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### Topic: buffer solution H3PO4 and HPO4--  (Read 25877 times)

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#### Albert ##### buffer solution H3PO4 and HPO4--
« on: May 08, 2005, 04:40:48 AM »
Hi, everybody! I'm an Italian student of pharmaceutical chemistry, so I apologize for my English.
However, here is my problem:
Calculate the concentration (mol/L) of HPO4-- in a buffer solution (pH: 7.30), which is 0.2 mol/L in H3P04.
I know it comes out a solution of H2PO4- , but I cannot understand how to manipulate the H&H expression in order to solve the problem. Maybe there's a formula to get the pH of such a strange buffer solution, but ,INCREDIBLE AS IT MAY APPEAR, it seems to be impossible to find (neither on the web nor on my book).

[H+]=6.32E-8*([H2PO4-]/[HPO4--])

7.11E-3=[H+][H2PO4-]/[H3PO4]=x^2/(0.2-x)

6.32E-8/7.11E-3=[H3PO4][HPO4--]/[H2PO4-]^2

This is everything I worked out.

#### Borek ##### Re:buffer solution H3PO4 and HPO4--
« Reply #1 on: May 08, 2005, 04:49:15 AM »
which is 0.2 mol/L in H3P04.

That's a shortcut - it probably means that total concentration of all forms (H3PO4, H2PO4(-), HPO4(2-) and PO4(3-)) is 0.2M. There is no form of HH equation that will help you if you assume [H3PO4] = 0.2M.

pKa2 = 7.199 so it is very close to 7.3. You may safely assume there are only H2PO4(-) and HPO4(2-) anions in the solution.
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#### Albert ##### Re:buffer solution H3PO4 and HPO4--
« Reply #2 on: May 09, 2005, 03:38:34 PM »
You may safely assume there are only H2PO4(-) and HPO4(2-) anions in the solution.

I agree with you. So it is:

0.2M = [H2PO4-]+[HPO4--]

pH=3.95        [H+]=1.124E-4

...and now? How can I solve the problem?

#### Borek ##### Re:buffer solution H3PO4 and HPO4--
« Reply #3 on: May 09, 2005, 06:28:47 PM »
0.2M = [H2PO4-]+[HPO4--]

OK

Quote
pH=3.95        [H+]=1.124E-4

It was 7.30 in your original post. If it is 3.95 situation is much more complex.

Assuming you were right at first you have a set of equations - mass balance (above) and HH equation. Just solve them.
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#### Albert ##### Re:buffer solution H3PO4 and HPO4--
« Reply #4 on: May 10, 2005, 04:59:11 AM »
Assuming you were right at first you have a set of equations - mass balance (above) and HH equation.

0.2= [H2PO4-]+[HPO4--]

that's what I have in the beginning. Then I've to add HPO4-- in order to change the pH from 3.95, which is the value in the beginning, to 7.30, which is the final value. In my opinion, I don't have enough equations.
Sorry, it's my fault: my explaination wasn't clear enough. Looking at the keys at the end of my text book, the result is:
[HPO4--] = 0.705 mol/L

#### Borek ##### Re:buffer solution H3PO4 and HPO4--
« Reply #5 on: May 10, 2005, 05:18:32 AM »
Please write down carefully the question, as I am lost and I don't want to loose time guessing what has to be calculated ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### Albert ##### Re:buffer solution H3PO4 and HPO4--
« Reply #6 on: May 10, 2005, 06:54:42 AM »
Please write down carefully the question, as I am lost and I don't want to loose time guessing what has to be calculated Ok, sorry. What they give me is a solution which is 0.2 in [H3PO4].
Now, I have to add an unknown concentration of HPO4-- in order to have a buffer solution, whose pH is 7.30.
I know the result is [HPO4--]=0.705 mol/L

Thank you very much. #### Borek ##### Re:buffer solution H3PO4 and HPO4--
« Reply #7 on: May 10, 2005, 07:56:21 AM »
It still doesn't hold water. If the initial concentration was 0.2M H3PO4, initial pH was 1.46, not 3.95.
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#### AWK

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« Reply #8 on: May 10, 2005, 08:12:03 AM »
May be  at pH 3.95 a buffer H3PO4/NaH2PO4 exists that contain 0.2 M H3PO4 and we should increase pH to 7.30 by adding Na2HPO4, but this is only my guess
AWK

#### Albert ##### Re:buffer solution H3PO4 and HPO4--
« Reply #9 on: May 10, 2005, 11:21:26 AM »
It still doesn't hold water. If the initial concentration was 0.2M H3PO4, initial pH was 1.46, not 3.95.

Forget about pH=3.95. You're right: initial pH was 1.46, according to this expression

7.11E-3 = [H2PO4-][H+]/[H3PO4] = x^2/(0.2-x).

#### Borek ##### Re:buffer solution H3PO4 and HPO4--
« Reply #10 on: May 10, 2005, 11:47:10 AM »
So finally assuming that the question is

How much HPO4(2-) have to be added to the0.2M H3PO4 solution to get pH = 7.3?

At pH = 7.3 you have in solution HPO4(2-) and H2PO4(-) only (or mostly, other forms concentrations are negligibile). Their ratio is described by the HH equation.

HPO4(2-) when added to H3PO4 will act as a base:

HPO4(2-) + H3PO4 -> H2PO4(-)

So you have to add enough HPO4(2-) to first neutralize H3PO4 and then to create high enough concentration of HPO4(2-) itself. You may assume that the neutralization reaction goes stoichiometrically.

If it is still not the right answer I give up. I am still not sure I know what the question is about.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### Albert ##### Re:buffer solution H3PO4 and HPO4--
« Reply #11 on: May 10, 2005, 03:00:56 PM »
Thank you VERY much Mr.Borek!

That's what I needed! Unfortunately There was a misundersting and I set on edge you. I apologize.

Now, it's time for me to do my part. I'm going to explain everyone the proceedure, so that every fellow who might get in touch with such a particular problem: that's, in my opinion, what we ALL have to do here.

Here we are:

[H3PO4]=0.2 M

We have two different kind of reactions. The former is a neutralisation
H3PO4 + HPO4-- = 2H2PO4-     =>  [H2PO4-]=0.4M and

[HPO4--]'=0.2M

The latter is a simple buffer solution

[H+] = Ka2 * [H2PO4-]/[HPO4--]

5.012E-8 = 6.32E-8 * 0.4/x     => [HPO4--]'' = 0.504M

So, in the end, the total amount of HPO4-- required is

[HPO4--] = [HPO4--]' + [HPO4--]'' = 0.2 + 0.504 = 0.704M

Well, it is easy, isn't? 