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#### Sona

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##### a buffer reated problem
« on: April 11, 2017, 08:11:37 PM »

Calculate the H30 + (hydronium ion) concentration
for a buffer solution of 2 M phosphonic
acid and 1.5 M potassium hydrogen phosphate
[Ka1= 7.11* 10^-3; Ka2=6.32x10^-8].

I found this problem.
Firstly I feel that phosphonic acid is wrongly written, it should be phosphoric acid.
Assumimng it as phosphoric acid I caculated pH form the equation
pH=1/2(pKa1 + pKa2 + log [HPO4-2]/[H3PO4] ), here [HPO4-2]= 1.2M and [H3PO4]=2M
from this pH I calculated [H+].

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#### Borek

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##### Re: a buffer reated problem
« Reply #1 on: April 11, 2017, 09:16:00 PM »

Firstly I feel that phosphonic acid is wrongly written, it should be phosphoric acid.

Agreed, must be a typo. Ka values are those of phosphoric acid.

Quote
Assumimng it as phosphoric acid I caculated pH form the equation
pH=1/2(pKa1 + pKa2 + log [HPO4-2]/[H3PO4] ), here [HPO4-2]= 1.2M and [H3PO4]=2M

No idea what you are doing here nor where the equation comes from.

In general H3PO4 and HPO42- will react till there is only only one pair of conjugate acid and base dominating the solution. You can think about it as if it was a limiting reagent problem. Once you have these concentrations calculated from stoichiometry it is time for Henderson-Hasselbalch equation.
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#### Sona

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##### Re: a buffer reated problem
« Reply #2 on: April 12, 2017, 04:55:35 PM »

Sir, I couldn't understand the second part. Could you tell me in more detail using the values given in the question?
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#### Borek

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##### Re: a buffer reated problem
« Reply #3 on: April 12, 2017, 10:07:22 PM »

Hint: if you mix 1 mole of H3PO4 and 1 mole of HPO4- (whatever the counterion is), they will react:

H3PO4 + HPO42- 2H2PO4-

and at equilibrium only negligible amounts of H3PO4 and HPO42- will be left.
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#### Sona

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##### Re: a buffer reated problem
« Reply #4 on: April 15, 2017, 05:04:33 PM »

Okay. From here I could find that [H3PO4]= 0.5M and [HPO4-2] =0; I calculated the alpha value( degree of dissociation) but it was small so I thought to ignore it, and [H2PO4-]= 2*1.5=3M. So now the Handerson equation pH=1/2(pKa1 + pKa2 + log [H2PO4-]/[H3PO4]). Using the above value of concentration I'll get pH and then [H+].
Is this okay?
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#### Borek

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##### Re: a buffer reated problem
« Reply #5 on: April 15, 2017, 09:47:34 PM »

Okay. From here I could find that [H3PO4]= 0.5M and [HPO4-2] =0

OK

Quote
[H2PO4-]= 2*1.5=3M

No, no idea how you got that.

Quote
the Handerson equation pH=1/2(pKa1 + pKa2 + log [H2PO4-]/[H3PO4])

That's not the Henderson-Hasselbalch equation. I told you earlier I have no idea where this equation comes from.
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#### XeLa.

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##### Re: a buffer reated problem
« Reply #6 on: April 16, 2017, 02:11:17 AM »

A pKa value describes the equilibrium shift between a conjugate acid-base pair. So if you've got HPO42- and PO43- then the pKa which describes the activity of this conjugate pair is 12.32. Also, there is only one Henderson-Hasselbalch equation and it's only a Google/textbook search away.
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