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Author Topic: HSC Chemistry Question Thread  (Read 1040720 times)  Share 

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jakesilove

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Re: Chemistry Question Thread
« Reply #270 on: April 23, 2016, 07:22:08 pm »
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Hi, just a bit stuck on this titration question - I don't really understand how to do it so any help would be appreciated.  :)

"The electrolyte in car batteries is sulfuric acid. A student decided to determine the concentration of this acid in a well-charged car battery by taking exactly 2 mL by pipette and titrating it with 1.16 mol/L sodium hydroxide solution. 17.1 mL was needed to reach the equivalence point. Calculate the molarity of the sulfuric acid in the battery."

Thanks!

Hey!

Below is the general approach for all titration questions. I would recommend becoming quite comfortable with the method. Hope this helps!



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Johny1234567

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Re: Chemistry Question Thread
« Reply #271 on: April 24, 2016, 02:03:46 am »
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What is the effect on solubility of adding concentrated sodium carbonate to the carbon dioxide equilibrium.

Na2CO3 + H3O^+ -> CO2(g) + 2Na^+ + H2O
So, since sodium carbonate reacts with hydronium, the disturbance is that it decreases the hydronium concentration. This means it needs to shift right to increase hydronium concentration, correct? While this would increase CO2(aq) thus increasing solubility, there would be an increase in CO2(g) as well due to the reaction as stated above. Would the increase in CO2(g) also be treated as a disturbance, thus leading to an increase in solubilty. Or, does it simply decrease carbon dioxide solubity since it's a gas. Need some help please !

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Re: Chemistry Question Thread
« Reply #272 on: April 24, 2016, 08:51:05 am »
+1
What is the effect on solubility of adding concentrated sodium carbonate to the carbon dioxide equilibrium.

Na2CO3 + H3O^+ -> CO2(g) + 2Na^+ + H2O
So, since sodium carbonate reacts with hydronium, the disturbance is that it decreases the hydronium concentration. This means it needs to shift right to increase hydronium concentration, correct? While this would increase CO2(aq) thus increasing solubility, there would be an increase in CO2(g) as well due to the reaction as stated above. Would the increase in CO2(g) also be treated as a disturbance, thus leading to an increase in solubilty. Or, does it simply decrease carbon dioxide solubity since it's a gas. Need some help please !

The carbon dioxide equilibrium is just this:
H2CO3(aq) ⇌ H2O(l) + CO2(g)

Therefore I believe either you started the question wrongly, or the question was given misleadingly.

Introduction of sodium carbonate (which is a basic substance) will automatically cause it to react with the carbonic acid. According to LCP, because the concentrations of carbonic acid falls down, the equilibrium will shift to the left to minimise this disturbance. This implies that more carbon dioxide will be converted to carbonic acid, and thus solubility should increase.

RuiAce

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Re: Chemistry Question Thread
« Reply #273 on: April 24, 2016, 08:54:00 am »
+1
hey, this is a really simply question but i can't manage to get my head around a few things:

The solubilities at 25C of three white barium salts, in grams per 100mL of water are:
Barium Sulfate 0.00025
Barium Nitrate 10.1
Barium Iodide 220

15g of a white powder was prepared by mixing 5.0g of each of these three barium salts. Your task to to obtain a sample of pure barium sulphate and a sample of virtually pure barium nitrate. Write the method for the separation process. Explain at each stage what happens to the mixture.

Firstly, when the question says to obtain PURE barium sulphate and VIRTUALLY PURE barium nitrate, how should that change the process you choose to use. Also, the answers specifically says that you must add more than 55mL of water, why is this so? And it also says after separating barium sulphate through filtration, barium nitrate will stay dissolved but barium iodide will crystallise out. Why is this so, doesn't barium iodide have a higher solubility. Thank you :)

Making an assumption that this is indeed preliminary chemistry.

