the initial reading on the burette.

One of the two reacting solutions must have a known concentration.

QUICK FACTS

pH

121314

11

10987654321

05045403530252015105

volume of 0,100 M NaOH added (mℓ)

equivalence point pH, 8,72

equivalence point pH, 7,00

titration of strong acid HCℓ

phenolphthalein pH range

bromothymol blue pH range

methyl orange pH range

titration of weak acid CH3COOH

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• The first run is a test run, and the acid is added to the solution in the Erlenmeyer flask quite quickly. Stop the addition of the acid as soon as the solution changes colour. The result of this run is inaccurate and should not be used.• Take note of the volume of acid that has been added to the solution. • Repeat the process at least three times, accurately, so that an average value can be determined for the added volume.• Add acid quite quickly to the base in the Erlenmeyer flask to approximately 2 cm3 from the end point, as determined in the test run. Then slowly add more of the acid. Near the end point the acid must be added drop by drop.• Add the acidic solution in the burette to the base, drop by drop, until the last drop causes the colour to change suddenly. This is the end point of the titration.• Read off the volume of acid that was added to the base to reach the end point from the burette. • Use this reading and subtract the initial reading on the burette.• Use the volume of the acid and the volume of the base together with the known concentration to calculate the unknown concentration.

Usually, the acid is added to the base and the burette is therefore filled with the acid. Although it is possible to add the base to the acid and fill the burette with the base, water soluble bases are ionic solutions and when the water evaporates from the solution, the formation of ionic salt crystals can make the burette’s tap get stuck, and block the thin supply pipe.

13.3 Calculations

The reaction between the strong acid HCℓ and the strong base NaOH occurs as follows: HCℓ(aq) + NaOH(aq) → NaCℓ(aq) + H2O(ℓ)

If the ions in solution are considered, the reaction occurs as follows: H+(aq) + Cℓ-(aq) + Na+(aq) + OH-(aq) → Na+(aq) + Cℓ-(aq) + H2O(ℓ)

The net neutralisation reaction is therefore:H+(aq) + OH-(aq) → H2O(ℓ)

We must be able to determine the mol ratio in which the acid and base react during the reaction, from the balanced reaction equation.

Remember:

The indicator itself

is an acid or base in

solution. Large

quantities can

influence the end

point of the

titration.

Examplehydrochloric acid + sodium hydroxide → sodium chloride + water 1HCℓ(aq) + NaOH(aq) → NaCℓ(aq) + H2O(ℓ)∴ 1 mol of acid reacts with 1 mol of alkali ∴ the mol ratio is 1:1.

sulfuric acid + sodium hydroxide → sodium sulfate + water

1H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + H2O(ℓ)∴ 1 mol of acid reacts with 2 mol of alkali ∴ the mol ratio is 1:2.

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The following formula can be used to calculate the unknown concentration:na caVa nb cbVb

nb = cbVb

or na

= caVa

where na and nb are the coefficients of the acid and base in the balanced equation (not the actual mol).c: concentration (mol⋅dm-3)V: volume (dm3)

ExampleDuring a titration, 16,5 cm3 of dilute H2SO4 neutralises 25 cm3 of a NaOH solution exactly. If the concentration of the H2SO4 solution is 0,4 mol⋅dm-3, calculate the concentration of the NaOH solution.• Write down the balanced reaction equation:

1H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + H2O(ℓ) mol ratio: 1 : 2• Write down the given information.

Acid: H2SO4 Unit Base: NaOH

na = 1 nb = 2

ca = 0,4 mol⋅dm-3 cb = ?

Va = 16,5 cm3 Vb = 25

• Do the calculation: na caVa

nb = cbVb

1 (0,4)(16,5) 2

= cb(25)

1⋅cb(25) = (0,4)(16,5)(2) cb = 0,528 mol⋅dm-3

Experiment 20 Date: ........................................

Aim: To prepare a standard oxalic acid solution with a concentration of 0,1 mol⋅dm-3.

Theory:

The formula for hydrated oxalic acid crystals is C2H2O4⋅2H2O or (COOH)2⋅2H2O.The 2H2O that is part of the formula is water of crystallisation. The mass thereof must be considered when weighing the crystals.