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March 28, 2024, 11:41:43 pm

Author Topic: VCE Chemistry Question Thread  (Read 2313115 times)  Share 

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soNasty

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Re: VCE Chemistry Question Thread
« Reply #1125 on: July 08, 2014, 12:13:26 pm »
+1
[H3O+]=[OH-] in water at 25 degrees celcius.
When you increase the temperature, there is a higher amount of [H+] ions, therefore causing the pH of the water to decrease and hence become acidic.

nhmn0301

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Re: VCE Chemistry Question Thread
« Reply #1126 on: July 08, 2014, 12:15:58 pm »
+2
hello
how would i do this question 14c) and 14d) i know 14 a and b are increase
for 14c) if both h30+ and OH- ions concerntration increases how does that equate to an overall ph decrease? cause i know the increase h30+ concerntration would account for the decreased ph, but wouldnt the increase oh- then counteract that?
for 14d) if 14c) is in fact ph decrease, wouldnt that mean the water becomes acidic, but then i know that h30+ concerntration = oh- concerntration therefore it has to be neutral, but then the ph decreased?

so confused????
A solution is considered as neutral when pH=7 at 25C . When you increase, or decrease the temperature outside 25C, a solution is considered neutral when [H+] = [OH-]. Hence, you can have a solution of pH = 6 at a given temperature but still called neutral.
Kw = [H30+] x [OH-] = 10^(-14) M at 25C
When you increase the temperature, forward reaction is favoured, hence, you obtained more of both [H30+] and [OH-]. This will change the value of Kw (that's why like all equilibrium constant, Kw is temperature dependent).
As mentioned above, because in a neutral solution, you have equal amount of hydronium and hydroxide, the Kw at any given temperature is
Kw= [H30+] x [OH-] = [H30+]^2
Since Kw is increasing when you apply heat, let's make up a larger Kw value and see how that changes the pH value.
New Kw = 51.3 x 10^(-14) M
       [H3O+] = *sqrt* 51.3 x 10^(-14)
       [H30+] = 7.16 x 10 ^ (-7)
Hence, pH = 6.14
In brief, even if you increase the temperature, the solution is still neutral despite the fact that it's pH is decreasing. pH 7 of a neutral solution only holds at 25C.
Correct me if I'm wrong.
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hyunah

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Re: VCE Chemistry Question Thread
« Reply #1127 on: July 08, 2014, 01:01:08 pm »
0
thank you

can you please also explain 14c) why there a ph decrease?
 if both h30+ and OH- ions concerntration increases how does that equate to an overall ph decrease? cause i know the increase h30+ concerntration would account for the decreased ph, but wouldnt the increase oh- then counteract that?

lzxnl

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Re: VCE Chemistry Question Thread
« Reply #1128 on: July 08, 2014, 02:04:28 pm »
+1
pH actually has no direct relationship with the concentration of hydroxide. It's JUST the H+ concentration; the corresponding increase in OH- with a higher temperature only means your solution is still neutral. As stated above, the increase in both H+ and OH- concentrations is a result of an increase in Kw. Normally we associate increasing [H+] with decreasing [OH-] because Kw is normally constant. It's not now.
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hyunah

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Re: VCE Chemistry Question Thread
« Reply #1129 on: July 09, 2014, 09:51:08 am »
0
thank you :)

can someone please explain the mistakes in my ans?

Sodium carbonate is readily soluble in water but calcium
carbonate is insoluble. A solution of sodium carbonate was
found to have a pH of 8. A quantity of calcium nitrate was
added to the solution. Explain the
effect the addition of calcium nitrate would have on the pH
of the solution.

na2co3 + ca(no3)2 (mixed with water ) --> caco3 + nano3
becasue caco3 are insoluble, the only ions left in the mixture is na+ and no3-
oh- ions from water react with na+ and H react with no3-
wheres the effect on ph?




lzxnl

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Re: VCE Chemistry Question Thread
« Reply #1130 on: July 09, 2014, 11:47:41 am »
+1
thank you :)

can someone please explain the mistakes in my ans?

Sodium carbonate is readily soluble in water but calcium
carbonate is insoluble. A solution of sodium carbonate was
found to have a pH of 8. A quantity of calcium nitrate was
added to the solution. Explain the
effect the addition of calcium nitrate would have on the pH
of the solution.

na2co3 + ca(no3)2 (mixed with water ) --> caco3 + nano3
becasue caco3 are insoluble, the only ions left in the mixture is na+ and no3-
oh- ions from water react with na+ and H react with no3-
wheres the effect on ph?

