VCE Chemistry Question Thread

[Chocolatepistachio]

How do you do this question I am so confused
In order to standardise an approximately 0.2molL-1 solution of hydrochloride acid an analyst first prepared an accurately know solution of sodium carbonate. To do this 5.404 g of pure anhydrous sodium carbonate was weighed out dissolved in water and the volume made up to 500.0 ml in a volumetric flask . 25.0 ml of this carbonate solution was placed in a conical flask(using a pipette) and titrated with the unknown hydrochloric acid solution from a burette; 26.8 ml of acid was needed to reach the equivalence point. The products of the reaction are water, aqueous sodium chloride and carbon dioxide. Calculate the concentration of the hydrochloride acid solution.

[keltingmeith]

How do you do this question I am so confused
In order to standardise an approximately 0.2molL-1 solution of hydrochloride acid an analyst first prepared an accurately know solution of sodium carbonate. To do this 5.404 g of pure anhydrous sodium carbonate was weighed out dissolved in water and the volume made up to 500.0 ml in a volumetric flask . 25.0 ml of this carbonate solution was placed in a conical flask(using a pipette) and titrated with the unknown hydrochloric acid solution from a burette; 26.8 ml of acid was needed to reach the equivalence point. The products of the reaction are water, aqueous sodium chloride and carbon dioxide. Calculate the concentration of the hydrochloride acid solution.

It may help to read this one sentence at a time, and draw pictures explaining the situation

[Chocolatepistachio]

Are you supposed to calculate the moles of sodium carbonate first and then the moles divided by 0.5

[keltingmeith]

Are you supposed to calculate the moles of sodium carbonate first and then the moles divided by 0.5

What brought you to this conclusion?

[Chocolatepistachio]

Idk there is a mass of sodium carbonate so maybe use that to calculate the moles and  Then maybe use the 500.0ml to calculate the concentration of sodium carbonate

Or maybe calculate the moles of sodium carbonate and then use the mole ratio to get the moles of HCl

[keltingmeith]

Idk there is a mass of sodium carbonate so maybe use that to calculate the moles and  Then maybe use the 500.0ml to calculate the concentration of sodium carbonate

Or maybe calculate the moles of sodium carbonate and then use the mole ratio to get the moles of HCl

Both of these are correct, sounds like you know what you're doing!

[Chocolatepistachio]

I tried doing it but then I just end up with the answer of 0.2

[keltingmeith]

I tried doing it but then I just end up with the answer of 0.2

Well, why not share your working so we can work through it? I can't help you if I don't know where you went wrong.

[Chocolatepistachio]

It won’t let me attach photos from my phone

n(Na2CO3)= 5.404/105.99
= 0.05098594mol

C=n/V= 0.05098594/0.5= 0.10197188

n(Na2CO3)= 0.10197188 x25.0/1000 = 0.0025493
n(HCl)= 2x 0.0025493= 0.0050986
C=n/v c= 0.0050986/0.0268 = 0.19024627

[keltingmeith]

It won’t let me attach photos from my phone

n(Na2CO3)= 5.404/105.99
= 0.05098594mol

C=n/V= 0.05098594/0.5= 0.10197188

n(Na2CO3)= 0.10197188 x25.0/1000 = 0.0025493
n(HCl)= 2x 0.0025493= 0.0050986
C=n/v c= 0.0050986/0.0268 = 0.19024627

I mean, pending rounding to the appropriate number of significant figures, this is correct. Remember that the answer SHOULD be close to 0.2, because we know that to 1 sig fig the answer should be 0.2. However, in the process of doing this experiment, we have used glassware and compounds with more than 1 significant figure of information - so our final answer is correct to three significant figures.

[Chocolatepistachio]

For this question I’m a bit confused
What volume of 0.125molL-1 nitric acid would be required to react with a solution made by dissolving 3.2 g calcium hydroxide in 75ml water. The products of this reaction are water, and aqueous calcium nitrate.

This is what I did but the answer is 69 ml
n(Ca(OH)2) =3.2/54.096= 0.0591541
n(HNO3)= 2x 0.0591541= 0.1183082
V= 0.1183082/0.125= 0.95L

[keltingmeith]

For this question I’m a bit confused
What volume of 0.125molL-1 nitric acid would be required to react with a solution made by dissolving 3.2 g calcium hydroxide in 75ml water. The products of this reaction are water, and aqueous calcium nitrate.

This is what I did but the answer is 69 ml
n(Ca(OH)2) =3.2/54.096= 0.0591541
n(HNO3)= 2x 0.0591541= 0.1183082
V= 0.1183082/0.125= 0.95L

For starters, your molar mass is wrong. But I think they might be off by a magnitude, because I'm getting an answer of 690 mL (to 2 sig figs, 691 to 3).

[Chocolatepistachio]

Oh ok thanks
Does the answer need to be in litres
Also in the question why is 75ml given if it is not used at all in the calculation

[keltingmeith]

Oh ok thanks
Does the answer need to be in litres
Also in the question why is 75ml given if it is not used at all in the calculation

Maybe they envisioned a different method to answering it, yours works fine though so lol. That or they were trying to throw you off - make sure you're understanding the circumstances of each value.

Also, units can be whatever you want as long as they're correct - but typically you'll be told what units to use. If you're not, I do recommend using the standard units, so as to not confuse your marker (for example, in high school music, I did all my exams in bass clef [because I played bass, it's the notation I was more familiar with] but everyone else did treble clef. My teacher accidentally gave me 0 for that section because she was on autopilot in marking lol. Also that example probably made no sense if you don't know music but oh well).

[Chocolatepistachio]

Made sense I did music and I used to play the clarinet