Post by: dinosaur93 on November 10, 2011, 07:22:38 pm
Calculate the pH of the solution.
m ( H2SO4 ) = 12.5g
N ( H2SO4 ) = 800 mL
n ( H2SO4 ) =
n ( H2SO4 ) =
n ( H2SO4 ) = 0.127395 mol
c ( H2SO4 ) =
c ( H2SO4 ) =
From this stage on, the solution pathway says:
pH = -log [H3O+ = -log [0.159] = 0.799
but isnt we still need to multiply the concentration by 2 (before solving for pH) because there is 2 H+ ions in the given equation? ???
Post by: Panicmode on November 10, 2011, 07:36:00 pm
12.5g of H2SO4 is dissolved in 800 mL of deionised water
Calculate the pH of the solution.
m ( H2SO4 ) = 12.5g
N ( H2SO4 ) = 800 mL
n ( H2SO4 ) =
n ( H2SO4 ) =
n ( H2SO4 ) = 0.127395 mol
c ( H2SO4 ) =
c ( H2SO4 ) == 0.159 M
From this stage on, the solution pathway says:
pH = -log [H3O+ = -log [0.159] = 0.799
but isnt we still need to multiply the concentration by 2 (before solving for pH) because there is 2 H+ ions in the given equation? ???
Although H2SO4 is a diprotic acid, the ionisaiton of the second proton only happens to a very small extent (almost negligible). Therefore, the pH of the solution is very unlikely to be greatly affected by ionastaiton of the second proton.
Post by: dinosaur93 on November 10, 2011, 07:46:35 pm
Could you pls rephrase?
Post by: REBORN on November 10, 2011, 07:47:02 pm
Post by: dinosaur93 on November 10, 2011, 07:53:29 pm
Post by: Panicmode on November 10, 2011, 07:56:12 pm
Step 1: H2SO4 donates a hydrogen to become HSO4- (This happens to almost 100% of the H2SO4 molecules)
Step 2: HSO4- donates a hydrogen to become SO4 2-
However step 2 only happens to a very small extent. Only a tiny fraction of HSO4- ions will further go on to donate another hydrogen and become SO4 2- ions..
In terms of calculating pH, since HSO4- is considered a weak acid (does not readily donate protons) and will only ionise to a very small extent, we do not take this into consideration when determining the pH. There are ways of determining to what extent the second acid will donate its proton, but these are beyond the scope of the Unit 1-2 course. In Unit 4, you learn about equilibrium reactions and Ka values; this is when you consider this.
For now, don't worry too much about it.
Post by: dinosaur93 on November 10, 2011, 07:59:58 pm
I kinda get what you mean now....
nice, nice!!
tnx bro, because apparently it came out of a practise exam and just wondering how the answer came to be.....
tnx heaps bro! I'll take not of what you said! :D
Post by: Panicmode on November 10, 2011, 09:07:13 pm
oh, ok, i see....
I kinda get what you mean now....
nice, nice!!
tnx bro, because apparently it came out of a practise exam and just wondering how the answer came to be.....
tnx heaps bro! I'll take not of what you said! :D
You're very welcome :). If you need any more help don't hesitate to ask/pm me :)
Post by: REBORN on November 10, 2011, 09:07:40 pm
Post by: Panicmode on November 10, 2011, 09:27:12 pm
For the purpose of [u2] we times by 2 right?
No, for the purpose of unit two I wouldn't even consider ionisation of the second proton. I think it's reasonable to understand that HSO4- is a very weak acid. It would be unrealistic of them to expect you to calculate the percentage ionisation of HSO4- and thereby calculate the absolute [H+] to calculate a more reliable pH.
In your working I'd figure out the pH (not doubling) and then write, "HSO4- only ionises very slightly and so for the purpose of this calculation is negligible". This is much more correct then doubling which will lead to a much less accurate pH.
This is what I was taught in year 11 Chem, your teachers may teach differently.
Post by: tea.squaredd on November 10, 2011, 09:56:19 pm
Not being critical, but i recall most u3 exams when calculating pH still 'doubled' the H+ concentration for H2SO4.For the purpose of [u2] we times by 2 right?
No, for the purpose of unit two I wouldn't even consider ionisation of the second proton. I think it's reasonable to understand that HSO4- is a very weak acid. It would be unrealistic of them to expect you to calculate the percentage ionisation of HSO4- and thereby calculate the absolute [H+] to calculate a more reliable pH.
In your working I'd figure out the pH (not doubling) and then write, "HSO4- only ionises very slightly and so for the purpose of this calculation is negligible". This is much more correct then doubling which will lead to a much less accurate pH.
This is what I was taught in year 11 Chem, your teachers may teach differently.
It wasn't until u4 that we had to take into consideration all the Ka and ionisation percentages.
Just my two cents.
But i agree, doing what panicmode said on your u2 exam will make you look pro az
Post by: Panicmode on November 10, 2011, 10:53:18 pm
Not being critical, but i recall most u3 exams when calculating pH still 'doubled' the H+ concentration for H2SO4.For the purpose of [u2] we times by 2 right?
