HSC Chemistry Question Thread

[RuiAce]

Hey, I think it’s because you have to add all the numbers and you pick the one with the lowest numbers.
Like for a:
There is a 1+1+2+2+2= 8

And for b:
There is a 2+2+1+1+1= 7

So that’s why it’s d because 7<8.

I’m not too sure tho, so just wait for one of the moderators to confirm

Yeah this is fine.
________________________

Will add a comment though. From what I now know, this technically is no longer how IUPAC does things. According to the actual IUPAC rules, you should just go by alphabetical, which would give you A. But we just have to suck it up

[3.14159265359]

Yeah this is fine.
________________________

Will add a comment though. From what I now know, this technically is no longer how IUPAC does things. According to the actual IUPAC rules, you should just go by alphabetical, which would give you A. But we just have to suck it up

If that was true, then what would be different from choosing A or D???

[RuiAce]

A is alphabetical. The 1 was automatically assigned to the first prefix that appeared, i.e. "bromo-".

[Abdul_k]

Hi, I'm just unsure for this question and there are no answers. Thanks in advanced
https://drive.google.com/file/d/1Kq0lbrCQJ2AluVs3kdTnQaZpkyK9-1FK/view?usp=drivesdk

[MisterNeo]

Hi, I'm just unsure for this question and there are no answers. Thanks in advanced
https://drive.google.com/file/d/1Kq0lbrCQJ2AluVs3kdTnQaZpkyK9-1FK/view?usp=drivesdk

Hey
Sodium acetate forms a basic solution in water because of the acetate ion forming an equilibrium with acetic acid.

Initially, when the sodium acetate is dissolved in water, there is no acetic acid in the products of the equilibrium, hence it will shift to the right to maintain that special ratio (K constant). This shift to the right will produce hydroxide ions in the process and thus increase the pH of the solution and become basic. We can leave out Na⁺ in the equation because it's a spectator ion.

For the Part B, I would start off by defining a buffer solution and how sodium acetate and acetic acid form a buffer. A buffer is a solution made from equal amounts of a weak acid/base and its conjugate base/acid, which experiences small changes in pH when a strong acid/base is added.

When NaOH is added, it neutralises the H3O⁺ ions, causing the buffer equilibrium to shift to the right (Le Chat's Principle) to replace the lost ions, hence the pH will only experience small changes when the strong base is added.

[skisso]

Hi guys, not too sure how to work this one out

Thanks!

[moxiao402]

Hello guys, I’ve got a titration question. It says a standard solution of Na2CO3 is prepared by dissolving 1.208g and 250ml water up the volumetric flask. 25ml aliquot of the Na2CO3 solution is titrated against HCl, and the average titration result is 22.1 ml of HCl. What would the concentration be? I got 0.103molL^-1. Please help!

[emily_p]

Hello guys, I’ve got a titration question. It says a standard solution of Na2CO3 is prepared by dissolving 1.208g and 250ml water up the volumetric flask. 25ml aliquot of the Na2CO3 solution is titrated against HCl, and the average titration result is 22.1 ml of HCl. What would the concentration be? I got 0.103molL^-1. Please help!

Yep! I also got 0.103 M

n(Na2CO3) = 1.208/2(22.99)+12.01+3(16)
                   = 0.011397...
C(standard solution) = 0.011397.../0.25
                                  = 0.045589206...
Na2CO3 + 2HCl => 2NaCl + CO2 +H2O
n(Na2CO3) = 0.45589206...x 0.025
                    = 0.0011397...
n(HCl) = 2 x 0.0011397... (stoichiometric ratio)
            = 0.00227946...
C(HCl) = 0.00227946.../0.0021
     = 0.10314300...
      = 0.103 M (3 sf)

[3.14159265359]

Hi guys, not too sure how to work this one out

Thanks!

Hey, basically what you would do is look at the question and figure what it’s asking. They question is asking which of the 4 acids that were tested is the weakest acid? In other terms which of them has the highest ph. From the table we can see that Acid X has a ph of 3.8 which is the highest ph meaning weakest acid. We know the ph “graph” is 0-7 acidic, 7 neautral and 7-14 basic. And since all options have a ph<7 we know they’re all acidic, but acid X is the weakest acid as it has the highest ph. Therefore the answer is b.

