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Offline Donaldson Tan

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pH of a buffering system
« on: May 23, 2005, 07:54:38 PM »
Here's a question from my past year chemistry exam. Did I do it correctly?

Quote
An aqueous solution containing 5ppm of FeSO4 is buffered with 0.5M of sodium dihydrogen phosphate at 25C under the exclusion of oxygen. Find an estimate for the pH of the buffer solution

this is how i did it.

assumption:
1. hydrolysis of Fe2+ is negligible

since we are dealing with a phosphoric acid derivative, here's the list of Ka values.
Ka1 = 7.1E-3
Ka2 = 6.3E-8
ka3 = 4.5E-13

since, i am given a solution of NaH2PO4, hydrolysis happens via:
(1) H2PO4- + H2O <-> HPO42- + H3O+
(2) H2PO4- + H2O <-> H3PO4 + OH-

the first step is govened by Ka2 and the second step is governed by Kb1.

Kb1 = Kw/ka1 = 1.408E-12
Ka2 = 6.3E-8

Since Ka2/Kb1 = 44730, then pH contribution by Kb1 is negligible. The pH of the buffer depends mainly on Ka2.

Initially,
[ H+] = 1E-7 M
[ H2PO42- ] = 0.5 M
[ HPO42- ] = 0 M

equilibrium,
[ H+] = 1E-7 + X M ~ X M (assume X >> 1E-7)
[ H2PO42- ] = 0.5 - X M
[ HPO42- ] = X M

Ka2 = X2/(0.5 - X)

solving quadratically,
X = 1.7745 (check X >> 1E-7)

estimated pH = -lg[ H+ ] = -lg x = 3.75



« Last Edit: May 23, 2005, 08:42:45 PM by geodome »
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Offline Borek

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Re:pH of a buffering system
« Reply #1 on: May 23, 2005, 08:26:41 PM »
Here's a question from my past year chemistry exam. Did I do it correctly?

First of all I will assume that the solution contains sodium dihydrogen phosphate, not sulphate ;)

Second - there is no buffer in classical meaning in this solution.

Quote
hydrolysis of Fe2+ is negligible

Good one. There are ppms vs 0.5M.

There is a mistake in the reaction equation and I will call the first dissociation, not hydrolysis, but otherwise - a good one.

Quote
Since Ka2/Kb1 = 44730, then pH contribution by Kb1 is negligible. The pH of the buffer depends mainly on Ka2.

Wrong one, but doing this question I will do the same mistake.

Quote
Ka2 = X2/(0.5 - X)

X << 0.5, so there is no need for quadratic equation.

Quote
X = 1.7745 (check X >> 1E-7)

You forgot exponential part, but that's math only.

Quote
estimated pH = -lg[ H+ ] = -lg x = 3.75

Well - this is correct result for the selected way of doing calculations. But it is a wrong one. pH = 4.68 and the hydrolysis can't be neglected... To be honest I am wise thanks to BATE, probably I shouldn't start it before thinking as now I know too much about the solution and my reasoning is skewed :(

It is 2:25 am, more tomorrow...
« Last Edit: May 23, 2005, 08:31:06 PM by Borek »
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Re:pH of a buffering system
« Reply #2 on: May 23, 2005, 08:29:58 PM »
Firstly, you switch from sulfuric acid to Phosphoric acid.  you should probably change all references to SO4 to PO4 so you don't confuse anyone else.


I could be wrong, but I think your first step should be to react the PO4-3 and the H2PO4- before dealing with the pH.  that would give you some H2PO4- and some HPO4-2.  then plug that into this equation:

pKa + log([base]/[acid])= pH.

Offline Donaldson Tan

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Re:pH of a buffering system
« Reply #3 on: May 23, 2005, 09:19:07 PM »
Firstly, you switch from sulfuric acid to Phosphoric acid.  you should probably change all references to SO4 to PO4 so you don't confuse anyone else.
LOL. I mistyped phosphate as sulphate.

Second - there is no buffer in classical meaning in this solution.Good one. There are ppms vs 0.5M.
the dihydrogen phosphate ion is amphoteric, so why can't it be a buffer?

