Post by: Andiio on January 26, 2011, 03:35:58 pm
b) What amount, in mole, of hydrochloric acid will have been used up when the reaction is complete?
I think I'm confusing myself by overthinking/overanalysing the question, but thanks!
Post by: thushan on January 26, 2011, 03:44:29 pm
This is a redox reaction, where H+ is reduced to H2 gas. Also, by balancing the equation we know that n(HCl)[reacted] = 2n(Mg).
So n(Mg) = 0.012/24.3 = 4.94 x 10^(-4) mol.
Hence n(HCl) = 9.9 x 10^(-4) mol.
Hence 9.9 x 10^04 mol of HCl would have been used up.
Post by: Andiio on January 26, 2011, 03:47:29 pm
We know that HCl (aq) is in excess.
This is a redox reaction, where H+ is reduced to H2 gas. Also, by balancing the equation we know that n(HCl)[reacted] = 2n(Mg).
So n(Mg) = 0.012/24.3 = 4.94 x 10^(-4) mol.
Hence n(HCl) = 9.9 x 10^(-4) mol.
Hence 9.9 x 10^04 mol of HCl would have been used up.
But when you balance the eq, isn't it:
2Mg (s) + 2HCl (aq) -> 2MgCl (aq) + H2(g) ?
thus wouldn't n(HCl) = n(Mg) = 4.94 x 10^-4 mol?
Post by: stonecold on January 26, 2011, 03:49:26 pm
We know that HCl (aq) is in excess.
This is a redox reaction, where H+ is reduced to H2 gas. Also, by balancing the equation we know that n(HCl)[reacted] = 2n(Mg).
So n(Mg) = 0.012/24.3 = 4.94 x 10^(-4) mol.
Hence n(HCl) = 9.9 x 10^(-4) mol.
Hence 9.9 x 10^04 mol of HCl would have been used up.
But when you balance the eq, isn't it:
2Mg (s) + 2HCl (aq) -> 2MgCl (aq) + H2(g) ?
thus wouldn't n(HCl) = n(Mg) = 4.94 x 10^-4 mol?
Nah.
It is this:
Mg (s) + 2HCl (aq) -> MgCl2 (aq) + H2(g)
Post by: Andiio on January 26, 2011, 03:49:54 pm
We know that HCl (aq) is in excess.
This is a redox reaction, where H+ is reduced to H2 gas. Also, by balancing the equation we know that n(HCl)[reacted] = 2n(Mg).
So n(Mg) = 0.012/24.3 = 4.94 x 10^(-4) mol.
Hence n(HCl) = 9.9 x 10^(-4) mol.
Hence 9.9 x 10^04 mol of HCl would have been used up.
But when you balance the eq, isn't it:
2Mg (s) + 2HCl (aq) -> 2MgCl (aq) + H2(g) ?
thus wouldn't n(HCl) = n(Mg) = 4.94 x 10^-4 mol?
Nah.
It is this:
Mg (s) + 2HCl (aq) -> MgCl2 (aq) + H2(g)
OMG WOOPS FORGOT ABOUT VALENCIES LOL
Post by: Andiio on January 26, 2011, 04:41:41 pm
The concentration of potassium ions in the resultant solution, in mole per litre, is?
I've written the equation - KOH + HCl -> KCl + H2O ; but mind-blanked and not sure what to do.
Thanks!
Post by: pi on January 26, 2011, 04:53:13 pm
KOH + HCl --> KCl + H2O
n(KOH) = cV = 0.30(0.0200) = 0.006 mol
V(final products) = 40.0 mL
c(K+) = n/V = 0.006/0.040 = 0.15 mol/L
Post by: luken93 on January 26, 2011, 04:59:16 pm
n(KOH) = 0.30 x 0.020 = 0.006 mol
According to the equation, the ratio of the reactants is 1:1, meaning that the KOH is in excess.
See if you can go from there...
Post by: thushan on January 26, 2011, 05:02:25 pm
So just find n(K+) by finding n(KOH), and from there you can find [K+].
Post by: pi on January 26, 2011, 05:05:48 pm
No need to even consider the reaction, because K+ is a spectator ion and would not take part in the reaction.
So just find n(K+) by finding n(KOH), and from there you can find [K+].
Does that mean I was on the right track?
