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Cy304 Practical write-up 2
Conductometric titrations
Introduction
One problem with using visible indicators to perform acid-base titrations is that the change in colour of the indicator rarely coincides with the actual end point of the titration, an indicator has to be chosen with a change as close as possible to the end point. One way of avoiding this problem is to perform a conductometric titration.
When a strong acid is titrated against a strong base, e.g. sodium hydroxide, the hydrogen ions react with the hydroxide ions in the base and are removed, since the hydrogen ions have a greater mobility than the other ions in solution, this will cause a reduction in conductivity. When all the hydrogen ions are neutralised, further addition of base will cause an increase in conductivity due to the presence of the highly mobile hydroxide ions. A plot of conductivity against volume of base added should show two straight lines, which intersect each other; the intersection of the two lines is the endpoint of the titration.
If a weak acid is titrated using this method the results will be somewhat different. Initially the conductivity will increase, this is due to the relatively low concentrations of the hydroxide ion, caused by the incomplete dissociation of the acid, not having much influence on the overall conductivity, instead the increase in the total number of ions will cause the conductivity to rise. In this case the end point can be determined when a sharp increase in conductivity is noted due to all of the acid being neutralised and an excess of hydroxide ions being present.
Conductometric titrations can also be used to determine the concentrations of a solution containing a mixture of strong and weak acids, as both end points will show on the graph.
Method
Hydrochloric acid (2ml, unknown concentration) was diluted to 100 ml; 20 ml of this solution was mixed with 100 ml of conductivity water in a beaker into which was placed a dipping cell of known cell constant. The cell was connected to a conductance meter and the conductivity of the acid solution was noted. The solution was then titrated against sodium hydroxide (0.5M) and after each 5ml addition of base the new value of conductivity was noted. This was repeated until the titration was well past its end point and the results were plotted.
The experiment was then repeated with the acetic acid solution and the solution of mixed acids, although the mixed acid solution was not diluted further as with the acetic and hydrochloric acid solutions.
Results
Vol NaOH added (ml) Conductivity of HCl solution (mS) Conductivity of Acetic acid solution (mS) Conductivity of mixed acid solution (mS)
0.0 15.68 0.58 7.47
0.5 14.94 0.57 6.86
1.0 14.29 0.60 6.22
1.5 13.61 0.72 5.46
2.0 12.83 0.87 4.78
2.5 12.17 1.02 3.94
3.0 11.47 1.20 3.28
3.5 10.77 1.34 2.53
4.0 10.04 1.52 2.31
4.5 9.35 1.72 2.36
5.0 8.70 1.90 2.52
5.5 7.99 2.10 2.68
6.0 7.35 2.27 2.83
6.5 6.70 2.45 2.97
7.0 6.06 2.63 3.11
7.5 5.38 2.79 3.26
8.0 4.86 2.95 3.44
8.5 4.31 3.08 3.77
9.0 4.34 3.31 4.24
9.5 5.06 3.46 4.73
10.0 5.47 3.64 5.22
10.5 5.90 3.78 5.71
11.0 6.47 3.95
11.5 6.81 4.11
12.0 7.19 4.27
12.5 7.59 4.41
13.0 4.43
13.5 4.71
14.0 4.82
14.5 4.98
15.0 5.26
15.5 5.69
16.0 6.13
16.5 6.58
17.0 7.04
Treatment of results
The plot of volume of base added versus conductivity for HCl gives an end point when 8.7ml of base were added.
Concentration of base = 0.5 moll-1
8.7ml = 0.0087 l
Number of moles of base = 0.00435 moles
0.00435 moles of base will react with the same number of moles of acid therefore there were 0.00435 moles of acid in 2ml of the original solution
0.00435 moles in 2ml
0.002175 moles in 1ml
2.175 moles in 1l
The original concentration of the HCl was 2.175 moll-1
The plot of volume of base added versus conductivity for acetic acid gives an end point when 14.7ml of base were added.
Concentration of base = 0.5 moll-1
14.7ml = 0.0147l
Number of moles of base = 0.00735 moles
0.00735 moles of base will react with the same number of moles of acid therefore there were 0.00735 moles of acid in 2ml of the original solution
0.00735 moles in 2ml
0.003675 moles in 1ml
3.675 moles in 1l
The original concentration of the acetic acid was 3.675 moll-1
The mixed acid solution gave 2 end points
The end point corresponding to HCl was at 3.75 ml
Concentration of base = 0.5moll-1
3.75ml = 0.00375 l
Number of moles of base = 0.001875 moles
The end point corresponding to acetic acid was at8.05ml
8.05ml = 0.00875 l
Number of moles of base =0.004025 moles
Therefore 20ml of the mixed acid solution contained 0.001875 moles of HCl
0.001875 moles in 20ml give 0.00009375 moles in 1ml and 0.09375 moll-1
The 20ml of the mixed acid solution also contains 0.00405 moles of acetic acid
0.00405 moles in 20ml give 0.0002025moles in 1ml and 0.2025moll-1
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