X << 0.5, so there is no need for quadratic equation.You forgot exponential part, but that's math only.
ok. X = 1.77E-4 (i omitted the exponential part accidentally)

But it is a wrong one. pH = 4.68 and the hydrolysis can't be neglected...
how come?
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

Offline Donaldson Tan

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Re:pH of a buffering system
« Reply #4 on: May 23, 2005, 10:11:55 PM »
Quote:
Since Ka2/Kb1 = 44730, then pH contribution by Kb1 is negligible. The pH of the buffer depends mainly on Ka2.

Wrong one, but doing this question I will do the same mistake.

why is this wrong?
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

Offline AWK

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Re:pH of a buffering system
« Reply #5 on: May 24, 2005, 03:50:55 AM »
Sodium dihydrogen phosphate shows buffering properties but its pH cannot be calculated from Haselbach equation.
pH of soluble dihydrogen phosphates is practically independent on concentration - best choose is
1/2 (pKa1+pKa2)
« Last Edit: May 24, 2005, 04:59:53 AM by AWK »
AWK

Offline Donaldson Tan

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Re:pH of a buffering system
« Reply #6 on: May 24, 2005, 05:27:01 AM »
Why the pH is independent of the salt concentration?

 :Confuzzled:
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

Offline AWK

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Re:pH of a buffering system
« Reply #7 on: May 24, 2005, 06:26:41 AM »
AWK

Offline Donaldson Tan

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Re:pH of a buffering system
« Reply #8 on: May 24, 2005, 07:40:22 AM »
(1) H2PO4- + H2O <-> HPO42- + H3O+
(2) H2PO4- + H2O <-> H3PO4 + OH-

the first step is govened by Ka2 and the second step is governed by Kb1.

I realised where i went wrong. cheers, awk

kb1 = Kw/Ka3 = 0.0222
Since Kb1/Ka2 = 352733, then Kb1 is primarily responisble for the pH, and not Ka2.

initially,
[ H2PO4- ] = 0.5M
[ H3PO4 ] = 0 M
[ OH- ] = 1E-7 M

at equilibrium,
[ H2PO4- ] = (0.5 - X) M
[ H3PO4 ] = X M
[ OH- ] = 1E-7 + X ~ X M (assume X >> 1E-7)

Kb1 = X2/(0.5 - X)

solving quadratically,
X2 + (Kb1)X - (0.5)(Kb1) = 0
X = 0.0948 ~ 0.095M (check X >> 1E-7)

pH = 14 - pOH = 14 + lg x = 12.977 ~ 13

I still don't get borek's answer. how come?
« Last Edit: May 24, 2005, 07:41:15 AM by geodome »
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

Offline Borek

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Re:pH of a buffering system
« Reply #9 on: May 24, 2005, 07:45:08 AM »
the dihydrogen phosphate ion is amphoteric

OK, I am back, after some fast hydraulic repairs I was forced to do at home instead of answering pH questions ;)

In a classical sense buffer is the mixture of acid and conjugate base in similar concentrations. So solution containing H3PO4/H2PO4-, or H2PO4-/HPO42- pairs - in comparable concentrations - are buffers. Solution containing only one of these ions is not.

It can be all deducted from HH ;)
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Re:pH of a buffering system
« Reply #10 on: May 24, 2005, 07:56:12 AM »
Hydraulic Repairs? It must be one of the byproducts of a married life.. LOL

borek: do you mean H2PO4- cannot be a buffer because the solution don't consist of the acid and its conjugate base. It's amphoteric, so it can react with both acid and alkali. I can't see how..

the PDF from AWK's link says that H3PO4/H2PO4- and H2PO4-/HPO42- can act as buffer each, as long Ka1/Ka2 > 1000 and Ka2/Ka3 > 1000
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

Offline Borek

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Re:pH of a buffering system
« Reply #11 on: May 24, 2005, 08:09:24 AM »
pH = 14 - pOH = 14 + lg x = 12.977 ~ 13

This result is much worse than the previous one - and it is so obvious I don't understand how you have overlooked it. H2PO4- is an acid - weak, but acid. Yet you have found the pH to be 13...

