Stoichiometric problem... (1 Viewer)

[CHUDYMASTER]

I did this experiment to find the concentration of various ions in a water solution as part of an independent investigation.

NOW...

It's kinda tricky with PO4 ions because of the 3- valency. This is because (in the experiment I did), I formed a compound of Barium Phosphate, meaning my end equation for the compound was

Ba2+ + (PO4)3- --> Ba3(PO4)2 (s).

I need to know the concentration of PO4 ions in the water, which was 0.05 L in volume (50 ml). My end precipitate of barium phosphate was 0.32 grams. Could someone do the stoichiometric calculations and tell me if you get 1010 ppm? I'm probably wrong in that it's half of that, but meh...

 

[McLake]

Ba2+ + (PO4)3- --> Ba3(PO4)2 (s).
NOT BALANCED!!!
3Ba2+ + 2(PO4)3- --> Ba3(PO4)2 (s).

n = m/FM

m[Ba3(PO4)2] = 0.32 g
n = m/(3*137.3 + 2*[30.97 + (16*4)])
n = 0.32/601.84
n = 5.32 x 10^-4

now n[Ba3(PO4)2] = 2n[(PO4)3-]

so n[(PO4)3-] = 2.67 x 10 ^-4
m = n * FM
m = 2.67 x 10 ^-4 * (30.97 + 16*4)
m = 0.025 g

so I get 1268 ppm.

Check my working please ...

 

[Lord Ac]

*runs from the forumn*



Ac

 

[CHUDYMASTER]

CRAP, sorry McLake, I meant to say 0.1 L or 100 ml. But according to your working, I think you are correct.

 

[McLake]

Originally posted by CHUDYMASTER
CRAP, sorry McLake, I meant to say 0.1 L or 100 ml. But according to your working, I think you are correct.

so ans is 2537 ppm?

 

[CHUDYMASTER]

Actually, how did you get that???

0.025 * 1000 / 0.1 = 250 ppm!!!??

 

[McLake]

Originally posted by CHUDYMASTER
Actually, how did you get that???

0.025 * 1000 / 0.1 = 250 ppm!!!??

Oops, sorry, that should be 253.7 ppm (250 rounded). Forgot to devide by 10 ...

 

[CHUDYMASTER]

Crap, that's unusually SMALL for phosphate concentration of POLLUTED WATER (?)

 

[McLake]

Originally posted by CHUDYMASTER
Crap, that's unusually SMALL for phosphate concentration of POLLUTED WATER (?)

Is it? I don't know what normal/high levels are.

 

[CHUDYMASTER]

Anything under 500 ppm phosphate is considered healthy.

 

[McLake]

So what was your source?

 

[CHUDYMASTER]

The old conquering chemistry textbook I think.
Why do you care...ex-HSC person? Are you considering a future in water monitoring and management?

 

[McLake]

Originally posted by CHUDYMASTER
The old conquering chemistry textbook I think.
Why do you care...ex-HSC person? Are you considering a future in water monitoring and management?

No, I meant water source.

Why do I care? Just trying to help ...

 

[CHUDYMASTER]

...Oh...

Well I obtained my water from Powell's Creek, in Flemington I think.

 

[McLake]

As someone unfimilar with this creek, why do you think it should be so polluted?

Also, did you take several samples and test them all?

 

[CHUDYMASTER]

I did for my TDS experiment, but for the anion experiment, I had to test for 3 anions - sulphates, chlorides and phosphates using the same sample, lest I killed myself with impatience. Carrying out such experiments are often long and tedius, so one was enough for the anions one.

And I consider it odd because GENERALLY, water classfied as "polluted" contains relatively high concentrations of phosphate ions.

 

[McLake]

So maybe it was a "freak" reading.

I hope I helped.

Gotta go.

Post anymore stoichiometric problems you have, cause they are often tricky (but they get easier ...)

 

[spice girl]

Just one thing,

how the hell are you certain that all the solid is Barium Phosphate? Some of the crap may be BaSO4.

 

[CHUDYMASTER]

Well that's part of my method, basically, I first get rid of the sulphates by acidifying the solution. Under these conditions, phosphate will not react. Later on, the water is made alkaline and then the Ba2+ is added to form barium phosphate.

 

[spice girl]

I beg to differ pplz. Chudymaster was right with 1010ppm in the first place.

Data: (NB: FW means formula weight)
FW(Ba3(PO4)2) = 601.84
FW(PO4) = 94.97

Mass phosphate / mass barium phosphate = (94.97 * 2) / 601.9

= 189.94 / 601.84 = .3156

Now mass Ba3(PO4)2 = 0.32g
so mass phosphate = .3156 * 0.32g = .1010g = 101.0mg

Now, ppm = mg/L
so 101.0mg / 100mL = 1010mg/L = 1010 ppm