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March 28, 2024, 10:57:24 pm

Author Topic: VCE Chemistry Question Thread  (Read 2313081 times)  Share 

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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #15 on: December 28, 2013, 03:46:31 pm »
+3
Please help!  :) (this thread kind of died, not sure where to post haha)
A solution of potassium permanganate can be standardised using pure iron wire. in a particular experiment 0.317 g of iron wire was dissolved such that Fe2+ ions were formed, and the resulting solution was made up to 250.0 mL.
20ml aliquots of this solution were then taken, with 11.72ml of the permanganate solution being required.
calculate the molarity of the permanganate solution.

So a Volumetric analysis question:
The reaction between the potassium permanganate and the iron wire can be written omitting spectator ions in the ionic form as follows (Due to the potassium permanganate being composed of K+ and MnO4-)

MnO4-  + Fe2+ ---> Fe(MnO4)2
If we balance this equation we put a 2 in front of the MnO4 and get a 2:1 mole ratio:
2MnO4-  + Fe2+ ---> Fe(MnO4)2
Keep this in mind for later.

Looking at the question lets convert everything to moles first as it is easier!
n(Fe2+)= 0.317/55.8 = 0.005681mol total in 250.0 mL of solution.

Thus in a 20mL aliquot, we use a down-scaling factor to find out the amount of Fe2+ in 20 mL
ie. n(Fe2+) in a 20mL aliquot = 20/250 x 0.005681 = 0.0004545 mol

Now remember the mole ratio worked out earlier, we can see that we need 2 times more MnO4- than Fe2+ for the reaction to go to completion (the 2:1 ratio).
Thus we can conclude:
n(MnO4-) required = 2 x n(Fe2+)
                                   = 2 x 0.0004545 = 0.0009090 mol

Now to find the concentration of the permanganate solution. We know that 11.72mL of the solution reacted to completion with the Fe2+ solution. We also know the mol which reacted from our working so far. So we are left with an arbitrary calculation:
C(permanganate solution) = 0.0009090/0.01172 = 0.07756M

The concentration of the permaganate solution is 0.07756M rounded to 4 sf.
Can anyone else confirm my answer by checking the caluculations?

(This should be right, disclaimer I haven't done volumetric analysis for ages so I may have errors but hopefully not ;))

Firstly...I don't quite get your first step, DJALogical. The reaction between iron(II) and potassium permanganate is not a precipitation reaction but a redox reaction. Permanganate is a pretty powerful oxidant, enough to oxidise iron(II) to iron(III).

MnO4-(aq) + 8H+(aq) + 5 Fe2+(aq) => 5 Fe3+(aq) + Mn2+(aq) +4H2O(l)

I can just write this equation out because I've dealt with this equation too many times in trial exams, and because you can just replace any electrons in the permanganate reduction equation with Fe(II) ions :D

So...0.317 g iron wire eh? Molar mass of Fe = 55.8 g/mol => n(Fe2+) = 0.317/55.8 mol = 5.681 mmol
As you make a 250 mL solution of this, and then take 20 mL of it, you really just have 20mL/250 mL*5.681 mmol = 0.45448 mmol
Mol ratio suggests that n(Fe2+)=5n(MnO4-) => n(MnO4-)=9.0896*10-5 mol

Concentration = n/V = 9.0896*10-5 mol / 0.01172 L = 7.76*10-3 M
Note that as the mass of the wire is to 3 s.f., the answer is only to 3 s.f.

Interesting. My answer differs from yours by a factor of ten exactly.
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DJA

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Re: VCE Chemistry Question Thread
« Reply #16 on: December 28, 2013, 03:52:23 pm »
0
Interesting. My answer differs from yours by a factor of ten exactly.

 :) lol this is what happens when I haven't done redox titrations since like midyear chem olympiad-totally missed on on the possibility of it being a redox reaction.
Yep since the mole ratio if you realise it is redox is 5:1 whereas if you incorrectly assume precip, you get a mole ratio of 1:2.
Otherwise theoretically the steps following the redox reaction are the same.

Thanks for the pick up!! I'm modifying my post so I dont screw over the next poor sod who comes along.
« Last Edit: December 28, 2013, 03:55:43 pm by DJALogical »
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DJA

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Re: VCE Chemistry Question Thread
« Reply #17 on: December 28, 2013, 03:59:53 pm »
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lzxnl just for my own understanding could you write out the two separate ionic equations for the permaganate and Iron (II) so I can see what they look like and balance out? I've gotten it written down on paper I just want to check if its right with you. :)
Its been ages since I did redox.
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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #18 on: December 28, 2013, 04:03:41 pm »
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lzxnl just for my own understanding could you write out the two separate ionic equations for the permaganate and Iron (II) so I can see what they look like and balance out? I've gotten it written down on paper I just want to check if its right with you. :)
Its been ages since I did redox.

MnO4-(aq) + 8H+(aq) + 5e- => Mn2+(aq) +4H2O(l)

5Fe2+(aq) => 5Fe3+(aq) + 5e-
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Re: VCE Chemistry Question Thread
« Reply #19 on: December 28, 2013, 04:06:07 pm »
+1
MnO4-(aq) + 8H+(aq) + 5e- => Mn2+(aq) +4H2O(l)

5Fe2+(aq) => 5Fe3+(aq) + 5e-

Cheers!