Let me know if I don’t make any sense, I’ll try explaining in different way. And I’m pretty sure I repeated myself a few times so excuse that.

[KT Nyunt]

Hi guys, not too sure how to work this one out

Thanks!

I just had a look at the answer above me, I think there is more to answering this question, especially since they also give you the concentration of each of the monoprotic acids. We need to remember the fact that weak acids only partially ionise, whereas strong acids ionise completely, and we also need to recall that pH is a measure of Hydrogen ion concentration. Thus (as described in the answer above me), weaker acids have a higher pH as there are fewer Hydrogen ions but also the calculated pH (using the formula -log[H+]) should also be much lower than the actual pH (the pH in the table) as there aren't as many hydrogens ionised. 


If we used the ph formula and substitute the concentrations from the table into [H+] we'll see that W and Z are strong acids as the calculated pH is either the same (or very close) to the the actual pH, meaning all the hydrogens ionised.
However, if X were to be a strong acid, its pH should be 2 (using the formula). Likewise, if Y were to be a strong acid, its pH should be 1. As the pH given in the table is much higher than this, this indicates that these acids have only partially ionised and thus are weak acids.
Since there is a greater difference in pH in Y than X, I believe the answer should be Y.

Hope this helps! Let me know if this clarifies things.

[alisoneom]

Can I please have some solutions / working out for both parts of this question? Thank you 

A household cleaning agent contains a weak base of general formula NaX. 1.00 g of this compound was dissolved in 100.0 mL of water. A 20.0 mL sample of the solution was titrated with 0.1000 mol L–1 hydrochloric acid and required 24.4 mL of the acid for neutralisation.
(a) What is the Brönsted–Lowry definition of a base? [1 mark]
(b) What is the molar mass of this base? [3 marks]

[meditatingwalrus]

heyy i having some trouble with this MC. can someone help me out? thanks and much appreciated

[kauac]

heyy i having some trouble with this MC. can someone help me out? thanks and much appreciated

Hi...
Answer would be A, I assume.

Using flame colour, you can eliminate calcium (a red flame) and iron (no characteristic colour).
Then by referring to the solubility rules, you will find that all sulfates are soluble, except for barium, calcium, silver and lead.
Therefore, because the ion forms a ppt with Na2SO4, Barium2+ is the only feasible option left.

I'm sure there are other ways of working it out too. 

[meditatingwalrus]

heyy im asking wayy too many questions on this thread hahah! but,, can someone help me out with this question?

A gas is produced when 10.0 g of zinc is placed in 0.50 L of 0.20 mol L−1 nitric acid.
Calculate the volume of gas produced at 25°C and 100 kPa. Include a balanced chemical equation in your answer.

why is it that we use the moles from HNO3 and not Zn? couldnt i go v = vc x n , v = 0.153 x 24.79?

[KT Nyunt]

heyy im asking wayy too many questions on this thread hahah! but,, can someone help me out with this question?

A gas is produced when 10.0 g of zinc is placed in 0.50 L of 0.20 mol L−1 nitric acid.
Calculate the volume of gas produced at 25°C and 100 kPa. Include a balanced chemical equation in your answer.

why is it that we use the moles from HNO3 and not Zn? couldnt i go v = vc x n , v = 0.153 x 24.79?

Hi!
So they give you this information because we need to find the number of moles of nitric acid (HNO3) in this reaction as well as the number of moles of Zinc (Zn) we have, in order to see which one is the limiting reagent...

From the answers:
n(HNO3) = 0.1 mol [we find this by doing n = Cv]
n(Zn) = 0.153 mol [we find this by doing mass/molar mass]

As there are fewer moles of nitric acid than Zinc, this is the limiting reagent, meaning only 0.1 moles of HNO3 reacted with 0.1 moles of Zn and some Zn was left in excess.
From the equation, we can deduce the molar ratios, which will help us find the volume of the gas produced, H2. Now that we know 0.1 moles of HNO3 reacted, and the molar ratio between HNO3 and H2 is 2:1, the number of moles of H2 produced is 0.05 mol.
Then we put this into the equation n x 24.79, giving us the answer: 1.24L as required.

Hope this helps!