Generalkorn12 Question Thread.

[generalkorn12]

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.

[SenriAkane]

1mole of [H30+] = 1Mol of  [Ol-] .  HIO(aq)<->H3O+(aq)+IO-(aq)

[generalkorn12]

Just one more question, how do we determine the appropriate electrodes to use in a Galvanic Cell for the Anode and Cathode?

[Hancock]

 

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.

[Graphite]

We assume they are the same because we know that the H+'s and OI's are coming from a single reaction.
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-.

[generalkorn12]

For an equation such as.

Pb(s) + PbO2(aq) + 4H+ (aq) + 2SO4 (aq) -> 2PbSO4 (aq) + 2H2O (l)

How do we split them into it's half equations?

[charmanderp]

For an equation such as.

Pb(s) + PbO2(aq) + 4H+ (aq) + 2SO4 (aq) -> 2PbSO4 (aq) + 2H2O (l)

How do we split them into it's half equations?

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.

[generalkorn12]

Ahhh, seems to make more sense now . 

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?

 

[Lasercookie]

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.

[generalkorn12]

That's got the lightbulb in my head clicking!

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.

[ICECOLD]

Given that the temperature is 25degC, the following relationship can be used:

Kw = [H+][OH-] = 10^-7 x 10^-7 = 10^-14 M^2

[generalkorn12]

Now I get it, thanks ICECOLD!

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... 

[charmanderp]

ICECOLD!

Is it wrong that this made me sing a certain song outloud? HEEEY YAAA.

Anyway. Your question. Where is the reaction arrow?

[generalkorn12]

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. -_-

[charmanderp]

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. -_-

No worries! I assumed that's where the arrow would be, just wanted to confirm.

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.