Question

The concentration of barium ions in any solution can also be determined via gravimetric analysis. An impure sample of barium nitrate with a mass of 1.234 g, is completely dissolved in water and the resulting solution is reacted with an excess of aqueous sodium sulfate. A precipitate forms, and after filtering and drying, it was found to have a mass of 0.848 g.

A) What is the relevance of adding eccess sodium sulfate?
B) Calculate the % of barium nitrate in the original 1.234 g sample.

Answer 1

The percentage of barium nitrate in the impure sample is 76.98%.

Step-by-step explanation:

`Ba(NO_3)_2+Na_2SO_4\rightarrow BaSO_4+2NaNO_3`

A) When excess of sodium nitrate solution added to barium nitrate solution to precipitate barium sulfate which is white in color. The relevance of an excessive sodium sulfate is to precipitate maximum amount of barium ions dissolved in the solution.

B) Mass of the precipitate i.e barium sulfate = 0.848 g

Moles of barium sulfate =
`(0.848 g)/(233 g/mol)=0.003639 mol`

According to reaction, 1 mole of barium sulfate is obtained from 1 mole of barium nitrate.

Then 0.003639 moles of barium sulfate will be obtained from :

`(1)/(1)* 0.003639 mol=0.003639 mol` of barium nitrate.

Mass of 0.003639 moles of barium nitrate:

`0.003639 mol* 261 g/mol=0.9499 g`

Percentage of barium nitrate in impure sample:

`(0.9499 g)/(1.234 g)* 100=76.98\%`

The percentage of barium nitrate in the impure sample is 76.98%.