Also I wanted to ask, is there any 'atmospheric chemistry' in Chemistry 3/4? Because I hated that stuff in year 11 and am sincerely hoping I don't have to do it again.
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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #20 on: December 28, 2013, 04:15:02 pm »
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Nope. None of that rubbish at all. I hated it too.
If you're keen, look through last year's exam for a guideline on what's on the course now.
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Re: VCE Chemistry Question Thread
« Reply #21 on: December 28, 2013, 05:04:32 pm »
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Ohhhh, I clearly need to practice more haha. Thanks both of youse ^^

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Re: VCE Chemistry Question Thread
« Reply #22 on: December 29, 2013, 10:56:56 am »
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Hi guys, I'm stuck with this question.
A solution of hydrochloric acid of concentration 0.1M has a pH of 1. Ethanoic acid of the same concentration has a pH of approximately 3. Explain how this difference arises.

psyxwar

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Re: VCE Chemistry Question Thread
« Reply #23 on: December 29, 2013, 11:22:46 am »
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Hi guys, I'm stuck with this question.
A solution of hydrochloric acid of concentration 0.1M has a pH of 1. Ethanoic acid of the same concentration has a pH of approximately 3. Explain how this difference arises.
Ethanoic acid is a weak acid and unlike hydrochloric acid (a strong acid), does not completely ionise in water. Ionising less = less H+ in solution = higher pH. In a solution of HCl, most of the molecules are Cl- and H+, whereas in a solution of ethanoic acid, most of the molecules are ethanoic acid molecules (CH3COOH) that have not ionised.
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DJA

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Re: VCE Chemistry Question Thread
« Reply #24 on: December 29, 2013, 11:24:50 am »
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Hi guys, I'm stuck with this question.
A solution of hydrochloric acid of concentration 0.1M has a pH of 1. Ethanoic acid of the same concentration has a pH of approximately 3. Explain how this difference arises.

Hydrochloric acid is a strong acid. A strong acid is one which ionises completely in water. We use a straight arrow to denote this
HCl + H2O -> H3O+ + Cl-

Ethanoic acid is a weak acid. A weak acid does not completely ionise in water, we use a reversible arrow to denote it.
CH3COOH + H2O ↔ H3O+ + CH3COO-

The H+ or H3O+ species is formed when an acid reacts with water. pH for practical purposes is the measure of the concentration of the hydronium ion H3O+ (the negative logarithm of the hydronium ion activity). Since ethanoic acid is not completely ionized, there is less than the maximum possible number of hydronium ions in the solution (there are un-ionized ethanoic acid molecules floating around) hence making it a higher pH than hydrochloric acid which is fully ionized and has more hydronium ions  in the solution than ethanoic acid giving it a lower pH.

(Edit:psyxwar beat me to it but ill post anyway for the sake of completeness)
« Last Edit: December 29, 2013, 11:28:03 am by DJALogical »
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Re: VCE Chemistry Question Thread
« Reply #25 on: December 29, 2013, 05:02:46 pm »
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Hi, can you guys help me with this question:
Write a balanced ionic equation for the reaction between iodine and sodium thisulfate (Na2S2O3)
I2 (aq) + 2Na2S2O3 (aq) -> 2NaI (aq) + Na2S4O6 (aq)
Thanks for help!!
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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #26 on: December 29, 2013, 05:44:37 pm »
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In this reaction, the sodium ions don't really do anything so your ionic reaction is I2(aq)+2S2O32-(aq) => 2I-(aq) + S4O62-(aq)
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Re: VCE Chemistry Question Thread
« Reply #27 on: January 02, 2014, 05:44:22 pm »
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Hi guys, I would love some help with this question. Ive already got the worked out solutions but i still dont understand it.
25.00ml of a 0.100 M solution of HCl is added to 25.00ml of a 0.180 M solution of naOH.
The concentration of OH-(aq) remaining in the solution, in M is

A. 0.0400

Thanks!  :)

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Re: VCE Chemistry Question Thread
« Reply #28 on: January 02, 2014, 06:04:38 pm »
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25.00mL of a 0.100 M solution of HCl is added to 25.00mL of a 0.180 M solution of NaOH. Find the concentration of OH-(aq) remaining in the solution in M.

The ionic equation for this reaction is:

H+(aq) + OH- (aq) --> H2O(l)

Note that Na+(aq) and Cl-(aq) are merely spectator ions.

n(H+) = n(HCl) = c*V = 0.02500*0.100 = 0.00250 mol
n(OH-) = n(NaOH) = c*V = 0.02500*0.180 = 0.00450 mol

It is clear that H+ is the limiting reagent. Hence, n(OH-) remaining after the reaction goes to completion is 0.00450 - 0.00250 = 0.00200.

This means that c(OH-) = n/V = 0.00200/0.05000 = 0.0400 M, as required.



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Re: VCE Chemistry Question Thread
« Reply #29 on: January 04, 2014, 10:35:39 am »
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When a certain non-metal whose formula is X8 burns in air XO3 forms. Write a balanced equation for this reaction. If 120.0g of oxygen gas is consumed completely, along with 80.0g of X8, identify element X.

Is this balanced equation correct: X8 (g) + 12 O2 (g) -> 8XO3 (g)
I'm not sure how to proceed with this question....

Your help is appreciated   :D