Post by: generalkorn12 on June 19, 2012, 06:42:14 pm
For a question sucha as, '0.1M of Hpoidodous Acid has a pH of 5.8, determine the Ka value',
for the numerators, [H3O] and [OI], how come we would assume that they are the same concentration
and not use the 10^-14 rule as we would for a question such as,
'Perchloric Acid has a concentration of 0.01M determine the concentration of [OH]?
Thanks.
Post by: SenriAkane on June 20, 2012, 08:10:52 pm
Post by: generalkorn12 on July 03, 2012, 10:02:23 am
Post by: Hancock on July 03, 2012, 10:22:28 am
I've just got one question that's been bugging me :-\,
For a question sucha as, '0.1M of Hpoidodous Acid has a pH of 5.8, determine the Ka value',
for the numerators, [H3O] and [OI], how come we would assume that they are the same concentration
and not use the 10^-14 rule as we would for a question such as,
'Perchloric Acid has a concentration of 0.01M determine the concentration of [OH]?
Thanks.
The reason you don't use the 10^-14 rule is because that is for water and H30+ and OH- at 25 degrees.
You assume that [H3O+] is equal to [OI-] because it is a weak acid. Therefore the acid will not ionize the water alot.
You therefore have Ka = [H3O+][IO-]/[HIO].
Since we assume that [H3O+] is equal to [OI-], that means that the amount reacted of HIO is so little that we assume that HIO initial is equal to HIO (after).
Which reduces to Ka = [H3O+]^2/[HIO] = (10 ^ -5.8 )^2/0.1=2.51 x 10^-11 M.
Post by: Graphite on July 10, 2012, 06:42:02 am
Although this is correct to call it an assumption because we know H+'s don't only come from that reaction in a solution, as there are small contributions made by the self ionisation of water. However, at 25 degrees celcius, we know that this contributes only 10^-7 moles of H+'s in 1 L of solution which is small enough to neglect.
With the question regarding "10^-14 rule" I think you are asking why it works for OH- and not OI-. It is actually somewhat complicated. When you are finding [OH-] using that value, you are focused on the effects of the production of H+ along with the consumption of water on the position equilibrium of the self ionisation of water (reducing OH-) and it does not actually concern the ionisation of the acid.
If this solution was at 25 degrees, we can indeed apply the "10^-14 rule" to find OH-.
Post by: generalkorn12 on August 08, 2012, 07:42:42 pm
Pb(s) + PbO2(aq) + 4H+ (aq) + 2SO4 (aq) -> 2PbSO4 (aq) + 2H2O (l)
How do we split them into it's half equations?
Post by: charmanderp on August 08, 2012, 07:50:25 pm
For an equation such as.You have to determine which is the reductant and which is the oxidant. In this case it's probably sulfuric acid but you can use oxidation numbers to confirm that.
Pb(s) + PbO2(aq) + 4H+ (aq) + 2SO4 (aq) -> 2PbSO4 (aq) + 2H2O (l)
How do we split them into it's half equations?
Post by: generalkorn12 on August 14, 2012, 04:33:45 pm
I've just got one more question in regards to changing pH levels. Why is it that, if an acid is fully ionised a change by a factor would cause it's pH to change considerably compared to if it wasn't fully ionised?
:-\
Post by: Lasercookie on August 14, 2012, 04:46:08 pm
I've just got one more question in regards to changing pH levels. Why is it that, if an acid is fully ionised a change by a factor would cause it's pH to change considerably compared to if it wasn't fully ionised?You know that pH is directly dependent on the concentration of H3O+ ions:
You can look at it in terms of equilibrium position, where an acid that fully ionises will shift all the way to the right.
If an acid is fully ionised, then it'll have donated a hydrogen ion completely. In other words, the concentration of H3O+ produced will be greater (and hence the change in H3O+ will be greater too).
When an acid only partially ionises, it only partially donates a hydrogen ion and hence it doesn't produce as much H3O+ as we had before.
Post by: generalkorn12 on August 17, 2012, 02:24:10 pm
For the attached image, Question 3 bi) I'm having trouble understanding why they're using 10^-14, I thought that only applied for strong acids and strong bases.