No, for the purpose of unit two I wouldn't even consider ionisation of the second proton. I think it's reasonable to understand that HSO4- is a very weak acid. It would be unrealistic of them to expect you to calculate the percentage ionisation of HSO4- and thereby calculate the absolute [H+] to calculate a more reliable pH.
In your working I'd figure out the pH (not doubling) and then write, "HSO4- only ionises very slightly and so for the purpose of this calculation is negligible". This is much more correct then doubling which will lead to a much less accurate pH.
This is what I was taught in year 11 Chem, your teachers may teach differently.
It wasn't until u4 that we had to take into consideration all the Ka and ionisation percentages.
Just my two cents.
But i agree, doing what panicmode said on your u2 exam will make you look pro az
Really? From the outset of unit two our teacher told us about diprotic and tripotic acids and that this didn't mean we doubled/tripled the [H+]. In fact, he said the opposite, that he would take marks away if we did.
Post by: Mao on November 11, 2011, 01:08:59 am
Not being critical, but i recall most u3 exams when calculating pH still 'doubled' the H+ concentration for H2SO4.For the purpose of [u2] we times by 2 right?
No, for the purpose of unit two I wouldn't even consider ionisation of the second proton. I think it's reasonable to understand that HSO4- is a very weak acid. It would be unrealistic of them to expect you to calculate the percentage ionisation of HSO4- and thereby calculate the absolute [H+] to calculate a more reliable pH.
In your working I'd figure out the pH (not doubling) and then write, "HSO4- only ionises very slightly and so for the purpose of this calculation is negligible". This is much more correct then doubling which will lead to a much less accurate pH.
This is what I was taught in year 11 Chem, your teachers may teach differently.
It wasn't until u4 that we had to take into consideration all the Ka and ionisation percentages.
Just my two cents.
But i agree, doing what panicmode said on your u2 exam will make you look pro az
Really? From the outset of unit two our teacher told us about diprotic and tripotic acids and that this didn't mean we doubled/tripled the [H+]. In fact, he said the opposite, that he would take marks away if we did.
:') He is a true scientist.
Post by: zool3 on November 12, 2011, 09:48:02 am
Post by: Graphite on November 12, 2011, 11:44:27 am
Post by: burbs on November 12, 2011, 11:57:18 am
Post by: Bismuth on November 12, 2011, 12:20:57 pm
Post by: Panicmode on November 12, 2011, 12:21:48 pm
What? I thought that second ionisation is a negligible change to pH for our calculations or something?
That's exactly what I just said x_x.
Post by: Mao on November 12, 2011, 12:53:29 pm
For Unit 4 though, you do not learn how to solve the concentrations of polyprotic acids properly. That you learn at uni. You just need to appreciate the fact that second/third ionization are weak.
For unit 2, you need to appreciate the fact that a polyprotic acid can donate up to twice or three times the amount of H+ than a monoprotic acid.
Post by: Graphite on November 12, 2011, 12:57:51 pm
What? I thought that second ionisation is a negligible change to pH for our calculations or something?Its not always negligible, its negligible normally in water.
Post by: burbs on November 12, 2011, 01:13:34 pm
Post by: Graphite on November 12, 2011, 01:18:31 pm
Post by: burbs on November 12, 2011, 01:20:25 pm
Post by: zool3 on November 12, 2011, 09:52:10 pm
Post by: Panicmode on November 12, 2011, 09:54:11 pm
Post by: burbs on November 12, 2011, 09:59:44 pm
So would, as someone said, blah blah blah blah blah"assuming complete ionization "be sufficient? Oooooooor have i got it wrong?
Yes, just say assuming complete ionisation: blah blah
However, if the question say Assume complete ionisation DO NOT IGNORE IONISATION OF THE SECOND PROTON!!!!!
now I confus
Post by: Panicmode on November 12, 2011, 10:02:56 pm
UNLESS the question states, "Assume full ionisation" in which case you assume both ionised completely.
Post by: taqi on November 12, 2011, 10:22:12 pm
Post by: Zebra on November 12, 2011, 10:23:34 pm
Post by: taqi on November 12, 2011, 10:54:59 pm
Post by: Mao on November 13, 2011, 05:24:00 am
okay, just say the question wants you to consider it. NOW WHAT? what do you have to do to find the H+ conc and all that jazz?It involves a lot of maths. Several simultaneous equations.
you need the Ka value to figure out the H+ concentration like any acid equilibrium question. And then add the concentration of the first ionization and the second one, and use that value to find the ph.Not that simple. You will need to set up equations which simultaneously satisfy both equilibrium. University level.
In summary:
1. 2nd/3rd/subsequent ionizations are always weak. Often negligibly weak. Ignore their effects when simply dissolving in water.
2. 2nd/3rd/subsequent protons can react with base. Assume full deprotonation in acid/base reactions.
3. If the question states 'assume full ionization', then assume 2nd/3rd/subsequent ionizations are complete. However, this is rare, unscientific and a generally idiotic thing to assume.
4. I reiterate, unless under explicit instruction, do not assume full ionization.
5. You will not be required to perform calculations to determine the extent of 2nd/3rd/subsequent ionization.