Volumetric analysis
exercise 6
|
Volume
|
|
| Start |
10.40
mls
|
| Finish |
22.22
mls
|
| Total |
11.82
mls
|
A 0.600gram sample of impure NaOH was dissolved in 20 mls of water. The solution was then titrated against a 0.12M HCl solution.
What is the percentage composition of NaOH in the sample.
( Atomic mass of Na = 23, H = 1, O =16 )
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 = 0.0118 X 0.12 = 0.00142
Step 2 Using
the equation calculate the number of mole of NaOH in the sample.
Mole of NaOH = 0.00142
Step 3 Find
the mass of NaOH present in the sample.
Mass of NaOH = 0.00142 X 40 = 0.0568grams
Step 4 Calculate the percent of the mass due to NaOH.
0.0568 / 0.600 X 100 = 9.47%