volumetric analysis

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A standard potassium carbonate solution is made by adding 1.227g of to a 250 mL volumetric flask and filled to the mark with water. 20.00 mL aliquots are taken and titrated against sulfuric acid, using methyl orange indicator. The average titre was 22.56 mL of sulfuric acid.
a) Write the equation for the reaction. DONE
b) Calculate the concentration of the solution. DONE
c) Calculate the concentration of the sulfuric acid solution. NOT DONE

Please help me with part c!! I don't really understand 'what's happening' in the original question either, since I missed a lesson on volumetric analysis so if someone could explain it that would be great. Thanks.

[Collin Li]

Basically, you have an aliquot (fixed amount) of base sitting in a clean beaker.

What you want to do is to react it with acid, to neutralise the base completely with sulfuric acid, so that the pH jumps from basic to acidic. The methyl orange indicator will be what allows you to experimentally see this. The point at which the indicator changes colour, is called the "end point." The "equivalence point" is when the number of moles of the base is equal to the number of moles of the acid (accounting for stoichiometry), so that this where a steep change in pH occurs. In experiments, we assume the end point is the equivalence point (since it roughly is -- the end point comes slightly after the equivalence point usually), and hence we can use the number of moles of the substance with a known concentration (in this case the base), and match it up with the number of moles of the substance with an unknown concentration (acid).

Basically, you know the number of moles of sulfuric acid you delivered, because you titrated it until you reached the end point (roughly the equivalence point), and hence you can substitute in the number of moles of the base. Then you use your titre (a volume) to divide through and get the concentration of the acid.

[danieltennis]

I just did this question just now, LOL. 

[Collin Li]

So, for this question, you would go:

Reaction:


So hence:



(Interpretation: this amount of acid was used to neutralise the amount of base from a 20mL aliquot)



(Interpretation: this amount of acid present in the 22.56mL of acid solution delivered reveals the concentration)

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Thanks coblin I get it now, you're a champ!
But I think

[Collin Li]

Hmm... yes, I accidentally calculated potassium sulfate's molar mass instead.

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Another question:

A particular brand of commercially available Cloudy Ammonia claims to contain not more than 4% ammonia by mass. To verify this claim, a 23.27g sample of the window cleaner was placed in a volumetric flask and diluted to 250 mL. A 20.0 mL aliquot of this solution was titrated with 0.0892 M hydrochloric acid. The volume of the acid used was 21.35 mL. The equation for the reaction is



a) Calculate the amount of hydrochloric acid, in mol, used in the titration.
b) Calculate the amount of ammonia, in mol, used in the titration.
c) Calculate the amount of ammonia, in mol, in the 23.27g sample of Cloudy Ammonia.
d) Calculate the mass of ammonia in the 23.27g sample of Cloudy Ammonia.
e) Calculate the percentage of ammonia by mass in Cloudy Ammonia. Is the manufacturer's claim about the ammonia content of his product verified.

I think I know how to do this, but my answers don't match up with the solutions! Can someone please check these thx.

[Pandemonium]

a) n(HCl) = c(HCl) x v(HCl) = 0.02135 x 0.0892 = 0.00190 mol
b) n(NH3)t = n(HCl) = 0.00190442 mol = 0.00190 mol
c) n(NH3)s = n(NH3)t x 250/20.00 = 0.02380525 mol = 0.0238 mol
d) m(NH3) = n(NH3)s x Mr(NH3) = 0.02380525 x 17 = 0.40468925 g = 0.405g
e) %(NH3) = m(NH3)/m(sample) x 100% = 0.405/23.27 x 100% = 1.74%

Verified.

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Thanks pandemonium

[methodsboy]

1) use C(K2CO3) = the answer from part b)
2) use V(K2CO3) = 20 mL (aliqot) = 0.02L
3) now work out the n(K2CO3)= .....mol
4) draw a dual number line to work out the moles for the H2SO4. (treat this as a normal stoi question)
5) as the average titre was 22.56 mL ----> use v(H2SO4) = 0.02256L
6) C(H2SO4) = n(H2SO4)/v(H2SO4) = n(H2SO4) / 0.02256L
u should now get the required answer.