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March 28, 2024, 11:30:32 pm

Author Topic: VCE Chemistry Question Thread  (Read 2313104 times)  Share 

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Edward21

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Re: VCE Chemistry Question Thread
« Reply #75 on: January 09, 2014, 08:55:23 pm »
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Just a little heads up, in the new 2013 study design, titration curves are mentioned explicitly in a new dot-point, I think you can interpret that as: they're trying to make it clear that titration curves need to be understood well, and you could potentially be asked to graph one. They'd be a little tricky though, they'd use a weak acid or base. Ie. Titration curve of NaOH and CH3COOH. Where at the equivalence point you have water and the CH3COO- ethanoate ion, this is the conjugate base of the acid CH3COOH which accepts H+ from H2O forming OH- ions (at least that's what I understood from the textbook..)

That's why the pH isn't 7 or "neutral" at 25 degrees, but above 7 at 25 degrees with more [OH-]>[H+] in the solution. It could be a 3 mark question where 1 is for the correct shape --> added base (NaOH) and the line keeps on rising or pH rises with more basic titre volume added, 2nd mark for the equivalence point being over 7 due to the reason I stated, and the 3rd mark for the correct aligning of the equivalence point (to neutralise the acid) with the volume of NaOH as per calculation.

I expected this to come up in the 2013 exam, so just familiarise yourself with doing this just in case for the 2014 exam :)
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Strawberrry

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Re: VCE Chemistry Question Thread
« Reply #76 on: January 11, 2014, 06:17:10 pm »
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Thank you everyone!!

When you dilute something, does the amount, in mol, stay the same?

DJA

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Re: VCE Chemistry Question Thread
« Reply #77 on: January 11, 2014, 06:57:59 pm »
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Thank you everyone!!

When you dilute something, does the amount, in mol, stay the same?

Yes.
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Snorlax

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Re: VCE Chemistry Question Thread
« Reply #78 on: January 11, 2014, 07:15:45 pm »
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On that note, what's the purpose of diluting a solution before like a titration?
Is it just to get enough volume?
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Yacoubb

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Re: VCE Chemistry Question Thread
« Reply #79 on: January 11, 2014, 07:20:05 pm »
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On that note, what's the purpose of diluting a solution before like a titration?
Is it just to get enough volume?

Well you don't have to dilute something. But the reason dilutions are made is usually because you have a stock of 12M HCl for example, and you want about 0.5M only. Therefore, you dilute! :)

Also, the smaller the concentration, the more credible your experimental results.

brightsky

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Re: VCE Chemistry Question Thread
« Reply #80 on: January 11, 2014, 09:06:45 pm »
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On that note, what's the purpose of diluting a solution before like a titration?
Is it just to get enough volume?

If the concentration of the titrant is too high, then the titre would, in all likelihood, be too small, since then a smaller volume of the titrant is needed for the equivalence point to be reached. We usually want the titre to be of moderate measure (not too small and not too big), so in order to achieve this, we dilute before the titration.
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Re: VCE Chemistry Question Thread
« Reply #81 on: January 11, 2014, 09:09:51 pm »
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If the concentration of the titrant is too high, then the titre would, in all likelihood, be too small, since then a smaller volume of the titrant is needed for the equivalence point to be reached. We usually want the titre to be of moderate measure (not too small and not too big), so in order to achieve this, we dilute before the titration.

So does that mean that if the concentration is too high, the titre value will be very low, and so the accuracy and credibility of results is less than if the titre value was higher, achieved by diluting your titrant??

Just need a clarification! Thanks brightsky!

brightsky

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Re: VCE Chemistry Question Thread
« Reply #82 on: January 11, 2014, 09:15:57 pm »
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So does that mean that if the concentration is too high, the titre value will be very low, and so the accuracy and credibility of results is less than if the titre value was higher, achieved by diluting your titrant??

Just need a clarification! Thanks brightsky!

Pretty much! We wouldn't want the concentration of the titrant to be too low either, because then the titre would be too large. Consider what would happen if you had a titre of 60 mL and a burette with a maximum capacity of 50 mL...
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Re: VCE Chemistry Question Thread
« Reply #83 on: January 11, 2014, 09:18:03 pm »
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Pretty much! We wouldn't want the concentration of the titrant to be too low either, because then the titre would be too large. Consider what would happen if you had a titre of 60 mL and a burette with a maximum capacity of 50 mL...

LOL!

Thanks brightsky.


PsychoT

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Re: VCE Chemistry Question Thread
« Reply #84 on: January 12, 2014, 06:09:34 pm »
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I'm baaaccck.

Anyone care to help me out with this one?

If 50cm3 of 0.1M HNO3 is mixed with 60cm3 of 0.1M Ca(OH)2, what volume of 0.050M H2SO4 would be required to neutralize the solution?

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PsychoT

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Re: VCE Chemistry Question Thread
« Reply #85 on: January 12, 2014, 06:19:52 pm »
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Got one more, the answers in this little booklet I have are pissing me off, I reckon most have mistakes.

What is the mass of the Phosphorus Oxide P4O10 that is produced from the reaction between 4g of P406 and 4g of I2?

5P406 + 8I2 --> 4P2I4 + 3P4O10
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SocialRhubarb

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Re: VCE Chemistry Question Thread
« Reply #86 on: January 12, 2014, 06:31:40 pm »
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If 50cm3 of 0.1M HNO3 is mixed with 60cm3 of 0.1M Ca(OH)2, what volume of 0.050M H2SO4 would be required to neutralize the solution?

50 cm3 is equivalent to 0.05 L. 0.05 L of 0.1M HNO3 contains 0.005 mol of H+.

60 cm3 is equivalent to 0.06 L. 0.06 L of 0.1M Ca(OH)2 contains 0.012 mol of OH-, as each mole of calcium hydroxide produces two moles of hydroxide ion.

Hence, there will be 0.007 mol of hydroxide ions left in solution.

0.050 M H2SO4 has a H+ concentration of 0.10 M. Hence, 0.07 L of solution is required, which is equivalent to 70 cm3.
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PsychoT

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Re: VCE Chemistry Question Thread
« Reply #87 on: January 12, 2014, 07:05:54 pm »
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50 cm3 is equivalent to 0.05 L. 0.05 L of 0.1M HNO3 contains 0.005 mol of H+.

60 cm3 is equivalent to 0.06 L. 0.06 L of 0.1M Ca(OH)2 contains 0.012 mol of OH-, as each mole of calcium hydroxide produces two moles of hydroxide ion.

Hence, there will be 0.007 mol of hydroxide ions left in solution.

0.050 M H2SO4 has a H+ concentration of 0.10 M. Hence, 0.07 L of solution is required, which is equivalent to 70 cm3.

Thankyou.
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eagles

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Re: VCE Chemistry Question Thread
« Reply #88 on: January 12, 2014, 07:28:32 pm »
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In the Heinemann 2 textbook, pg32, there is a table of sample titration results.
I'm not sure why the initial burette reading changes for each titration number.

Can someone please explain? Thank you  :D

brightsky

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Re: VCE Chemistry Question Thread
« Reply #89 on: January 12, 2014, 07:32:49 pm »
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We expect the initial burette readings to change, because during each titration, a certain volume of solution is dispensed from the burette. For this reason, it is natural that the initial burette reading for the next titration is lower.
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