AWK referred rather to the page 29 of the pdf.

And here is my $.02, similar to the approach from the pdf, but different in one important aspect:

assume H2A is our acid, HA, A and H being ions (charges neglected to speed up writing):

Ka1 = [H][HA]/[H2A]
Ka2 = [H][A]/[HA]

three unknowns, two equations, it doesn't look good, but lets forget about the problem for now. Solve first equation for [HA]

[HA] = Ka1[H2A]/[H]

insert into second equation:

Ka2 = [H]^2[A]/(Ka1[H2A])

solve for [H]

[H] = sqrt(Ka1 Ka2 [H2A]/[A])

What is important here is the fact, that this is very similar to the equation from the page 29 of the pdf, but it also clearly shows that assumptions done in pdf boil down to one - [H2A] = [A]. If so

[H] = sqrt(Ka1 Ka2)

or

pH = (pKa1+pKa2)/2

For the phosphoric acid situation is exactly the same - third proton behaves like nailed and Ka3 doesn't disturb above equlibrium.

How far is this from the real results?

Not too far. pH = 4.7 if it is calculated as avearge of pKa1 and pKa2. If calculated using BATE for different NaH2PO4 concentrations:

concpH
0.1M4.69
0.01M4.79
0.001M5.13

and so on.
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Re:pH of a buffering system
« Reply #12 on: May 24, 2005, 08:09:25 AM »
Hydrogen phosphates and dihydrogen phospates act as buffers, but calculation pH of its solutions cannot be done by Haselbach equation, which is to simplified to be true in this case.
Use equation which is given in my link, or its simplified form given by me previously.
Borek BATE program use activities, which in fact give correct results, but students use concentrations and the difference betwen such calculations and reality is quite large for  for diluted solution of multicharged  ions.
« Last Edit: May 24, 2005, 08:16:55 AM by AWK »
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Offline Borek

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Re:pH of a buffering system
« Reply #13 on: May 24, 2005, 08:21:27 AM »
Hydraulic Repairs? It must be one of the byproducts of a married life.. LOL

You are not far from the truth, but in this case it was rather effect of hiring lousy plumbers three years ago.

Quote
borek: do you mean H2PO4- cannot be a buffer because the solution don't consist of the acid and its conjugate base. It's amphoteric, so it can react with both acid and alkali. I can't see how..

For the buffer to be efficient it must be able to compensate for acid or base addition. It is described by the parameter called buffer capacity or something like that (no idea about English term). If you look at the HH equation it is obvious that the largest buffer capacity is when the concentration of both acid and conjugate base are the same - that's why you should choose a buffer with pKa as close to desired pH as possible. If you take a solution of H2PO4- it will not act as a buffer - small additions of acid or alkali will result in large changes of pH. Funny thing is, adding acid or base you will effectively increase the buffer capacity, as you will get closer with pH to pKa1 or pKa2.

Quote
the PDF from AWK's link says that H3PO4/H2PO4- and H2PO4-/HPO42- can act as buffer each, as long Ka1/Ka2 > 1000 and Ka2/Ka3 > 1000

Not exactly if I recall correctly - it rather states that if differences between pKa values are large enough you can use HH equation for calculation independently for both buffers. If the Ka values are too close, HH equation is no longer valid. However, such substances (TRIS, citric acid) are very good for buffer creation. They can be used in much wider pH range, cause the buffering effects of different dissociation steps overlap.
« Last Edit: May 14, 2006, 08:43:26 AM by Borek »
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Offline Borek

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Re:pH of a buffering system
« Reply #14 on: May 24, 2005, 08:23:49 AM »
Borek BATE program use activities, which in fact give correct results, but students use concentrations and the difference betwen such calculations and reality is quite large for  for diluted solution of multicharged  ions.

No, BATE allows you to select whether activities are calculated or not. All results posted here were for calculations done with the ionic strength neglected. So no difference between my results and average student results.
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