"N" in fertiliser - standard solution (1 Viewer)

[mitch_f1]

nitrogen in fertiliser

Hey
I have to perform an experiment to test the nitrogen content in fertiliser. I am doing this by a titration. So far all the experiments I have read have stated that I am using 1.03g or 1.3g of fertiliser. So, my question is: why 1.03/1.3 why not 1g or 5g? Is there any difference?

Thanks
Mitch

 

[insert-username]

I would assume this has something to do with the molar mass of fertiliser, and that using 1.3 or 1.03g of fertiliser will give you simple and convenient numbers to work with in the calculation stage (i.e. moles = mass/molecular mass). It makes no real difference, it just might be easier to work out the nitrogen content later on.


I_F

 

[mitch_f1]

Makes sence.

I'll expand the question. Here is a summarised procedure for one of the experiments:
*grind up 1.3g of fertiliser
*transfer to volumetric flask, add deionised water until the calibration line. Shake thoroughly to dissolve
*pipette 20mL into 3 conical flasks
*pipette 20mL of 0.1M NaOH into each flask
*one flask add 50mL deonised water
*boil for 10min, add water to replace that lost, to remove ammonium
*test neck with litmus, if blue keep boiling until all ammonium gone
*repeat with other flasks
*fil burette with 0.1M HCl
*add methyl red indicator and titrate


So, in the initial stages of the experiment, why does deionised water get added, as opposed to NaOH?

thanks

 

[Commerce Freek]

Mitchy,

You add deionised water to the fertiliser because the water will not affect the pH.

 

[mitch_f1]

But won't it affect the concentration? Hence the final readout?

 

[Commerce Freek]

no because it is just diluting it it isnt changing the actual pH

 

[mitch_f1]

Can somebody please confirm if deionised water will affect the overal concentration of nitrogen molecules???? Because as far as I can tell, by adding water, you are diluting the nitrogen, which, in the end, will affect the final result.

thanks

 

[Dreamerish*~]

no because it is just diluting it it isnt changing the actual pH

By diluting a solution, you're changing the concentration of the solute. 0.5 moles of HCl in 100 mL of water is different to 0.5 moles in 200 mL.

And since pH = -log₁₀[H⁺], it's affected by concentration.

 

[mitch_f1]

Might as well post this is the same topic.

I am have done the experiment, which i now realise was a back titration, and am now calculating the concentration of Nitrogen. And because it was my first time doing a back titration, I kind of have very little idea what to do. I have found these steps on the internet:

1. Calculate the amount of NaOH in mol added to the fertiliser before heating
2. Ammonium compounds react with the NaOH: NH4+ + OH- --> NH3 + H2O . After heating, all the ammonium ions will have reacted and the amount of unreacted OH- can be determined by titrating the solution with standard HCl.
Use the concentration of HCl and average titre to find the amount of HCl used in the titration
Find the amount of NaOH that did not react with the fertiliser solution
3. Calculate the amount of NaOH that reacted with the fertiliser solution in each flask
4. find the amount of present in 60mL aliquot of fertiliser solution
5. calculate the amount of ions originally present in the 250mL volumetric flask
6. find the mass of nitrogen present in the sample of fertiliser
7. calculate the percentage by mass of nitrogen in the sample of fertiliser.

I am up to step 3, I have calculated that 0.0443moles of NaOH remain unreacted (NaOH + HCl --> NaCl + H2O There is 0.006 moles NaOH, and 0.00157 moles of HCl I subtracted the two. I also used the titration formula to calculate the volume of NaOH, using the volume of HCl used un the reaction, to get back to where I started -15.7mL- the amount of HCl used in the titration....but it can't be equal volumes used in the reaction...can it?????). And I am now stuck on step 3. The only thing I can think of doing is subtracting the ammount of unreacted NaOH, from the orginal amount....but then that just gives me the moles of HCl.

If i've left anything out ( I prolly have) please tell me, and i'll put it up, or if you just want more info.

