Through the previous chapter we had concerned our experiments to the techniques of direct titration. That is, wherever a standard solution directly reacts by the substance being determined. In this chapter, we shall give only one experiment, to reveal another of the techniques that is popularly referred to as Back - titration. This is gained whenever in an acid - base reaction, the amount of base added to a certain amount of the acid is in excess of the stoichmetric amount of the acid needed leaving the base in excess and the resulting solution basic.(If it is the acid that is excess, the remaining solution will be acidic).
The amount of base in excess can now be determined via titrating the excess basic solution against another standard acid solution different from the initial acid that reacted through the base.
Acid - base titrations. Determination of the concentration of a substance that is in excess after a chemical reaction (back titration).
Sample A is a solid anhydrous
Sample B is a standardisation
solution of HCl containing 0.303 1 litre mole/dm3
Sample C is a standardized solution of NaOH containing 0.131 mole/ dm3
As listed in experiment 2
1 litre standard flask
1. 1000cm3 of solution B was put into a one litre (dm3) standard flask.
2. Pour into the solution, all of the Na2CO3 sample.
3. Mix methodically. Warm gently if necessary. We might notice slight bubble. Allow to cool if warming was executed.
4. Rinse a burette twice with a few cm3 of solution C and next fill it through similar solution above the zero mark and drain to the mark making sure the burette tip is full.
5. Using a pipette, transfer 25cm3 of the solution B into a 250cm3 conical flask. Add 2 to 3 drops. Situate the conical flask onto a white sheet of paper or tile.
6. Run solution C into solution B in the conical flask whilst swirling the flask to ensure adequate mixing of the content until a permanent pink coloration is examined.
7. Recur the titration using fresh portions of solution B until at least two titration values are within the accuracy of 0.2cm3 of the titrant.
8. Find the average of our titre values.
From our average titre value, determine
a) The concentration in moles per dm3 of solution C that has reacted through solution B
b) The mass of solid sample A in grams that was dissolved in solution B (Na = 23, 0= 16, H =1)
Volume of solution of C
Average titre reading, = + +/3 = 22.50cm3 (assuming)
Treatment of results:
In solving this problem, we must understand that whenever sample A, was added to solution B, HCl.
The reaction is represented via
2HCl+ Na2CO3 → 2NaCl + H2CO3 (H2O + CO2)
Afterward when solution C containing NaOH was now titrated against solution B, the reaction must have been through the excess amount of the HO for example: Back titration. The reaction can be symbolized via
HCl + NaOH → NaCl + H2O
Since there was no evidence of the 1st reaction, it is only the 2nd reaction that can be utilized to determine the amount of the first reactant. We might wish to note as well that, the average titre value that will be utilized in solving this problem might not coincide through the titre value we gained in the laboratory. This is just a demonstration. We shall begin from the titre value
The amount in moles of NaOH
= Molar concentration x Volume
= 0.131 moldm3 χ / 1000
= 0.000131 χ mol of NaOH
(Where, χ is the average titre value obtained from the experiment)
From the stoichiometric equation. χ
1 mole NaOH = 1 mole HCl
0.000131 χ moles NaOH =0.000131 χ moles HCl
That is, 25cm3 of HCl contained 0.000131 χ moles HCl
1000cm3 of HCl will contain 0.0001231 x 1000 x χ moldm-3 / 25
= 0.000524 χ moldm-3
This is the molar concentration of HCl that was in excess and that reacted through the NaOH solution. From the data provided, the original molar concentration of the HCl acid was 0.303 mole dm3 Out of this, 0.00524 χ reacted as excess with NaOH
= The molar amount of HCl that must have reacted with the solid Na2CO3 sample
= 0.303 - 0.00524 χ mol dm-3 = γ mol/ dm3
To find the amount in moles, the original molar concentration of HCl was 0.303 moldm-3
For example 1000 cm3 will contain 0.303 x 25 moles / 1000
= 0.0076 moles.
Out of this 0.000131 χ moles HCl reacted by the NaOH
The mole of HC1 that must have reacted with the Na2CO3
= 0.0076 - 0.000131 χ moles.
= Z moles of HCl.
From equation 1
2HCl + Na2CO3 → 2NaCl + H2O + CO2
That is 2 moles HCl= 1Z mole NaCO3
Z Mole HCl = 1/2 Z moles Na 2CO3
Amount in moles = mass/ Molar mass (Remember of a solid particle)
Molar mass of Na2CO3 = (2 x 23) + 12 + (3 x 16) = 106g mole -1
Amount in moles = 1/2 Z mole = mass / 106g mole -1
Mass of Na2CO3 added to solution B = 1/2 Z x 106
I hope we have not been confused via such calculations. This is the reason why i did not use a titre value for the first time so that we can really work ourselves into the experiment. If I might add though, the following values were attained via me.
Z = 0.0038 χ moles HCl
1/2 Z = 0.0019 moles Na2CO3
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