Post by: ICECOLD on August 17, 2012, 02:32:04 pm
Kw = [H+][OH-] = 10^-7 x 10^-7 = 10^-14 M^2
Post by: generalkorn12 on August 20, 2012, 05:53:51 pm
I've just got one more question that confuses me when it comes to deriving the half equations:
Pb (s) + PbO2 (s) + 4H+ (aq) + 2 SO4 (aq) + 2PbSO4 (s) + 2H2O (l)
How do I derive the half equations from this? I get up to assigning oxidation numbers, but have trouble understanding where exactly we should split them... :-\
Post by: charmanderp on August 20, 2012, 05:55:42 pm
ICECOLD!Is it wrong that this made me sing a certain song outloud? HEEEY YAAA.
Anyway. Your question. Where is the reaction arrow?
Post by: generalkorn12 on August 20, 2012, 07:19:57 pm
Pb (s) + PbO2 (s) + 4H+ (aq) + 2 SO4 (aq) -> 2PbSO4 (s) + 2H2O (l),
Sorry about that :), copying and pasting seemed to have removed that arrow. -_-
Post by: charmanderp on August 20, 2012, 07:36:38 pm
No worries! I assumed that's where the arrow would be, just wanted to confirm.
Pb (s) + PbO2 (s) + 4H+ (aq) + 2 SO4 (aq) -> 2PbSO4 (s) + 2H2O (l),
Sorry about that :), copying and pasting seemed to have removed that arrow. -_-
Pb(s) + SO4/2-(aq) --> PbSO4(s) + 2e-
PbO2(s) + SO4/2-(aq) + 4H+(aq) + 2e- --> PbSO4(s) + 2H2O(l)
In the first Pb goes from 0 to +2 ergo oxidised and in the second Pb goes from+4 to +2 ergo reduced.
Post by: generalkorn12 on August 21, 2012, 09:14:47 am
No worries! I assumed that's where the arrow would be, just wanted to confirm.
Pb (s) + PbO2 (s) + 4H+ (aq) + 2 SO4 (aq) -> 2PbSO4 (s) + 2H2O (l),
Sorry about that :), copying and pasting seemed to have removed that arrow. -_-
Pb(s) + SO4/2-(aq) --> PbSO4(s) + 2e-
PbO2(s) + SO4/2-(aq) + 4H+(aq) + 2e- --> PbSO4(s) + 2H2O(l)
In the first Pb goes from 0 to +2 ergo oxidised and in the second Pb goes from+4 to +2 ergo reduced.
Ahh!! I get it now, thanks! I've just got one query about this, for the sulphate ions, when is it alright to place them in both half equations?
Post by: charmanderp on August 21, 2012, 05:22:19 pm
I don't understand your question, sorry!No worries! I assumed that's where the arrow would be, just wanted to confirm.
Pb (s) + PbO2 (s) + 4H+ (aq) + 2 SO4 (aq) -> 2PbSO4 (s) + 2H2O (l),
Sorry about that :), copying and pasting seemed to have removed that arrow. -_-
Pb(s) + SO4/2-(aq) --> PbSO4(s) + 2e-
PbO2(s) + SO4/2-(aq) + 4H+(aq) + 2e- --> PbSO4(s) + 2H2O(l)
In the first Pb goes from 0 to +2 ergo oxidised and in the second Pb goes from+4 to +2 ergo reduced.
Ahh!! I get it now, thanks! I've just got one query about this, for the sulphate ions, when is it alright to place them in both half equations?
Post by: generalkorn12 on September 02, 2012, 10:22:40 am
I'm having trouble understanding why the answer is, 'lower and heat of reaction should be lower'.
Post by: charmanderp on September 02, 2012, 04:00:28 pm
CF is equal to E/dT. Energy is calculated by SHC*m*dH. Since in this case we're using a mass of 105g rather than 100g we're going to have a greater CF than what we should.
Therefore when we try to calculate energy for heat of reaction (HoR=E/n) for the HCL reaction we're going to obtain a larger value for E than what we should, when we multiply an overestimated CF by dT. Hence when you divide your overestimated E by n you'll get a higher value for heat of reaction that you should.
Post by: generalkorn12 on September 03, 2012, 06:16:20 pm
Another question, is there a reason we can electroplate objects using Silver nitrate solution? I'm looking at the Electrochemical Series and assumed water would be preferentially reduced.
Post by: stephanieteddy on September 03, 2012, 09:21:55 pm
Since galvanic cells require a negative gradient on the electrochemical series for a reaction to occur,
Does that mean electrolysis requires a positive gradient?
