Method
- Collect equipment
- Make sure there are no safety hazards in their way.
- Pour solution 1(HCI- Hydrochloric acid) into the burette up to 0 ml on the visual measurement reading on the burette.
- Add 25 ml of sodium hydroxide into a flask.
- Add 10 drops of Universal indicator solution into the sodium hydroxide using a dropper.
- Open the tap and run the acid, from the burette drop by drop.
- Wait until the conical flask has turn Green (neutral, The end point, which happens when the titrant and the amount of stuff being titrated is equal), while dropping the solution.
- Note the reading down and then repeat the process with each solution.
Measurements
The volume of acid is measured in ml. We will take measurements of the compound of the ending product in the conical flask. In addition, we will measure how much solution is left in the burette.
To help with accuracy we will do two tests for each solution so that we can get a more accurate result.
Research
All this research are secondary sources and were found in books, Internet, etc.
To analyse, we mean in chemistry, to separate something into its component parts in order to learn more about the nature of these components.
Sometimes we need to know only which substances are present in a solution. E.G., which elements are present in a sample?
To find the answers to these questions we use qualitative analysis. If we need to now the precise quantity of one or more of its components we use quantitative analysis. This might entail finding the percentage of nickel ore or the number of parts per million of mercury in a fish. Qualitative analysis can be done by means of volumetric analysis, in which reactions take place inside the solution.
Volumetric analysis is a means of finding out the concentration of a solution. He method is to add a base in a careful way until there is enough acid to neutralise the base. This method is called titration. In a titration, method, the titrant, is added to another slowly. As it is added, a chemical reaction occurs until one of the solutions in the compound is exhausted. This experiment is an acid based titration.
The concentration of one of the two substances must be known so that you can calculate the other concentration.
All acid based titration reactions are simple exchanges of protons.
How to work out the concentration of any solution?
E.G.
A solution of hydrochloric acid it titrated against a standard sodium hydroxide solution. What is the concentration of hydrochloric acid?
Solution
1. Write out the equation.
Hydrochloric acid + sodium hydroxide → sodium chloride + water
HCI (aq) + NaOH (aq) → NaCL (aq) + H2O (I)
This equation tells you that 1 mole of HCI neutralises 1 mole of NaOH.
2. Work out the amount of moles of the base.
Amount (mol) = Volume (I) x concentration (mol/l)
Amount (mol) of NaOH = volume (25.0cm3) x concentration (0.100 mol/l)
= 25 x 10-3 x 0.100 mol/l = 2.50 x 10-3 mol
3. Now work out the concentration of acid
Amount (mol) of HCI = Amount (mol) of NaOH = 2.5 10-3 mol
ALSO
Amount (mol) of HCI = volume of HCI (aq) x concentration of HCI (aq)
Therefor, if C = concentration of HCI
2.50 x 10-3 mol = 15.0 x 10-3 x c
C = 2.5 x 10 –3 mol
15.0 x 10-3
= 0.167 mol/l
Calculations for the real results.
Solution 1 – NaOH & HCI titration
- Equation for the reaction
Hydrochloric acid + Sodium hydroxide → sodium chloride + water
HCI (aq) + NaOH (aq) → NaCL (aq) + h20 (1)
From the above equation, we can say 1 mole of HCI neutralises 1 mole of NaOH.
- Amount in moles of NaOH
Amount (mol) = volume (I) x concentration (mol/l)
Amount of (mol) of NaOH = 25.0 cm3 x 1.0 mol/l
= 25/1000 1 x 1.0 mol/l
= 0.025 mol
- Concentration of HCI
Amount (mol) of HCI = Amount (mol) of NaOH
= 0.025 mol
Also
Amount (mol) of HCI = Volume (mol) of HCI (aq) x concentration of HCI (aq)
= 12.5/ 1000 I x C
Therefor
12.5/1000 I x c = o.o25 mol
C= 0.025/12.5 x 1000 mol/l
= 2.0 mol/l
=2.0 m
Solution 2 – NaOH & H2SO4 titration
- Equation for the reaction
Sulphuric acid + Sodium hydroxide → sodium chloride + water
H2S04 (aq) + 2NaOH (aq) → NaCL (aq) + h20 (1)
From the above equation, we can say 1 mole of H2SO4 neutralises 2 moles of NaOH.
- Amount of moles in NaOH
Amount in (mol) = volume (1) x concentration (mol/l)
Amount (mol) of NaOH = 25.0 cm3 x 1.0 mol/l
= 25/1000 x 1.0 mol/l
=0.025
-
Concentration of H2SO4
Amount (mol) of NaOH = 2 x amount (mol) of H2SO4
Amount (mol) of NaOH = 25.03 x 1.0 mol/l
= 25/1000 x1.0 mol/l
= 0.025 mol
Also
2 x Amount (mol) of H2SO4 = 2 x Volume of H2SO4 (aq) x Concentration of H2SO4
= 2 x 12.4/1000 x C
Therefor
2 x 12.4/ 1000 x c = 0.025 mol
C = 0.025 / 24.8 x 1000 mol/l
= 1.01 mol/l
= 1.0 m
Solution 3 – NaOH & HNO3
- Equation of the reaction
Nitric acid + sodium hydroxide → sodium nitrate + water
HNO3 (aq) → NaNO3 (aq) + H2O
From this equation you can tell that 1 mole of HNO3 neutralises 1 mole of NaOH.
- Amount in moles of NaOH
Amount in (mol) = volume (1) x concentration (mol/l)
Amount of NaOH = 25.0 cm3 x 1.0 (mol/l)
=25/1000 x 1.0 mol/l
=25/1000 x 1.0 mol/l
=0.025 mol
-
Concentration of HNO3
Amount (mol) of HNO3 = Amount (mol) of NaOH
= 0.025 mol
Therefor
25.0/1000 x C = 0.025 mol
C = 0.025/25 mol x 1000
C= 1.00 mol/l
C = 1 mol
Conclusion
I have found that: -
Hydrochloric acid- 2.00mol/l (2 m)
Sulphuric acid- 1.00 mol/l ( 1m)
Nitric acid- 1.00 mol/l (1m)
Evaluation
Due too accurate and the amount of tests, I say that these results are quite accurate. The main concern about the experiment was spillage. If anything had spilt, then it would had changed the readings. In addition, I was concerned about the idiotic behaviour of some pupils that felt that they could add more drops of a solution to the sodium hydroxide.
To improve this experiment I would get a more controlled environment and also tried other acids and bases.
Research 2
Sulphuric acid
Sulphuric acid is a corrosive, oily, colourless liquid. It melts at 10.36ºC. It boils at 340ºC. It is soluble and when mixed with water, considerable amount of heat is released.
Nitric acid
Nitric acid is a colourless corrosive liquid that melts at –42ºC and boils at 83ºC.