Acid standardization question

[Tokspor]

Hi guys, I have a question from a passage. I'll try to keep this as condensed as possible. Here it is.

Passage information:

A chemist is trying to determine how much nitrogen is in a certain sample. The nitrogen is in the form of ammonia.

1. The chemist standardizes HCl with a certain concentration of THAM. HCl donates 1 proton and and THAM accepts 1 proton.
2. Having just measured the concentration of HCl, the chemist reacts the ammonia (NH3) sample with an excess of HCl, creating the following reaction:
NH3 + HCl (excess) -> NH4+
3. The remaining unreacted HCl is titrated with NaOH. This indicates how much HCl reacted with NH3, and therefore how much NH3 there was in the sample of interest.

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The question is this:

What would happen if the chemist recorded a lower concentration of THAM than there actually was?

A. The results would be unaffected because nitrogen is titrated with HCl and not THAM.
B. The calculated nitrogen content in the sample would be too high.
C. The calculated nitrogen content in the sample would be too low.
D. The results would be unaffected because THAM is a primary standard.

The answer is C but I don't understand the explanation given by the text.

The text's explanation states that "when the back titration with NaOH is performed, less NaOH would be used for the titration, and a low value would then be calculated for the unreacted acid." But in this experiment, isn't the step involving NaOH (I think it's with a known concentration) used to determine the concentration of leftover HCl?

A miscalculation in the concentration of THAM might have led to a miscalculation of the original HCl concentration. However, assuming the NaOH concentration is correct, we would still be able to determine the correct concentration for the leftover HCl since we are titrating this leftover HCl with a known concentration of NaOH. Why would less NaOH be used if, as it seems to me, the standardization of leftover HCl with NaOH is independent of the original standardization of HCl with THAM?

Note: It's not explicitly stated in the passage whether the NaOH concentration is known beforehand, but I don't see how it would be possible to titrate the unknown concentration of leftover HCl using also an unknown concentration of NaOH.

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[loveoforganic]

The equation for HCl standardization:

[HCl] = [THAM]*(L THAM)*(1 mol HCl / 1 mol THAM) / (L HCl)

So true [THAM] > used [THAM].

If you plug a smaller value into the [THAM] variable, the calculated [HCl] is going to be low.

Then, net equation for the back titration:

mol ammonia = (mol HCl) - (mol NaOH) = [HCl]*(L HCl) - [NaOH]*(L NaOH)

So if the [HCl] is low, the calculated amount of ammonia is going to be low.


I'm trying to remember why you must know the concentration of the HCl to perform this back titration. Blanking out however!

 

[Tokspor]

Then, net equation for the back titration:

mol ammonia = (mol HCl) - (mol NaOH) = [HCl]*(L HCl) - [NaOH]*(L NaOH)

So if the [HCl] is low, the calculated amount of ammonia is going to be low.

I understand that the calculated amount of ammonia would be lower, but I don't understand why the explanation in the text states that the amount of NaOH used for the titration would be lower.

The "(L NaOH)" part of the above equation is determined by how much remaining HCl is left after it reacts with NH3, right?

If we use the equation M(HCl)V(HCl) = M(NaOH)V(NaOH) for the back titration with NaOH, we already have V(HCl) and M(NaOH).

By running the titration, we determine V(NaOH). Finally, we can determine M(HCl) from this, and this M(HCl) represents the concentration of remaining HCl after its reaction with NH3.

It seems to me that the amount of NaOH used is independent of whether or not the original concentration of HCl was miscalculated. From how I understand it, this is because NaOH is used to titrate the remaining HCl concentration, which is like a new titration experiment independent of the THAM titration. Am I misunderstanding something here?

 

[loveoforganic]

I understand that the calculated amount of ammonia would be lower, but I don't understand why the explanation in the text states that the amount of NaOH used for the titration would be lower.

The "(L NaOH)" part of the above equation is determined by how much remaining HCl is left after it reacts with NH3, right?

Yes, the 'back titration' part of a back titration measures the excess of the initial reagent.

If we use the equation M(HCl)V(HCl) = M(NaOH)V(NaOH) for the back titration with NaOH, we already have V(HCl) and M(NaOH).

By running the titration, we determine V(NaOH). Finally, we can determine M(HCl) from this, and this M(HCl) represents the concentration of remaining HCl after its reaction with NH3.

There were x moles NH3 in solution. You react with 75 L HCl. You add 25 moles NaOH. 25 moles HCl = excess. How do you know what % of the 75 L those 25 moles HCl made up? I think this answers my question from earlier.

It seems to me that the amount of NaOH used is independent of whether or not the original concentration of HCl was miscalculated. From how I understand it, this is because NaOH is used to titrate the remaining HCl concentration, which is like a new titration experiment independent of the THAM titration. Am I misunderstanding something here?

This part is definitely correct. Moles NaOH will be the same regardless. Bad explanation.

.

 

[fizzgig]

"when the back titration with NaOH is performed, less NaOH would be used for the titration, and a low value would then be calculated for the unreacted acid"

ok now i'm confused - if THAM is recorded low, then my HCl is recorded low, right, since i'm assuming they're 1:1?

so you have more acid than you thought, ammonia mols is constant, so you end up with more excess acid than if you'd added the lower conc you thought you had.

since you have MORE excess acid, wouldn't it require more NaOH to titrate, and then your conclusion would be oh, i titrated with 200 mols NaOH. so i added 400 mols of HCl (when really i added 500), and there's 200 left over, so i had 200 mols of NH3 (when really there were 300), hence the lower ammonia estimate?

mmmr?

 

[loveoforganic]

Regardless of what your calculations are, you're going to use the same amount of NaOH. Your calculations don't effect the number of moles actually in solution or the number of moles required to cause the indicator change at equivalence point.

 

[FlipDoc2Be]

Hey Tokspor is this passage from EK Gchem?

 

[fizzgig]

thanks organic for saying it the third time. for some reason my brain got stuck.

i'm on board now.

 

[jpatel]

Lower THAM than actual would give us lower HCl concentration than usual.

Since we are reacting excess HCl with NH3, all ammonia will be protonated. remaining left over H+ are titrated with OH-. lets say we added x moles of HCl to react with y moles of NH3

then x - y = z = moles H + will be left over = moles of NaOH you will use to titrate

Therefore, y = x - z. Since x ( HCL)is lower than actual, rocorded y(NH3) value will be lower than usual.

Ez as cake