Post by: Howzat on September 07, 2012, 12:49:31 pm
(Just adding more stuff to make my post seem more complete ^_^)
Think about if the oxidant was below the reductant, then a spontaneous reaction would not occur.
Applying electricity to this will force a reaction to occur and hence, the oxidant will be forced to undergo reduction whilst the reductant will be forced to undergo oxidation!
;D , It definitely makes sense now that I think about it. :)
Post by: generalkorn12 on September 12, 2012, 08:51:59 pm
CO(g) + 2H2(g) -> CH3OH(g), decreasing the pressure would have what effect on the mole/concentration of H2(g)?
I always thought that if we decreased pressure, system moves backwards to increase pressure, hence there will be more mol of H2 gas and a higher concentration. Yet the ans says that the mol would increase but the concentration decreases, I always based my poor logic (:)) on the formula c=n/v, and if mole is increased, concentration increases.
Post by: charmanderp on September 12, 2012, 10:20:12 pm
Post by: Bhootnike on September 12, 2012, 10:31:39 pm
if pressure decreases, then volume decreases.
if pressure or volume decrease, then n increases.
thats how i do it.
btw did the answer mention anything about which reaction would be favoured?
in response to you saying it moves backwards, i thought that if n(h2) increases, then system acts as to partially offset this initial increase in (h2) by subsequently decreasing (h2), which is achieved by favouring the forward reaction, since h2 is consumed.
Post by: generalkorn12 on September 19, 2012, 07:25:56 pm
Post by: Hancock on September 22, 2012, 03:22:13 am
Post by: generalkorn12 on October 09, 2012, 08:07:53 am
Post by: nisha on October 09, 2012, 11:49:39 am
Post by: generalkorn12 on October 17, 2012, 08:19:02 am
a) Steady current is passed on three cells, containing solutions, Pb(NO3), AgNO3 and Al(NO3), the molar ratio of n(Pb):n(Ag):n(Al) deposited at the negative electrode is:
b) 0.5mol of Cu2+ and 1mol of Ag+ are added to an electrolytic cell, the quantity of electricity to deposit all of the copper and silver are?
(I've always thought, we'd determine the electron requirements for both, and add, as that would be the total required)
c). 400mL of 0.2M Sulfuric and 400ml of 0.2 KOH released 4.56kJ. The energy released for 400mL of 0.2M Sulfuric Acid is mixed with 800mL of 0.2M KOH is...?
(For this question, would it be 4.56kJ, as Sulfuric Acid is the limiting reagent?)
Post by: charmanderp on October 17, 2012, 03:33:27 pm
Quote
a) Steady current is passed on three cells, containing solutions, Pb(NO3), AgNO3 and Al(NO3), the molar ratio of n(Pb):n(Ag):n(Al) deposited at the negative electrode is:
Should actually be Pb(NO3)2 and Al(NO3)3 but that's not strikingly important. Anyway, if you look at the electrochemical series you'll see that for each mole of electrons you get 0.5 mole of lead, 1 mole of silver and 1/3 mole of aluminium, hence the molar ratio of n(Pb):n(Ag):n(Al) deposited is 2:3:1.
BUT WAIT!!!!!! A wild standard reduction potential appeared! Al3+ is a weaker oxidant than water and hence H2O will always be reduced preferentially and you will not be able to obtain any aluminium. Hence the molar ratio is 1:2:0 (n(Pb):n(Ag):n(Al)).
Quote
0.5mol of Cu2+ and 1mol of Ag+ are added to an electrolytic cell, the quantity of electricity to deposit all of the copper and silver are?
(I've always thought, we'd determine the electron requirements for both, and add, as that would be the total required)
Since Ag is a stronger oxidant that will be reduced first. Calculate Q for Ag (96500C) and then calculate Q for Cu (also 96500C) and then add them together.
Quote
c). 400mL of 0.2M Sulfuric and 400ml of 0.2 KOH released 4.56kJ. The energy released for 400mL of 0.2M Sulfuric Acid is mixed with 800mL of 0.2M KOH is...?The first step is to write a balanced equation, like so:
(For this question, would it be 4.56kJ, as Sulfuric Acid is the limiting reagent?)
H2SO4(aq) + 2KOH(aq) --> K2SO4(aq) + 2H2O(l)
n(H2SO4)=CV=0.08mol
n(KOH)=CV=0.08
KOH is actually the limiting reagent here. So divide 4.56 by 0.08/2 which is 0.04 which will give you the heat of reaction, 114kJ.