If the solubility of barium sulfate is as poor as THAT compared to the other substances, then filtration is automatically the best way to go. This is because you will need TONS of water for all of that barium sulfate to dissolve, whereas the other substances only require few. Thus, filtration should be used to seperate barium sulfate without any further thought.

Filtration only separates the barium sulfate. The barium nitrate and barium iodide remain dissolved in solution. I have absolutely no clue where they gathered that barium iodide will crystallise out.

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Re: Chemistry Question Thread
« Reply #274 on: April 24, 2016, 09:08:06 am »
+1
Help with this ques thanks
0.132g of pure carboxylic acid (R-COOH) was dissolved in 25ml of water and titrated with 0.120mol/L NaOH solution
A volume of 14.80ml was required to reach the endpoint of the titration.
The carboxylic acid could be:
A)HCOOH
B)CH3COOH
C)C2H5COOH
D)C3H7COOH
ans:C


question 2: Oxalic acid C2O4H2
A 25.0ml solution of oxalic acid reacts completely with 15.0ml of 2.50mol/L of NaOH
what is the concentration of the oxalic acid solution?
ans: 0.750M

For Q2, oxalic acid is indeed, diprotic.
2 NaOH(aq) + C2O4H2(aq) → Na2C2O4(aq) + 2 H2O(l)

Hence the mole ratio is given
n(NaOH) = 2n(C2O4H2)
Thus, using n = CV
[NaOH] * V(NaOH) = 2* [C2O4H2] * V(C2O4H2)

2.50 mol L-1 * 15.0 mL = 2 * [C2O4H2] * 25.0 mL
[C2O4H2] = 0.75 mol L-1 as the answers requested
______________________________________

Q1 is one of those typical rigorous questions that the HSC puts in, where you essentially have to do up to 4 calculations to find the answer.
We wish to determine if methanoic acid (formic acid), ethanoic acid (acetic acid), propanoic acid or butanoic acid was used in the titration. We firstly note that ALL of the carboxylic acids are monoprotic.

Acid + NaOH(aq) → Sodium salt + H2O(l)

In every case, the mole ratio is 1 to 1.
Thus, we have
n(Acid) = n(NaOH)

We will apply n=CV to the sodium hydroxide, but n=m/M to the acid:
m/M = CV
0.132g / M = 0.0148 L * 0.120 mol L-1
M = 74.32432... g mol-1

Now that we know the molar mass, we can simply use the periodic table to find out which acid is the correct acid.

It turns out that propanoic acid is correct:
M = 2*12.01 + 5*1.008 + 12.01 + 16.00 + 16.00 + 1.008 = 74.078

The tiny amount of inaccuracy should've been anticipated since the start.
« Last Edit: April 24, 2016, 09:17:19 am by RuiAce »

Johny1234567

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Re: Chemistry Question Thread
« Reply #275 on: April 24, 2016, 12:50:49 pm »
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The carbon dioxide equilibrium is just this:
H2CO3(aq) ⇌ H2O(l) + CO2(g)

Therefore I believe either you started the question wrongly, or the question was given misleadingly.

Introduction of sodium carbonate (which is a basic substance) will automatically cause it to react with the carbonic acid. According to LCP, because the concentrations of carbonic acid falls down, the equilibrium will shift to the left to minimise this disturbance. This implies that more carbon dioxide will be converted to carbonic acid, and thus solubility should increase.
I thought the CO2 equilibrium consisted of 4 equations in the following sequence:
1. CO2(g) ⇌ CO2(aq)
2. CO2(aq) + H2O(l) ⇌ H2CO3(aq)
3. H2CO3(aq) ⇌ H+(aq) + HCO3-(aq)
4. HCO3-(aq) ⇌ H+(aq) + CO3^2-(aq)

RuiAce

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Re: Chemistry Question Thread
« Reply #276 on: April 24, 2016, 02:55:07 pm »
+1
The first step is trivially assumed. Writing that out is pointless.

We are talking about the carbon dioxide equilibrium, whereby the second equation (with carbon dioxide as a gas) is predominant. The third and fourth equations technically do occur as a result of Brønsted-Lowry acid base theory.

