Volumetric analysis
exercise 10
Irene is given 200ml of 0.2M HCl solution. Her friend accidentally drops in some solid NaOH which quickly dissolves. A 20ml aliquot is taken and titrated against a 0.1M NaOH solution. The results are shown on the right.
Calculate the mass of NaOH accidentally dropped in to the 200ml solution.
Atomic mass Na =23, O =16, H=1
|
Volume
|
|
| Start |
12.40
mls
|
| Finish |
32.22
mls
|
| Total |
19.82
mls
|
The equation for the reaction is shown below.
Step 1 Calculate
the mole of sodium hydroxide added
to the aliquot
Mole = volume X concentration
Mole of NaOH = 0.0198 X 0.1 = 0.00198
Step 2 Using
the equation calculate the number of mole of acid required to react
completely
According to the equation 0.00198 mole of HCl is needed to react.
Step 3 Find
the mole of acid that should have been present in the 20ml aliquot of
a 0.2M solution
Mole= volume X concentration
Mole of HCl that should be present in the 20ml
aliquot.
Mole = 0.02 X0.2 = 0.004
Step 4 Find the mole of acid that reacted with the NaOH which was accidentally
dropped in to the solution. This will equal the mole of NaOH that was
added.
Mole of NaOH added to the 20ml aliquot = 0.004 - 0.00198 = 0.002
The mole of NaOH added to the 200ml solution = (200/20) X 0.002 = 0.02
Step 5 Calculate
the mass of NaOH added.
Mass = Mole X formula mass = 0.02 X 40 = 0.8 grams