Acidic Solutions.

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Acidic Solutions

INTRODUCTION:

The purpose of this assignment is to find the accurate concentration of a solution of sulphuric acid acid of which concentration is thought to be between 0.05 and 0.15moldm-3 . I am provided with solid anhydrous sodium carbonate and a range of indicators. I have to plan and carry out an experiment to determine the accurate concentration of the sulphuric acid in order to meet the requirement for the assessment of my experimental skills in unit 2852/02 courswork.

BASIC PLAN:

* I will make a standard solution of sodium carbonate and calculate its concentration.

* After making the sodium carbonate solution of known concentration, I will use it to determine the concentration of the sulphuric acid. This will be possible through an acid-alkali titration. Sodium hydroxide is a weak alkali and will neutralise the acid in the reaction to produce salt, water and carbon dioxide. The data obtained from this quantitaive analyses will then be used to calculate the concentration of the sulphuric acid.

MAKING A STANDARD SOLUTION OF SODIUM CARBONATE

APPARATUS:

* Safety spectacles

* Weighing bottle

* Spatula

* Solid anhydrous sodium carbonate

* A balance capable of weighing to within 0.01g

* A beaker, 250cm3

* Wash bottle of distilled water

* Stirring rod with rubber end

* Volumetric flas with stopper and labe, 250cm3

* Dropping pipette

* Mortar and pestle.

REACTING QUANITITES:

Na2CO3(s) + H2SO4(aq) Na2SO4 (s) + H2O(l) + CO2(g)

: 1

mole : 1mole

In the titration, I will be using a 25cm3 pipette therefore the volume of the Na2CO3 used will be 25cm3. From the chemical equation above, it can be seen that the stiochometry of the reaction is 1:1, therefore the volume of the H2SO4 will also be aimed at 25.00cm3 .

The concentration of the H2SO4 is thought to be between 0.05 and 0.15 moldm-3 . Assuming the concentration of the acid to be 0.05 moldm-3, the concentation of Na2CO3 can be calculated as follows:

Conc of H2SO4 (c) = 0.05 moldm-3

Volume of H2SO4 (v) = 25cm3 = 0.025dm3

Number of mole (n) = cv = 0.05 x 0.025 = 0.00125mols

0.025dm3 of Na2CO3 solution will contain 0.00125mols of H2SO4

concentration of Na2CO3 = n/v = 0.00125/0.025 = 0.05 moldm-3

n = mass/ relative molecular mass

number of moles (n) of Na2CO3 = 0.00125mols

Relative molecular mass of Na2CO3 = (23 x 2)+ 12 + (16 x 3) = 106g

therefore, mass of Na2CO3 = n x relative molecular mass

= 0.00125 x 106 = 0.1325g

If 25cm3 contain 0.1325g of Na2CO3,

250cm3 will contain 0.1325 x 250

25 = 1.325g ? 1.33g

Assuming the concentration of H2SO4 to be 0.15 moldm-3 , the concentation of Na2CO3 can be calculate as follows:

Conc of H2SO4 (c) = 0.15 moldm-3

Volume of H2SO4 (v) = 25cm3 = 0.025dm3

Number of mole (n) = cv = 0.15 x 0.025 = 0.00375mols

0.025dm3 of Na2CO3 solution will contain 0.00375mols of H2SO4

concentration of Na2CO3 = n/v = 0.00375/0.025 = 0.15 moldm-3

n = mass/ relative molecular mass

number of moles (n) of Na2CO3 = 0.00375mol

relative molecular mass of Na2CO3 = (23 x 2)+ 12 + (16 x 3) = 106g

therefore, mass of Na2CO3 = n x relative molecular mass

= 0.00375 x 106 = 0.3975g

if 25cm3 contain 0.3975g of Na2CO3,

250cm3 will contain 0.3975 x 250

25 = 3.975g ? 3.98g

From the calculations above, I must weigh accurately a sample of Na2CO3 beteeen 1.33 and 3.98g and use it to make a solution of Na2CO3 of concentration between 0.05 and 0.15moldm-3 .
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PROCEDURE:

. I will tranfer some quantity of solid anhydrous Na2CO3 into a mortar and grind it using a pestle till fine powder is obtained.

2. I will transfer between 1.33 and 3.98g of solid anhydrous Na2CO3 from the mortar into a weighing bottle and weigh it to the nearest 0.01g.

3. I will put about 50cm3 of distilled water into the 250cm3 beaker and then carefully transfer the bulk of Na2CO3 from the weighing bottle into the beaker of water.

4. I will reweigh the weighing bottle with any remaining Na2CO3 to the ...

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