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### Topic: Mass of salt to add to make buffer  (Read 10932 times)

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#### Forsh

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##### Mass of salt to add to make buffer
« on: April 09, 2010, 01:59:41 PM »
Hi.
I've been given a question on buffer solutions and i'm having major issues completing it!
I've got most of it done, just I can't work out the last bit.

You want to make up 100ml of a 0.25M phosphate buffer pH 7.2 and have available the following three salts: Na3PO4, Na2HPO4 and NaH2PO4

The pKa values associated with the three dissociations of H3PO4 are: 2.12, 6.8, 13.32

Which two salts will you use?
I'm pretty sure I want Na3PO4 and Na2HPO4 because I start with the salt closest with the pKa closest the the pH required (Na2HPO4) and add small amounts of the salt with the higher pKa to increase the pKa from 6.8 to 7.2

Its the next question which i'm finding difficult:

Now calculate how much of each you need to weigh out to make this solution, show your working.

Now using the Henderson-Hasselbalch equation i've got:

pH = pKa + log (A-)/(HA)

substituting in numbers, taking the negative log then taking pKa away from pH gives me:

0.4 = log (A-)/(HA)

multiply both sides base 10

2.512 = (A-)/(HA)

multiply both sides by (HA)

2.512 (HA) = (A-)

And thats as far as I can get. I need to use the information given to find the number of moles of both HA and H- and from that work out the mass of each.

Can anyone help?

#### Borek

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##### Re: Mass of salt to add to make buffer
« Reply #1 on: April 09, 2010, 03:18:04 PM »
Salts you want to use would be OK for much higher pH buffer. You want to make buffer with pH close to pKa2, that means using ions present in Ka2.

You have one equation describing ratio of concentrations of HA and A-. You are told what should be the sum of their concetrations - that will give you the second equation. Two equations, two unknows, solve, done.
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#### Forsh

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##### Re: Mass of salt to add to make buffer
« Reply #2 on: April 09, 2010, 04:07:28 PM »
So the conc of A- plus the conc of HA is equal to the molar concentration of the buffer, in this case 0.25?

#### Borek

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##### Re: Mass of salt to add to make buffer
« Reply #3 on: April 09, 2010, 05:58:38 PM »
Yes, thats what "0.25M buffer" means.
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#### Forsh

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##### Re: Mass of salt to add to make buffer
« Reply #4 on: April 09, 2010, 08:08:21 PM »
I'm still having a bit of trouble with this. I have the feeling the answer is staring me right in the face but I can't quite grasp it.

I now have:

1.58[HA]+[A-]=0.25

But i'm not sure how I can use that to work out the values of [HA] and [A-].

#### Borek

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##### Re: Mass of salt to add to make buffer
« Reply #5 on: April 10, 2010, 04:19:16 AM »
What are two equations that you get from the information given in the question.
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#### Forsh

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##### Re: Mass of salt to add to make buffer
« Reply #6 on: April 10, 2010, 07:31:07 AM »
Ok.
The first equation tells me the ratio of HA to A- I need to make the solution: 1.58[HA] = [A-]

The second equation. I'm not sure. I'm told the molarity of the solution (0.2M) and the total amount of solution (100ml) so I could use that to find the total number of moles of solute are added to the solution. I don't know how that'll help me though, or what else the second equation could be.

#### Borek

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##### Re: Mass of salt to add to make buffer
« Reply #7 on: April 10, 2010, 09:19:19 AM »
So the conc of A- plus the conc of HA is equal to the molar concentration of the buffer, in this case 0.25?

Can you write this exact phrase in form of equation?
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#### Forsh

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##### Re: Mass of salt to add to make buffer
« Reply #8 on: April 10, 2010, 09:27:13 AM »
So the conc of A- plus the conc of HA is equal to the molar concentration of the buffer, in this case 0.25?

Can you write this exact phrase in form of equation?

1.58[HA] + [A-] = 0.25 ?

#### Borek

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##### Re: Mass of salt to add to make buffer
« Reply #9 on: April 10, 2010, 10:04:44 AM »
No. Concentration of one plus concentration of the other, why do you multiply by something? The simplest approach will do.
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#### Forsh

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##### Re: Mass of salt to add to make buffer
« Reply #10 on: April 10, 2010, 10:36:41 AM »
Sure that makes sense.
I added the multiply by 2.51 cause I know that the concentration of [HA] is going to be 2.51 times stronger for the pKa to equal the pH.

But the statement [HA]+[A-] = 0.25 makes more sense.

I don't know to to link the two equations though.

#### Borek

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##### Re: Mass of salt to add to make buffer
« Reply #11 on: April 10, 2010, 10:56:53 AM »
You have no idea how to solve two equations in two unknowns? Sue you school, this is a basic algebra skill.

Solve one equation for one of the variables, substitute into the other equation, solve for the remaining unknown.
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#### Forsh

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##### Re: Mass of salt to add to make buffer
« Reply #12 on: April 10, 2010, 11:38:33 AM »
If you've got X + Y = 0.25 then X or Y could be anything.

Equally with 2.51X = Y then X or Y could be anything.

Maybe i'm looking at it in the wrong way?

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