And also, I have another q:
A chemist analysed aspirin tablets for quality control. The initial step of the analysis was the standardisation of a NaOH solution. Three 25.00 mL samples of a 0.1034 mol L–1 solution of standardised HCl were titrated with the NaOH solution.The average volume required for neutralisation was 25.75 mL.
(a) Calculate the molarity of the NaOH solution.
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Three flasks were prepared each containing a mixture of 25 mL of water and 10 mL
of ethanol. An aspirin tablet was dissolved in each flask. The aspirin in each solution
was titrated with the standardised NaOH solution according to the following equation:
C9H8O4(aq) + NaOH(aq) → C9H7O4Na(aq) + H O(l) 2
The following titration results were obtained.
Tablet Volume (mL)
1 16.60
2 16.50
3 16.55
(b) (i) Calculate the average mass (mg) of aspirin per tablet.
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(ii) Why was it necessary to include the ethanol in the mixture?
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a)
NaOH
(aq) + HCl
(aq) -> H
2O
(l) + NaCl
(aq)The reaction immediately shows that the mole ratio between sodium hydroxide and hydrochloric acid is one to one.
n(NaOH) = n(HCl)
Using n=CV
[NaOH] * V(NaOH) = [HCl] * V(HCl)
[NaOH] * 25.75mL = 25.00mL * 0.1034 mol L
-1[NaOH] = 0.100388... mol L
-1Note that for this part of the question you would round to 4 significant figures and give your answer as 1.004*10
-1 mol L
-1 but for the remainder of the question you use the UNROUNDED answer in your calculator
b) (i) This can obviously be done by inspection, but as the HSC requires proper calculations the average is obviously
V(NaOH) = (16.60+16.50+16.55)mL/3 = 0.01655L
Important aside: It's not that easy to copy and paste a table. In the actual HSC question, we assume that the tablet is the same, but rather human error caused the differences in the calculated volumes for NaOH that it was titrated against. So the average volume is the volume of NaOH used for titration.Observe that the mole ratio is once again one-to-one. This means
n(C
9H
8O
4)
= n(NaOH)
= [NaOH] * V(NaOH)
= answer from part a) * 0.01655
= 1.6614278...*10
-3 mol
Using the fact that m = n*MM
m = 1.6614278...*10
-3 * 180.154
= 0.2993g (correct to 4 s. f. here)
b) (ii) The question was generous here to make this a one mark question. And so they should, because they didn't give us a diagram of what aspirin looks like on the molecular level.
So the only logical answer to using ethanol in the mixture is to
promote the solubility of aspirin (that's the answer). Just by looking at C
9H
8O
4 we can make a few assumptions:
a) There's quite a fair few carbons. Expect some kind of a chain, or benzene ring (tbh, yes benzene ring in the actual molecule) which ethanol will easily dissolve.
b) The 4 oxygens could all form -OH groups for all we know but obviously we don't know. Even then, both water AND ethanol interact through hydrogen bonding and dipole-dipole interactions when hydroxyl groups are present, so you can still safely assume that ethanol increases solubility.
But basically, it's cause there's obviously going to be carbon chains, which essentially makes ethanol a more effective solvent.
Heyyyaaa there!! I have a Question that has been bugging me.. would love to get your advice / help with it!
Q: Why would the neutralisation of HCl and Na2CO3 require the addition of methyl orange indicator? I thought the best suited indicator would be bromothymol blue because this titration involved a strong base and a strong acid - hence a neutral salt? (or is Na2CO3 a weak base)?
Got me reeaaallll caught up. Thanks in advance!
Na
2CO
3 is of course, a weak base. The only strong bases in the course are essentially NaOH, KOH and all the other group 1 hydroxides.