[danzmutt]

Hello everyone,

I was hoping some you would be able to assist me with this question (below). I know, or think from doing my own reading so far, that in order to find moles of NH4= (volume of acid * Molarity of acid) - (volume of base * Molarity of base)

From my understanding, you use the moles of NH4 to find the grams of Nitrogen and then divide that by the grams of sample to find the percent.

However, I'm having trouble with this problem since it has two Molarities and only one volume.

In the Kjeldahl method for analysis of nitrogen in protein, the sample is digested in concentrated sulfuric acid, converting the nitrogen to ammonium sulfate, (NH4)2SO4. NaOH is then added, converting the ammonium sulfate to ammonia gas, NH. The ammonia is reacted with a known amount of HCL to form ammonium chloride. The unneutralized acid is then determined by titration. Find the percent nitrogen by mass in a sample of protein with a mass of 5.503 grams if the HCL concentration in 500 mL of acid solution is reduced from 0.2000M to 0.0742M

I'm completely lost. Can someone help me make more sense of how to get started..

Thank you!

[Arkcon]

From my understanding, you use the moles of NH4 to find the grams of Nitrogen and then divide that by the grams of sample to find the percent.

Good.  Now, how will you accurately measure moles of ammonia gas?  That's what the second molarity  (of the HCl ) is for.

[danzmutt]

Okay, that makes sense. But the problem I am running into is finding the second volume. To find the moles of NH3, M=moles/volume, a second volume is needed (however not given), correct?

Thank you for your reply Arkcon!

So far what I have done is:

moles HCL = (.500L)(.2000M) = .1000moles
moles NH3 = volume * .0742M) = ??

.1000moles - moles NH3 = moles of NH4

grams of nitrogen = (moles of NH4)(14.00g)

%nitrogen= grams nitrogen/5.503g *100%

[Borek]

the HCL concentration in 500 mL of acid solution is reduced from 0.2000M to 0.0742M

How many moles of HCl initially?

How many moles of HCl after the reaction?

What happened to the missing HCl?

[danzmutt]

Initially there are  (0.2M X 0.5L) = 0.1 moles of HCL

After the reaction, I am unsure of the volume to use for HCL because of "The unneutralized acid is then determined by titration".  Would I not assume the volume has increased?  Not knowing how much was required for titration is what is confusing me.

If I should assume volume is unchanged then (0.5L * 0.0742M) =  0.0371 moles

(0.1 - 0.0371) = 0.0629 moles is the difference.

"What happenned to the missing HCl?" - This I am unsure of.  This is how I understand where we are now. 

I calculated the original concentration of HCl to be 0.1 moles, the concentration after titration being 0.0371 moles.

(0.1 - 0.0371) = 0.0629 moles is the difference.

grams of nitrogen = (0.0629moles)(14.00g)= 0.8806 g of Nitrogen

%nitrogen= 0.8806g/5.503g *100% = 16% of Nitrogen in the sample of protein?

For some reason I feel my calculations are missing something ie. the new volume after titration, or something similar.  My professor usually tries to trick us, so this would be unlike her to make it doable on the first try 

Again, thank you all for your continued assistance!!   

[plankk]

Everything is correct! It will be even better if you insert the equatation of reaction between chloride acid and ammonia. But it is only a cosmetic matter.

[Borek]

"What happenned to the missing HCl?" - This I am unsure of.

It was neutralized by the ammonia.

Note, that information you are given is just a step closer to the final solution. If you will be given volume of base used for titration you will use it to calculate amount of the hydrochloric acid left. But in this case you don't have to calculate it - it is already given, just not in terms of number of moles, but in terms of volume/concentration.