MeshPL
Hazard to Others
Posts: 329
Registered: 20-4-2015
Location: Universe
Member Is Offline
Mood: No Mood
|
|
Buffer pH problem
Hello everybody!
Can I rely on buffer solutions compositions posted on this site?:
http://delloyd.50megs.com/moreinfo/buffers2.html
I'm mainly intrested in the following ones:
100ml 0.1M KH2PO4 with 11.2ml 0.1M NaOH for pH=6.0
100ml 0.1M KH2PO4 with 58.2ml 0.1M NaOH for pH=7.0
100ml 0.1M KH2PO4 with 93.4ml 0.1M NaOH for pH=8.0
100ml 0.05M NaHCO3 with 21.4ml 0.1M NaOH for pH=10.0
I tried putting them into pH calculators on the Internet, mainly this one:
http://www.webqc.org/phsolver.php
but they don't seem to work properly, nor be suitable for such calculations.
However the results were only ~0.3 pH "incorrect". Results from buffer compositions found on wikipedia were less accurate.
I do not need pH to be exactly equal to that 6, 7, 8 or 10, I don't have aparature for such precision.
I'm making natural pH indicators (red cabbage, red onion, red grapes, carrots, rhubarb, curcuma, maybe cherries, red gooseberries, red currant,
aronia) and I would like to test them in those specific pH's (but I do not need high precusion of course)
I'll show you results next week or so.
[Edited on 3-7-2015 by MeshPL]
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
Quote: Originally posted by MeshPL  |
I'm mainly intrested in the following ones:
100ml 0.1M KH2PO4 with 11.2ml 0.1M NaOH for pH=6.0
100ml 0.1M KH2PO4 with 58.2ml 0.1M NaOH for pH=7.0
100ml 0.1M KH2PO4 with 93.4ml 0.1M NaOH for pH=8.0
100ml 0.05M NaHCO3 with 21.4ml 0.1M NaOH for pH=10.0
|
Especially on phosphate buffers there are lots of reliable resources.
But have you tried calculating these values from theory? It's not hard to do.
Calculating values for the first three, I get 6.30; 7.30 and 8.35 for instance.
[Edited on 3-7-2015 by blogfast25]
|
|
|
MeshPL
Hazard to Others
Posts: 329
Registered: 20-4-2015
Location: Universe
Member Is Offline
Mood: No Mood
|
|
I would like to calculate them, but I'm not sure how. How am I supposed to implement water autodisdociation? Or should I remember about water being
used up in formation of H3O+ or OH-. That are the main problem for me. Otherwise
calculating them wouldn't be too hard for me (especially with some Wolframalpha system of equation solver  )
The reason I'm interested in those buffer solutions is the fact that only phosphate I currently have is KH2PO4 (around 400g). I also don't have KOH,
but >500g of NaOH. Those compositions were useful because of the use of chemicals possesed by me.
I could try to convert other compositions into ones using chemicals available to me (NaOH, KH2PO4, citric acid, NaHCO3, borax, distilled vinegar 10%),
but I'm afraid they would be inaccurate. Or would they? Nevertheless converting percentage concentration into molar or whatever may require density
data and so on.
Do you have any particular compositions you would advise me to try? (Including those requiring conversion)
[Edited on 4-7-2015 by MeshPL]
[Edited on 4-7-2015 by MeshPL]
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
| Quote: Originally posted by MeshPL | I would like to calculate them, but I'm not sure how. How am I supposed to implement water autodisdociation?
The reason I'm interested in those buffer solutions is the fact that only phosphate I currently have is KH2PO4 (around 400g). I also don't have KOH,
but >500g of NaOH. Those compositions were useful because of the use of chemicals possesed by me.
I could try to convert other compositions into ones using chemicals available to me (NaOH, KH2PO4, citric acid, NaHCO3, borax, distilled vinegar 10%),
but I'm afraid they would be inaccurate. Or would they? Nevertheless converting percentage concentration into molar or whatever may require density
data and so on.
Do you have any particular compositions you would advise me to try? (Including those requiring conversion)
|
No, water dissociation is generally not taken into account, because the concentration of the buffer reagents tends to be soooo much larger than that
effect. It also makes the calculation much harder.
The buffer formula is:
pH<sub>buffer</sub> = pK<sub>a</sub> - log[C<sub>A</sub>/C<sub>B</sub>] (= pK<sub>a</sub>
+ log[C<sub>B</sub>/C<sub>A</sub>], if you prefer)
(A is acid, B is base)
For phosphate buffers, the acid is H<sub>2</sub>PO<sub>4</sub><sup>-</sup> and the base is
HPO<sub>4</sub><sup>2-</sup>. The pK<sub>a</sub> is the pK<sub>a2</sub> = 7.2 of phosphoric acid
(second dissociation).
