pH Calculation / ionization of diprotic and polyprotic acids in pure water

[dinosaur93]

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?

[Panicmode]

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.

[dinosaur93]

I don't understand!

Could you pls rephrase?

[REBORN]

I thought you doubled it :S

[dinosaur93]

yeah, I thought we were meant to multiply the concentration by 2 because there is 2 H⁺ ions in the given.

[Panicmode]

Hmm okay, I'll try it this way:

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.

[dinosaur93]

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! 

[Panicmode]

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! 

You're very welcome . If you need any more help don't hesitate to ask/pm me

[REBORN]

For the purpose of [u2] we times by 2 right?

[Panicmode]

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.

[tea.squaredd]

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.

Not being critical, but i recall most u3 exams when calculating pH still 'doubled' the H+ concentration for H2SO4.
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

[Panicmode]

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.

Not being critical, but i recall most u3 exams when calculating pH still 'doubled' the H+ concentration for H2SO4.
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.

[Mao]

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.

Not being critical, but i recall most u3 exams when calculating pH still 'doubled' the H+ concentration for H2SO4.
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.

[zool3]

so...are we meant to double or even triple or leave it as single?

[Graphite]

I think in VCE, you always count the number of  H's and double/triple as required. This is from what I see in solutions, but yeah if you wanna be a smarty then you gotta explain why and I don't see why that is wrong