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Topic: Acid and buffer  (Read 14781 times)

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strewart

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Acid and buffer
« on: June 01, 2005, 04:31:30 AM »
The question is:
20mL of 0.1 M NaOH is added to 50mL of 0.1 M KH2PO4

Write an equation for the reaction which occurs. (Do not include spectator ions) Identify the acid/base conjugate pairs in the equation.

Calculate the pH of the final solution. (The solution is a buffer)

(Ka = 6.3 x 10-8 for H2PO4-)

I hardly get what the question means... A buffer is a weak acid or base and the salt but sodium hydroxide is a strong base and the potasium thing looks like maybe a salt only, so I don't know where to start with this one...

Maybe:
OH + H2PO4- --> H3O+ + PO42-

It did say ignore the spectator ions. I don't see how that one helps though...

Thanks for any help.

Oh yeah, one other thing about blank solutions in spectrophotometry... What would you use as a blank to analyse a solution of indicator in .001 M HCl? I guessed .001 M HCl without the indicator, or would it be pure water with the indicator but no acid? Which is the species of interest?
« Last Edit: June 01, 2005, 04:37:48 AM by strewart »
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Offline Borek

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Re:Acid and buffer
« Reply #1 on: June 01, 2005, 05:24:15 AM »
H2PO4- has acidic protons, right?

It can act as an acid, right?

Quote
OH + H2PO4- --> H3O+ + PO42-

It is completely, awfully and disastrously wrong, but at least you are trying to move into the right direction.

What happens when you mix base and acid?

Will be any difference between any other acid/base reaction when you mix NaOH and H2PO4-?
« Last Edit: June 01, 2005, 05:37:07 AM by Borek »
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strewart

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Re:Acid and buffer
« Reply #2 on: June 01, 2005, 06:10:09 AM »
It makes a salt and water, but there are 2 protons on the acid one so does that mean I double the amount of NaOH?

2NaOH + H2PO4- --> 2H2O + NaPO4

Does that mean the K+ is the only spectator ion? I can't see the conjugate pairs either, though I would assume NaOH would go with water as conjugate base since NaOH is a proton acceptor therefore the other pair must be the acid ones.

Then can I use the formula of pH = pKa - log([NaPO4/H2PO4-])

I can use mole ratio of whatever has the smallest amount of moles on the left to work out the number of moles of NaPO4 then get its concentration with C=n/V (volume being 70mL?) to use in the equation to get pH. Does that sound right?

My only problem with that is it just seems like an acid base reaction and seems to have nothing to do with buffers, but maybe it doesn't need to..
« Last Edit: June 01, 2005, 06:11:52 AM by strewart »
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Offline Borek

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Re:Acid and buffer
« Reply #3 on: June 01, 2005, 07:11:13 AM »
Quote from: strewart on June 01, 2005, 06:10:09 AM
It makes a salt and water, but there are 2 protons on the acid one so does that mean I double the amount of NaOH?

2NaOH + H2PO4- --> 2H2O + NaPO4

More acidic protons react first. Compare amount of base and H2PO4-. How far the neutralization reaction can go?

And please stop creating nonexisting compounds in your reactions >:( There is no such thing as NaPO4.

Quote
Does that mean the K+ is the only spectator ion? I can't see the conjugate pairs either, though I would assume NaOH would go with water as conjugate base since NaOH is a proton acceptor therefore the other pair must be the acid ones.

You are forgetting about dissociation. You are adding solution of NaOH, it contains Na+ and OH- ions. NaOH is not a proton acceptor, OH- is. But it was already used for partial neutralization of H2PO4-, so it will be not present in the buffer equation.

Quote
My only problem with that is it just seems like an acid base reaction and seems to have nothing to do with buffers, but maybe it doesn't need to...

Reading through your posts I can tell you that's not your only problem. Seems to me you don't know very basic stuff, yet you are assigned a much more advanced question.

Sorry if it sounds rough but you are making so many mistakes in your reasoning that I have no idea where to start explaining :(
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Offline xiankai

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Re:Acid and buffer
« Reply #4 on: June 01, 2005, 08:50:52 AM »
to ignore spectator ions, that would be basically an ionic equation, and common knowledge about neutralisation reactions give us:

H+ + OH- -> H2O

or

H3O+ + 2OH- -> 2H2O

though i prefer the former :D

oh yes, remember the valency of PO4. it is -3, so it shoudl read Na3PO4. its not the same as NaPO4, which would give sodium a valency of 3(impossible  ::))
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strewart

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Re:Acid and buffer
« Reply #5 on: June 07, 2005, 02:35:01 AM »
Quote
Reading through your posts I can tell you that's not your only problem. Seems to me you don't know very basic stuff, yet you are assigned a much more advanced question

Right that would be from trying it so late at night when I'd hardly been listening in lectures.. Two equations not one eh, OH- and H2PO4- going to HPO4- and water to start with. Thats not the buffer equation I want though.. I think I have it now.

H2PO4- + H2O <--> HPO42- + H3O+

That a bit better? For the first equation I can work out (using that initial > change > final stuff) concentrations, then with the second equation work out the conc of H30+ with the Ka equation since I know Ka...
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Offline Borek

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Re:Acid and buffer
« Reply #6 on: June 07, 2005, 04:30:17 AM »
Quote from: strewart on June 07, 2005, 02:35:01 AM
H2PO4- + H2O <--> HPO42- + H3O+

Perfect this time.

Quote
That a bit better? For the first equation I can work out (using that initial > change > final stuff) concentrations, then with the second equation work out the conc of H30+ with the Ka equation since I know Ka...

No idea what you mean bye the first/second equation... You have a right reaction equation now. At this moment it is an limiting reagent question - check what will be left in the solution after neutralization. Then think about buffer stuff - acid, conjugate base, HH equation.
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arnyk

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Re:Acid and buffer
« Reply #7 on: June 09, 2005, 06:45:43 PM »
Stupid question: Why would H2PO4- give up a proton when it has a negative charge?
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Offline xiankai

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Re:Acid and buffer
« Reply #8 on: June 10, 2005, 03:47:50 AM »
sounds like a good question... it may be that the equilibrium is generally favoured towards H2PO4-, not towards HPO4 2-.
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