Other useful and ‘easy’ buffers are acetate buffers: acetic acid + NaOH. For CA = CB, pH = pK<sub>acetic acid</sub> = 4.76.
On eBay you can also buy sachets for 250 ml buffer sols, usually 4, 7 and 10 pH.
There’s also Cream of Tartar, not strictly a buffer system but very OTC:
https://en.wikipedia.org/wiki/Potassium_bitartrate#Chemistry
To test home made indicators you can also simply use solutions of NaOH or HCl of specified pH:
pH = 14 + logC<sub>NaOH</sub> (for pH 14 to 8) [Edit: + sign, not minus. Grrr!]
or:
pH = - logC<sub>HCl</sub> (for pH 1 to 6)
[Edited on 4-7-2015 by blogfast25]
|
|
|
MeshPL
Hazard to Others
Posts: 329
Registered: 20-4-2015
Location: Universe
Member Is Offline
Mood: No Mood
|
|
I made an ugly system of equations for calculating phosphate buffer ph:
a+b=c; a*x/c=d; b*y/c=z; f=d-z; p=7.2-log10(f/z); a=0.1; b=0.0582; x=0.1; y=0.1
Values of a, b, x and y can be changed. a is KH2PO4 volume, b is for NaOH volume, x is for molarity of KH2PO4 solution and y is molarity of NaOH
solution.
Results are very simillar to yours, so I guess I made a correct one. Here is the equation solver I used:
https://www.wolframalpha.com/examples/EquationSolving.html
Thanks for your help! I'll figure out good buffer compositions and I'll post pictures of indicators soon (after a week or so  ).
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
That's because what you've done is the same but more elaborately. But you really don't need and equation solver for that! You only need to plug the
calculated values for C<sub>A</sub> and C<sub>B</sub> into the buffer equation.
E.g. 100ml 0.1M KH2PO4 with 11.2ml 0.1M NaOH for pH=6.0
CA = (0.1 l x 0.1 M - 0.0112 l x 0.1 M)/(0,1 l + 0.0112 l)
CB = (0.0112 l x 0.1 M)/(0,1 l + 0.0112 l)
Thus CA/CB = (100 - 11.2)/11.2 = 7.93
Plug in and you get pH = 6.3
One can of course generalise that for different volumes and molarities but still only a pocket calculator is needed.
|
|
|
smaerd
International Hazard
Posts: 1262
Registered: 23-1-2010
Member Is Offline
Mood: hmm...
|
|
It really all depends. Calculations are great and all but when it comes to systems with multiple components its a question of "how much math do you
want to do before making the solution and measuring it's pH?"
There are techniques for solutions containing many MANY pH effecting species that can be solved via a simple spread sheet. This is important when one
takes into account solvated carbon dioxide (and the carbonate 'cycle'). As well as other things. Also factors such as ionic strength effect acid base
equilibria in certain cases. If you or anyone wants to know how this is done(it's not typically taught at an undergraduate level) let me know and I
will make a thread detailing the methodology in plain english.
What I typically do for general lab work is prepare the solution in accordance with the good old Henderson-Hasselbalch (described above by
blogfast25). Well, I'll make 90-95% of the solution in a beaker(slightly concentrated| ex: 90mL soln. rather then the full 100mL). Then I measure it
with a pH meter.
If my pH is a too low and I used sodium salts to prepare it I'll titrate the solution to the desired pH with a solution of sodium hydroxide.
If my pH is too high, I will add the acidic component (skews the molar concentration), or a bit of a strong acid (skews the ionic strength slightly).
Then I dilute the solution up to the desired concentration in a volumetric flask. Remember buffer solutions tend to resist changes in pH even on
dilution, so doing this is A-Okay. Acid strength can vary, and manufacturers (especially in the amateur acquisition setting) are not always on point
with their concentrations.
Hope this gives some real world advice on preparing buffers.
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
smaerd:
MeshPL guy wants some buffers to test home made indicators (fruit and veg juice, basically) with. We're not talking 3 significant digit pH buffers
here. Of course uncorrected Hasselbalch is only approx., that should go w/o saying, really.
Using activities from ionic strength calcs would fly right over 99 % of heads here: you'd be preaching to the converted. 
[Edited on 5-7-2015 by blogfast25]
|
|
|
MeshPL
Hazard to Others
Posts: 329
Registered: 20-4-2015
Location: Universe
Member Is Offline
Mood: No Mood
|
|
Juices or not (infusions, extracts... I'm actually going to boil theese things to extract dyes and to concentrate them) using buffers has 2
advantages:
-sometimes preparing non-buffer solution with pH~6 (or whatever non-extreme pH) is close to impossible (I stumbled upon: dilute 1ml vinegar in 200ml
of water and take 1ml of that solution into 250ml of water. Sounds like homeopathic drug recepture. )
-buffering effect will help negate indicators own pH.
|
|
|
![[*]](images/xpblue/default_icon.gif){kind=icon}
