Search for Answers to Your Chemistry Questions

Theory for a formal report on the precipitation of BaSO4

Hi, i have to do a formal report for the MST and i realised tht i cant answer most of the questions (i feel like a gone-case student :( , then i found this cool website which hopefully can help me progress and understand my chemistry)

My question is: why do i have to do the steps highlighted in red? What does it do?

Procedures

      (A) Precipitation of BaSO4

(1) Pipette 25.0 mL of the given sulphate solution into a 250 mL beaker.

(2) Add about 50 mL of water and 5 drops of concentrated HCl.

    Caution:  Concentrated HCl is very corrosive, add in the fume hood.

(3) Heat to boiling and, with vigorous stirring, add dropwise from a measuring cylinder 10 mL of 10% barium chloride solution.

(4) Cover the beaker with a watch glass and digest (heat to just below boiling) for 20 minutes.

(5) Test for complete precipitation by adding a few drops of BaCl2 to the clear supernatant liquid.

      (B) Washing and Filtration of BaSO4 precipitate

(1) Decant the clear supernatant by filtration (use a weighed filter paper) at the vacuum pump, using a pre-weighed crucible.

(2) Wash and swirl the precipitate with about 20 mL of warm deionised water.

(3) Use a 'rubber-policeman' to dislodge any particles on the beaker.

(4) Allow to settle.

(5) Decant the liquid through the filter paper.

(6) Repeat the washing and decanting at least 2 more times.

(7) Discard the filtrate.

      (C) Drying and weighing of BaSO4 precipitate

(1) Dry the crucible in the oven at 150oC for about 1/2 hour.

(2) Cool the crucible with BaSO4 precipitate in a desiccator for about 10 minutes.

(3) Weigh the crucible when it is cooled down.

(4) The difference between this weight and the empty crucible (plus filter paper) is the weight of BaSO4 precipitate. (Note: You may repeat Steps 1 to 4 until a constant weight of the precipitate is obtained.)

Hey check out this post! We have answerd a lot of question, so searching can really help

http://www.mychemistrytutor.com/forums/ap-chemistry/gravimetric-analysis-t1697

Add about 50 mL of water and 5 drops of concentrated HCl.  This would prevent the formation of Ba(OH)2, thus decreasing the amount of BaSO4 present.

(3)  Heat to boiling and, with vigorous stirring, add dropwise from a measuring cylinder 10 mL of 10% barium chloride solution.
The vigorous stirring and dropwise addition of the BaCl2 solution allows for complete mixing of the Ba+2 ions with the solution.  This will result in any dissolved SO4-2 ions forming BaSO4 (s).

4)  Cover the beaker  - the beaker is covered to prevent loss of the material in the beaker due to spattering.

B = 2)  Wash and swirl the precipitate with about 20 mL of warm deionised water.  You want to make sure that any other compounds are able to dissolve and not be included in the mass of your ppt.

C-(1)  Dry the crucible  If you don't dry the crucible completely, it will contain water which will make the mass of the ppt appear to be larger than it really is.

hey thanks for the link and answers!

oh and wht should i be able to observe from this experiment? (the experiment was done long ago and i just cant remember the steps)

I nid this for a section of my report:
(F) Discussion (20%)
Discuss the results and observations, possible reasons behind it. Comment on your results from the practical viewpoint with the help of reference books. Do not copy directly any irrelevant materials from books.

the results are:

Weight of crucible with the precipitate:                 22.58400g
Weight of crucible (with a piece of filter paper):  22.3813g
Weight of BaSO4 precipitate:                                   22.58400g – 22.3813g = 0.20265g

(Ba2+) + (SO42-) = BaSO4 (precipitate)

Moles of SO42- / Moles of BaSO4 = 1 / 1
[Molarity of SO42- x Volume of SO42-] / [Weight of BaSO4 / Mol weight of BaSO4]                   = 1 / 1                          
(Mol Weight BaSO4 = 233.33 g/mol)

Molarity of SO42-                             =              0.20265g / 233.33g/mol
                                                           =               8.685124073 x 10-4
                                                           =               (8.685124073 x 10-4) / 0.025L
                                                           =               0.034789699M
                                                           =               0.03M

Concentration of SO42- in g/L      =               Molarity of SO42- >                                                            =               3.339811159g/L
                                                           =               3.34g/L




If you find this answer useful please share it with other students.