I have forgotten all this stuff completely :(
Post by: luken93 on January 26, 2011, 05:13:00 pm
Yeah yours is correct, as you've found the n(K+) to start with / total volume = concentration in solutionNo need to even consider the reaction, because K+ is a spectator ion and would not take part in the reaction.
So just find n(K+) by finding n(KOH), and from there you can find [K+].
Does that mean I was on the right track?
I have forgotten all this stuff completely :(
Moderator action: removed real name, sorry for the inconvenience
Post by: Andiio on January 28, 2011, 10:17:06 pm
Here's another question - just want to see how everyone here would approach it.
To determine the empirical formula of an ester (containing only carbon, hydrogen and oxygen), 1.02g of the ester was burnt completely in excess oxygen. The only products formed were 2.20 g of carbon dioxide and 0.90 g of water vapour.
Calculate the mass of carbon in 1.02g of the compound.
Thanks!
Post by: samiira on January 28, 2011, 11:03:49 pm
mass of CO2 = 2.20g
mole of CO2 = mass/Mr = 2.2/44 = 0.050 moles
mole of Carbon = mole of CO2 = 0.050 moles
therefore mass of Carbon = 0.05 x 12 = 0.60g
i guess i cud be wrong ??
Post by: man0005 on January 28, 2011, 11:20:05 pm
with luken's way, don't think you can assume that all the oxygen in the products will be present in the ester since it is burned in excess oxygen.
i could be wrong though
Post by: david10d on January 28, 2011, 11:33:25 pm
I thought you couldn't assume that the mole ratio is correct without knowing the full equation and empirical formula of the ester (may or may not be correct).
Post by: luken93 on January 28, 2011, 11:37:54 pm
i though it would be :Using that method, add up all the components and see if it equals the mass of the ester?
mass of CO2 = 2.20g
mole of CO2 = mass/Mr = 2.2/44 = 0.050 moles
mole of Carbon = mole of CO2 = 0.050 moles
therefore mass of Carbon = 0.05 x 12 = 0.60g
i guess i cud be wrong ??
Shouldn't luken's method be correct as he has used the limiting reactant? That's how I would've worked it out.Yeah I found out the empirical formula of the ester, and then inserted it into the chemical equation before balancing it....
I thought you couldn't assume that the mole ratio is correct without knowing the full equation and empirical formula of the ester (may or may not be correct).
(not saying my approach is correct until it is proven, but it seemed right to me haha)
Post by: luken93 on January 28, 2011, 11:42:57 pm
Post by: david10d on January 28, 2011, 11:48:48 pm
Post by: luken93 on January 28, 2011, 11:49:55 pm
mol ratio is incorrect. oxygens don't balance.yeah fixed it up now
Post by: luken93 on January 28, 2011, 11:54:42 pm
Source: 2002 VCAA Exam 1
Post by: Pixon on January 29, 2011, 12:02:18 am
luken, you were on the right track, but to avoid the excess oxygen problem you want to find the mass of the carbon and hydrogen from the amounts you calculated. Then subtract those two masses from the 1.02g of ester to find the mass of oxygen and thus the amount of oxygen. Then everything else you did was correct. :)
If you're curious it's C5H10O2.
Post by: Andiio on January 29, 2011, 12:45:20 am
THANKS ANYWAYS :D
Post by: luken93 on January 29, 2011, 08:32:46 am
Yeh that's a rather silly question. It's supposed to waste time and trick people into finding the EF (which is more typical question).yeah i realised later on that you could d that, what an annoying q though haha
luken, you were on the right track, but to avoid the excess oxygen problem you want to find the mass of the carbon and hydrogen from the amounts you calculated. Then subtract those two masses from the 1.02g of ester to find the mass of oxygen and thus the amount of oxygen. Then everything else you did was correct. :)
If you're curious it's C5H10O2.
Post by: Andiio on February 14, 2011, 10:07:49 pm
Not sure if i'm right :P
Q: Some rocks were thought to consist of insoluble silica (SiO2) and calcium carbonate (CaCo3). The fraction of CaCO3 in an 8.64g sample of the crushed rock was determined by mixing the sample with excess hydrochloric acid. The acid reacts with CaCO3 according to the following equation.
2HCl(aq) + CaCO3(s) -> CaCl2(aq) + H2O (l) + CO2(g)
b) Excess ammonium oxalate solution was added to the filtered solution. The calcium ions present precipitate as CaC2O4 . H2O.