I don't quite get what you're trying to do.

You have a few equilibria at play here.
Na2CO3(aq) + H2O(l) <=> NaHCO3(aq) + NaOH(aq)
Ca2+(aq) + CO32-(aq) <=> CaCO3(s)

Add calcium nitrate => increase [Ca2+] => second reaction goes more towards the right => decrease carbonate concentration => first reaction goes backwards to reform carbonate => hydroxide ions concentration drops => pH decreases
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soNasty

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Re: VCE Chemistry Question Thread
« Reply #1131 on: July 09, 2014, 05:39:04 pm »
0
so i just did a practise exam and stumbled upon this question, which i didnt really know how to do. Could someone please give me insight as to how to tackle these sort of questions where you have an acid thats mono/di/triprotic etc and how mole calculations would differ?


here's the answers, sorry for the pics being so big lol

« Last Edit: July 09, 2014, 05:41:02 pm by soNasty »

Reus

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Re: VCE Chemistry Question Thread
« Reply #1132 on: July 09, 2014, 06:06:42 pm »
0
Would it be correct to say, that buffers ONLY occur with weak acids and its conjugate base, as if it were strong, it would ionise completely?

Thanks.
2015: Bachelor of Science & Bachelor of Global Studies @ Monash University

alchemy

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Re: VCE Chemistry Question Thread
« Reply #1133 on: July 09, 2014, 06:12:39 pm »
+1
so i just did a practise exam and stumbled upon this question, which i didnt really know how to do. Could someone please give me insight as to how to tackle these sort of questions where you have an acid thats mono/di/triprotic etc and how mole calculations would differ?

What I would do is pick any mono/di/tri protic acid that you know and calculate the concentration of them for the appropriate section in the table.
For example, for the first row of the table ("if the acid is monoprotic"), choose HCl to be the unknown acid in the question (since it's a monoprotic acid).
We can then write it's reaction with sodium carbonate in a balanced chemical equation as 2HCl + Na2CO3 --> 2NaCl + H2O + CO2
Therefore, the number of moles of the monoprotic acid (HCl) is 2*n(Na2CO3). From there, you can proceed to find the concentration of the diluted acid as well as the concentration of the original acid.
For the next row, pick any old diprotic acid you know (e.g. H2SO4) and repeat the above steps. Same applies for when it is a tri protic acid ;)

soNasty

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Re: VCE Chemistry Question Thread
« Reply #1134 on: July 09, 2014, 06:17:53 pm »
0
Thanks alchemy!!  ;D

hyunah

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Re: VCE Chemistry Question Thread
« Reply #1135 on: July 10, 2014, 03:46:47 pm »
0
can someone please explain to me how to do part c) and e)?

thank you

BLACKCATT

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Re: VCE Chemistry Question Thread
« Reply #1136 on: July 10, 2014, 07:38:37 pm »
0
Just wondering whether its better to finish all unit 3 practice exams now or save it for later? What do you guys think?

Yacoubb

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Re: VCE Chemistry Question Thread
« Reply #1137 on: July 10, 2014, 09:03:50 pm »
0
Just wondering whether its better to finish all unit 3 practice exams now or save it for later? What do you guys think?

I'm doing a few now, here and there for a scope of questions. But, I'm saving the majority for later.

Reus

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Re: VCE Chemistry Question Thread
« Reply #1138 on: July 10, 2014, 11:25:13 pm »
0
Could someone please explain why halving the volume (doubling the pressure) of an equilibrium will have the effect of a net forward reaction. Whereas when increasing pressure on the equilibrium reaction, there will be no effect?

Thanks :)
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soNasty

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Re: VCE Chemistry Question Thread
« Reply #1139 on: July 11, 2014, 12:18:11 am »
+1
When there is a reaction with more molecules on the LEFT as opposed to the right, increasing the pressure will cause a net forward reaction, to the side where there are less particles, to oppose that change in pressure.
eg. 2A+4B <===>2C+D

When there are equal numbers of particles on either side of a reaction, increasing the pressure will have no effect as the system cannot oppose that change.
eg. A+B<====>C+D