Volumetric analysis
exercise 5
|
Volume
|
Volume
|
Volume
|
|
| Start |
11.40
mls
|
22.43
mls
|
13.40
mls
|
| Finish |
21.22
mls
|
32.22
mls
|
23.22
mls
|
| Total |
9.82
mls
|
9.79
mls
|
9.82
mls
|
The average titre is (9.82 + 9.79 + 9.82) / 3 = 9.81mls
An alumina refinery puts out waste water into the local stream with a suspected high concentration of NaOH. 25ml aliquots were titrated against a 0.085M HCl solution. The results are shown on the left. Calculate the concentration of NaOH in the waste water.
The equation for the reaction is shown below.
Step 1 Calculate
the mole of substance delivered from the burette. In this case HCl solution
Mole = Average Volume(litres)
X Concentration
Mole of HCl = 0.0098 X 0.085 = 0.000833 mole of HCl
Step 2 Using
the equation calculate the number of mole of substance in the test sample.
In this case NaOH
For every mole of HCl
present one mole of HCl reacts.
0.000833 mole of NaOH are present in the 25 ml aliquot.
Step 3 Find the concentration of the substance in the test sample. In this case find the concentration of NaOH in the 25 ml aliquot.
Concentration of NaOH in the waste water = mole / volume(litres) =
0.00083 /0.025 = 0.033 M