The CaC2O4 . H2O was collected by filtration, washed and dried. It was then heated to convert it to CaO (molar mass 56.1 g/mol) and a mass of 3.87 g was obtained.
Using this mass of CaO, calculate the percentage of CaCO3 in the rock sample.
Thanks!!
Post by: m@tty on February 14, 2011, 10:24:10 pm
Post by: Andiio on February 14, 2011, 11:40:26 pm
Just another one:
A lawn fertiliser contains mainly ammonium sulfate ((NH4)2SO4) and potassium sulfate (K2SO4). A 1.87 g sample of the fertiliser was dissolved in water and the sulfate precipitated as barium sulfate (BaSO4). The dried BaSO4 had a mass of 2.41g.
a) Write a balanced ionic equation for the precipitation reaction.
Really noob question, but i'm not sure >_<
THanks!
EDIT: Is it just Ba ions + sulfate ions = BaSO4? cept the balancing.. hmm
Post by: Water on February 14, 2011, 11:47:41 pm
A lawn fertiliser contains mainly ammonium sulfate ((NH4)2SO4) and potassium sulfate (K2SO4). A 1.87 g sample of the fertiliser was dissolved in water and the sulfate precipitated as barium sulfate (BaSO4). The dried BaSO4 had a mass of 2.41g.
a) Write a balanced ionic equation for the precipitation reaction.
Really noob question, but i'm not sure >_<
THanks!
EDIT: Is it just Ba ions + sulfate ions = BaSO4? cept the balancing.. hmm
Somehow, I recognize this problem, from Heineman?
Yup :)
Post by: Andiio on February 14, 2011, 11:49:35 pm
Just another one:
A lawn fertiliser contains mainly ammonium sulfate ((NH4)2SO4) and potassium sulfate (K2SO4). A 1.87 g sample of the fertiliser was dissolved in water and the sulfate precipitated as barium sulfate (BaSO4). The dried BaSO4 had a mass of 2.41g.
a) Write a balanced ionic equation for the precipitation reaction.
Really noob question, but i'm not sure >_<
THanks!
EDIT: Is it just Ba ions + sulfate ions = BaSO4? cept the balancing.. hmm
Somehow, I recognize this problem, from Heineman?
Yup :)
Haha yep!
So it's basically just Ba2+ (aq) + SO4 2- (aq) -> BaSO4 (s) ?
Bit paranoid about any numbers in front since the fertiliser has 2 'S' atoms if you catch my drift :|
Post by: Water on February 14, 2011, 11:54:36 pm
PS: The NH4 and The K would be spectators, if you just needed that extra clarity
Post by: Andiio on February 26, 2011, 01:15:05 pm
What is the significant of H+ and H3O+ in acid-base/general chem questions? Does it make much of a difference in an actual question?
Also, if not specified whether the substance is an acid or base, and the pH > 7, do you automatically assume that it is a base?
So say for example an unknown substance attributed a pH of 10; does that mean it's concentration of H+, i.e. C[H+] = 10^-10?
Post by: pi on February 26, 2011, 01:16:54 pm
Also, if not specified whether the substance is an acid or base, and the pH > 7, do you automatically assume that it is a base?Yes, but only at 25 degrees Celsius
So say for example an unknown substance attributed a pH of 10; does that mean it's concentration of H+, i.e. C[H+] = 10^-10?
But be logical. If you have 0.000000000000001M HCl, that does not mean its pH = -log(0.000000000000001). Because then its pH would be >7, which is absurd.
Post by: nacho on February 26, 2011, 01:28:16 pm
Just got another few random questions that have been bugging me.Be careful with this, there is a difference between whether a substance is a base and whether it is Basic
What is the significant of H+ and H3O+ in acid-base/general chem questions? Does it make much of a difference in an actual question?
Also, if not specified whether the substance is an acid or base, and the pH > 7, do you automatically assume that it is a base?
So say for example an unknown substance attributed a pH of 10; does that mean it's concentration of H+, i.e. C[H+] = 10^-10?
Post by: Andiio on February 26, 2011, 01:35:55 pm
That means that the C[H+] of NaOH = 10^-10, and thus the C[OH-] = 10^-4?
Post by: pi on February 26, 2011, 02:29:35 pm
So if you have, say, a solution of sodium hydroxide with a pH of 10.