Any help would be apprecieated. Thanks.

 

[Riviet]

Well if you do have more information, could you please post it up for the sake of helping us help you?
You seem to have a lack of information of volumes.

 

[mitch_f1]

Well what else do you want??

I used 15.7mL of HCl in the titration, against 25mL of the boiled fertiliser mixture. In the fertiliser mixture was 60mL of dissolved fertiliser, 60mL of NaOH, and 150mL of water. The HCl was 0.1M, and the NaOH was 0.1M.

Anything I missed?

 

[Riviet]

I had a go at it and this what I've come up with:
n(NaOH) unreacted = 0.0443 mol. (what you had before was correct)
0.0443 mol of NaOH = 0.0443 mol of OH (check this if you're not convinced, I believe this applies for any selected element in a compound)
From the equation, NH₄⁺ + OH^- -> NH₃ + H₂O
0.0443 mol of OH^-1 reacts with 0.0443 mol of NH₄⁺
So m(N) = nM
=0.0443 x 14
=0.6202g
I believe this is the mass of N in 25g of the fertiliser mixture.
.'. m(N) in volumetric flask = (60+60+150)/25 x 0.6202
=6.69816g
.'. % N in fertiliser = 6.69816/60 x 100
= 11.16%
I think that's how you do it and correct me if I've made any errors.

 

[mitch_f1]

Makes sence....wish I read it earlier...I handed it in yesterday. I worked it out to be roughly 25%, which is horribly wrong, but hey it's better than nothing.

I pretty much lost you in this step:

.'. m(N) in volumetric flask = (60+60+150)/25 x 0.6202
=6.69816g

I think what you worked out was the amount of nitrogen in the 60mL solution...which has a fraction of the original 1g.

This is how I (and a fellow student) worked it out:

0.00157moles of HCl (using n=m/M)
0.006 moles NaOH (same deal)

HCl + NaOH --> NaCl + H2O

.'. n(HCl)= 0.00157 moles, which = n(NaOH)=0.00157 moles
meaning 0.00157 moles reacted with HCl, therefore there was 0.00157 moles left over from the reaction with ammonium.

.'. 0.006 - 0.00157 moles = 0.00443 moles NaOH
meaning 0.00443 moles reacted with ammonium.

NH4^+ + OH^- --> NH3 + H2O
n(OH) = n(NH4)
0.00443 = 0.00443 moles

Weight of N atoms in NH4 is
n=m/M
0.00443=m/14.00674
m=0.620498582 grams N reacted with 60mL NaOH in 60mL of fertiliser soln

.'. 0.620498582/60 x 250 = 0.2585410758g on N atoms in 250mL with 1g of fertiliser.

.'. 0.2585410758/1 x 100 = 25.85%(w/w) N in 1g of fertiliser.


I went wrong somewhere in there....but in my discussion I put it down to bad experimental procedure (I didn't have enough time to let all the NH3 boil out of the solution, hence the pH of the solution i was titrating into might have been affected, hence distorting the results)

Thanks heaps anyway
Mitch

P.S. How do you write in sub/super script in these windows???

 

[Riviet]

Yeah I wasn't sure about the last two lines of my working but I was figuring since you said in your second last post that you used 25mL of the boiled fertiliser mixture in the titration and in the fertiliser mixture was 60mL of dissolved fertiliser, 60mL of NaOH, and 150mL of water, which is a total of 270mL of solution. Then I assumed the 25mL reacted with the 6.69816g of nitrogen, so I multiplied the mass by 270/25, which is how I got to my answer.

How do you write in sub/super script in these windows???

(sup)insert text here(/sup) for a superscript
(sub)insert text here(/sup) for a subscript, but both with square brackets [ ] instead. Quote my previous post if you're still unsure.

 

[mitch_f1]

Yeh, ok, fair enough....so the bad was on my part; sorry.

Thanks for that as well.