Mass of sulfate in fertilizer (1 Viewer)

[stargaze]

Hi peoples,

This is a heaps commonly asked prac, and I just realized that I didn't complete my calculations on it when we did it in class..

Could someone pls help by just going thru how u calculate the mass of sulfate in fertilizer, perhaps using q27 HSC 2003 as the question (http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2003exams/pdf_doc/chemistry_03.pdf)

Thanks a lot!~

 

[Jumbo Cactuar]

Q27 a

BaSO4: M = 233.4 g/mol
m = 1.8g
n = 0.00771 mol

SO4 2-: M = 96.1 g/mo
n = 0.00771 mol
m = 0.741 g

Percentage Sulfate = 0.741/1 *100% = 74.1%

b

This reaction has several limitations. The scale of the reaction and the slight solubility of BaSO4 makes it useless for detecting trace quantities. The nature of the filtration also increases the amount of analyte lost.

There are however several procedures that can improve accuracy and provide decisive analyte eludication. The use of deionized water would eliminate contamination. The results can be confused with BaCO3 and so acidifying the solution slightly should be carried out before filtration. The filtrate should be treated with silica vacuum desication for a week before the filter is reweighed.

This method is however long and impractical. If the appropriate equipment is available a better and more precise method would be to filter only a very small amount and test the filtrate with flame AAS to determine the residual Ba2+ concentrations. Alternatively the remaining Ba2+ could be precipitated with anhydrous sodium carbonate and titrated to the bicarbonate endpoint.

 

[stargaze]

ahh icic thanks, just another q if u could

I'm not too confident/get confused on how u derive the the oxidation/reduction movements...

e.g.
1/2 H2 + OH- --> H2O + e- [sorry dont know how to do subscripts etc]

what oxidation or reduction reactions are happening and pls explain..

thanks

 

[Jumbo Cactuar]

I'm not too confident/get confused on how u derive the the oxidation/reduction movements...

e.g.
1/2 H2 + OH- --> H2O + e- [sorry dont know how to do subscripts etc]

what oxidation or reduction reactions are happening and pls explain..

thanks

The formal oxidation numbers (ON) are;

1/2 H2 + OH- --> H2O + e- E = 0.83V
.......0.....-2.1........1..-2

So the ON of one H is increased by one; so this is an oxidation half reaction. To be an actual reaction we need to pair it with a reduction half equation;

Cu 2+ + 2e- --> Cu(s) E = 0.34V
2........................0

Here the ON is decreased, so it is a reduction half equation.

Cu 2+ + 2e- --> Cu(s) E = 0.34V reduction
1/2 H2 + OH- --> H2O + e- E = 0.83V oxidation
_______________________________________
Cu(OH)2 + H2 --> Cu + H20 E = 1.17V

Since E is positive the reaction will happen spontaneously and could be designed to do work. Note that even though the equations have been multiplied to balance the electrons the E value stays the same.