That means that the C[H+] of NaOH = 10^-10, and thus the C[OH-] = 10^-4?
At 25 degrees Celsius, yes.
Post by: Andiio on February 26, 2011, 03:37:45 pm
20.0 mL of 0.10 M HCl is added to 20.0 mL of an unknown solution. The pH of the resulting mixture is measured and found to be 2.0. The unknown solution could have been:
A) 0.20 M NaOH
B) 0.10 M KOH
C) 0.04 M Ca(OH)2
D) 0.0010 M HCl
Post by: thushan on February 26, 2011, 04:31:22 pm
=> n(H+) final = 0.01 x 0.040 = 0.0004 mol
=> n(H+)react = 0.0020 - 0.0004 = 0.0016 mol
=> n(OH-) = 0.0016 mol
20 mL of 0.04 M Ca(OH)2 has 0.04 x 0.020 x 2 = 0.0016 mol.
Hence, C.
Post by: thushan on February 26, 2011, 06:55:30 pm
Post by: vea on February 26, 2011, 07:14:16 pm
Post by: Andiio on March 02, 2011, 07:26:33 pm
A student titrated an aliquot of standard sodium carbonate solution with hydrochloric acid in a burette. State whether the concentration determined for the hydrochloric acid would be likely to be higher, lower or unchanged compared with the actual value if the student had previously washed with water, but not dried, the following apparatus:
a the pipette used to deliver the aliquot of sodium carbonate solution
Wouldn't the presence of water within the pipette increase V(Na2CO3) and decrease C(Na2CO3) whilst having no effect on n(Na2CO3)?
Post by: Water on March 02, 2011, 07:30:34 pm
And we use the standard solution of sodium carbonate, combined with water, this would mean that there's lesser amount of mols than what should be there, as it is diluted, when titred against HCL.Say the titre of HCL was 23ml .
Then we'd have more mols within the HCL, than what would otherwise should be there.
Sorry about previous post: I read the question the other way around, ): , and didn't imagine it properly, thought sodium carbonate was unknown :2funny:
Post by: Andiio on March 02, 2011, 07:42:46 pm
I believe, that if the pipette still had water within it, that would mean that when you were going to deliver the sodium carbonate, the moles would be lesser.
Therefore, you would have a decrease lesser amount of HCL, titrating against it >>>And because , the n(NA2CO3) is unknown, we'd be using the HCL's titre and concentration to figure out sodium carbonate.
And if the water is further diluting the amount of mols within the sodium carbonate, that would mean, you would need less HCL to find the equivalence point.
Therefore underestimation (lower) of concentration than the actual value?
The addition of water doesn't affect the amount of moles in the sodium carbonate solution however..?
The answer is a higher concentration than the actual value
Post by: man0005 on March 02, 2011, 07:50:06 pm
therefore less HCl needed for equivalence point, so there would be a smaller titre
c = n / v
because v is smaller, c will be higher
so overestimation
Post by: luken93 on March 02, 2011, 07:57:51 pm
So if you delivered 20ml of 1.0M Na2CO3 as an aliqout, you would have 0.02 mol
Hence the concentration of HCl would be 0.02/0.02 = 1M HCl
If you delivered 19ml of 1.0M Na2CO3 as an aliqout (and 1ml leftover water in the pipette), you would have 0.019
Hence the concentration of HCl would be 0.019/0.02 = 0.95M HCl
But because you thought you had 0.2 mol of Na2CO3 to start with, you would think that the concentration is actually 1.0M instead of the real value of 0.95M
Post by: nacho on March 02, 2011, 09:13:59 pm
You can see that we need two mols of acid to react with 1 mol of base, hence acid will be the limiting reagent and base the excess
Post by: Andiio on March 03, 2011, 10:34:52 pm
C6H8O6(aq) + I2(aq) → C6H6O6(aq) + 2H+(aq) + 2I(aq)
If the maximum concentration of vitamin C is likely to be 0.00050 g mL1, describe how you would perform the analysis. You should mention:
the volume of fruit juice used
the maximum titre of iodine you would expect to obtain.
Not quite sure how one would find the specified volume of fruit juice used?
Thanks!