However, their presence is negligible to this question, as the carbonic acid equilibrium is only specific to the interaction of carbon dioxide with water. This produces a substance that is mostly acidic, not basic.

Worthwhile mention: the hydrogen carbonate ion buffer makes greater use of the generation of the carbonate ion, and the carbonic acid equilibrium
« Last Edit: April 24, 2016, 02:57:07 pm by RuiAce »

Johny1234567

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Re: Chemistry Question Thread
« Reply #277 on: April 24, 2016, 06:03:36 pm »
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Thank you :)

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Re: Chemistry Question Thread
« Reply #278 on: April 26, 2016, 10:56:20 pm »
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esters are polar but why are they non-water soluble?

RuiAce

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Re: Chemistry Question Thread
« Reply #279 on: April 27, 2016, 06:56:00 am »
+2
esters are polar but why are they non-water soluble?

Whilst the components of esters are quite appreciably polar, that is, the alkanol and alkanoic acid seperate, the ester itself is only a tiny bit polar. The polarity of the ester occurs as as the carboxylic acid group (-COOH) does contain one extra polar oxygen.

However, this is the only site of polarity. That OH part was dismantled when the ester was formed (note that water is a product of esterification). As this is the only site, the solubility of the substance will not be that great as soon as the chains start becoming longer in length. Much more of the molecule consists of carbon chains which are non-polar and do not readily dissolve into water.

(Some of the very short chained esters still somewhat dissolve into water.)

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Re: Chemistry Question Thread
« Reply #280 on: April 27, 2016, 02:04:24 pm »
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Whilst the components of esters are quite appreciably polar, that is, the alkanol and alkanoic acid seperate, the ester itself is only a tiny bit polar. The polarity of the ester occurs as as the carboxylic acid group (-COOH) does contain one extra polar oxygen.

However, this is the only site of polarity. That OH part was dismantled when the ester was formed (note that water is a product of esterification). As this is the only site, the solubility of the substance will not be that great as soon as the chains start becoming longer in length. Much more of the molecule consists of carbon chains which are non-polar and do not readily dissolve into water.

(Some of the very short chained esters still somewhat dissolve into water.)

Essentially there is still partial solubility but as the chain length increases this solubility involving -COOH carboxylic group can be considered negligible
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RuiAce

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Re: Chemistry Question Thread
« Reply #281 on: April 27, 2016, 08:17:27 pm »
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Essentially there is still partial solubility but as the chain length increases this solubility involving -COOH carboxylic group can be considered negligible
An ester doesn't have a -COOH functional group that's an alkanoic acid

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Re: Chemistry Question Thread
« Reply #282 on: April 27, 2016, 08:26:51 pm »
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An ester doesn't have a -COOH functional group that's an alkanoic acid

I think he's referring to the COO section when an ester is formed through the condensation reaction between an alkanoic acid and alkanol.
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Re: Chemistry Question Thread
« Reply #283 on: April 27, 2016, 09:07:53 pm »
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I think he's referring to the COO section when an ester is formed through the condensation reaction between an alkanoic acid and alkanol.

The "COO" group (obviously it isn't really one) loses most of it's impact mainly cause it's right in the middle of the molecule. Partial solubility is definitely present but if it's more towards the middle it is slightly limited from impacting as much as if it were on the end.

If it were on the end, an entire tail (or head) would be dissolved. Kinda the principles behind how soaps work (industrial chemistry option for the HSC). But being in the middle it will hardly dissolve a bit as the molecule would have to bend too much for the solubility to be appreciable.

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Re: Chemistry Question Thread
« Reply #284 on: April 30, 2016, 12:30:18 pm »
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Hey!
My second 3/4 chem sac is regarding aspirin and oil of wintergreen (methyl salicylate).
Is they key structural difference between the two the fact that aspirin lacks a distinct hydroxyl functional group or the positioning of the methyl functional group on aspirin?