Post by: samiira on March 03, 2011, 10:44:27 pm
2.8 x 10^-6 x 1000 = 0.0028 mol/L of C6H8O6
If we take 50.0 mL of this solution there are :
0.050 L x 0.0028 mol/L = 0.00014 moles
Moles I2 needed = 0.00014
V of I2 = 0.00014 / 0.0100 M =0.014 L = 14 mL
Post by: Andiio on March 03, 2011, 11:08:06 pm
moles = 0.00050 g / 176 g/mol =2.8 x 10^-6
2.8 x 10^-6 x 1000 = 0.0028 mol/L of C6H8O6
If we take 50.0 mL of this solution there are :
0.050 L x 0.0028 mol/L = 0.00014 moles
Moles I2 needed = 0.00014
V of I2 = 0.00014 / 0.0100 M =0.014 L = 14 mL
Mm, but I'm just not sure why we take 50.0 mL? o_O
Post by: Andiio on March 15, 2011, 08:10:04 pm
Q: A student titrated an aliquot of standard sodium carbonate solution with hydrochloric acid in a burette. State whether the concentration determined for the hydrochloric acid would be likely to be higher, lower or unchanged compared with the actual value if the student had previously washed with water, but not dried, the following apparatus:
a) The pipette used to deliver the aliquot of sodium carbonate solution
b) The burette.
For part a), what I reasoned was that, the presence of water would lower the concentration of sodium carbonate solution, and as a result, the V(HCl) needed in the titration to reach the equivalence point would be lower, thus yielding a lower titre - V(HCl). And because C = n/V, since V(HCl) is lower, it would yield a higher concentration. I am not really sure if this is correct as I was thinking that maybe the student would have delivered a lower amount of HCl (and thus a lower amount of moles of HCl) in the pipette as there would be water 'taking up' a few mL's of space.
For part b), I wrote that the water would dilute the HCl (but with no effect on the amount of moles of HCl) which would then give a lower concentration of HCl.
Thanks!
Post by: luken93 on March 15, 2011, 08:18:35 pm
b) If the burette has some water left over, then more HCl will be delivered to neutralise the NaOH. Because c = n/V, a higher V reading will result in a lower concentration than there actually is.
I haven't read your answers btw, but hopefully they match up :)
It always works better with formulas :P
Post by: Andiio on April 25, 2011, 04:57:51 pm
How would one analyse a colourless compound using UV-Vis? (Is it merely to convert the colourless compound into a coloured one?)
What can samples be held in, apart from quartz? (UV-Vis)
Also, do we have to know about the double-beam spectrophotometer?
Thanks!
Post by: Greatness on April 25, 2011, 08:22:49 pm
Well UV analyses molecules so i dont think you would be able to analyse metals.
How would one analyse a colourless compound using UV-Vis?
You would do it the same way, you would just have to compare the absorption spectrum with a known compound.
Also, do we have to know about the double-beam spectrophotometer?
I doubt it.
Post by: Water on April 25, 2011, 08:28:53 pm
Post by: Greatness on April 25, 2011, 08:41:15 pm
UV Vis can be used to analyze transition metalsReally? Link me pls :D
Post by: Andiio on April 25, 2011, 08:43:16 pm
Can UV-Visible spec be used to analyse metal ions at all? (Transition metals?)
Well UV analyses molecules so i dont think you would be able to analyse metals.
How would one analyse a colourless compound using UV-Vis?
You would do it the same way, you would just have to compare the absorption spectrum with a known compound.
Also, do we have to know about the double-beam spectrophotometer?
I doubt it.
Yep as Water said, UV-Vis spec can be used to analyse transition metals; but can't analyse metal ions? What's the diff/significance? :\
I agree with you; analysing colourless compounds quantitatively should be fine, but how about qualitative? :\ You'd have to change it to a coloured compound, wouldn't you?
Post by: Water on April 25, 2011, 08:54:38 pm
Quote
Yep as Water said, UV-Vis spec can be used to analyse transition metals; but can't analyse metal ions? What's the diff/significance? :\
No no. Sorry, you can find out metal ions, through UV VIS.
Sorry Andio..when transition metals go in solution, they automatically become ionized. I was working on that basis.
@swarley you react the metal ions with another component, and you work from there :)
Post by: thushan on April 26, 2011, 03:53:20 pm
Post by: luken93 on April 26, 2011, 04:48:03 pm
Post by: luken93 on April 26, 2011, 04:49:33 pm
http://vce.atarnotes.com/forum/index.php/topic,39634.msg415664.html#msg415664