alzabo said:
hello and thanks,
I would very much like to go into the calculations. My strongest math skills are linear rate equations. Although I used to understand more complicated things such as integration/differentiation/minima/maxima & titration, that was several years ago and i remember very little of it.
This is related to my
"Questions on cold water technique" thread (post #2).
I have a very conrete example:
If one wished to create a solution of phosphoric acid and water at a ph of 2.5, how would they go about it without using PH paper?
Assuming that one has the following materials on hand.
1.) Water.
2.) A measuring cup to measure liters.
3.) An eye dropper to measure milliliters.
4.) Phosphoric acid solution at 25%.
5.) A milligram scale.
Eeew, you went for a triprotic acid!!!
That is an "extreme" example, because phosphoric acid has 3 pKa numbers. Phosphoric acid dissociates like this to provide protons (or H+, the stuff that makes a solution acidic) in the water:
H3PO4 <-> H2PO4- and H+ :: pKa=2.15 (as shoe said)
H2PO4- <-> HPO4-- and H+ :: pKa=7.2
HPO4-- <-> PO4--- and H+ :: pKa-12.38
Phosphoric acid is a weak acid and it does not fully dissociate in solution. Hence the "
<->". If it were a strong acid, like hydrochloric acid I'd write: HCl
-> H+ and Cl. The pKa means the pH at which half of the species will be dissociated and half will be non-dissociated, e.g at a pH of 2.15 50% of the acid will be H3PO4 and 50% will be H2PO4-
As one dissolves the acid in water, the acid will be mostly dissociating and the more the pH goes down the less phosphoric acid will dissociate, until it reaches the pH of 2.15, where 50% will be dissociated.
Now, if you want to calculate how much acid to add in water to get a pH of 2.5 it will be easier to consider only the first dissociation reaction: H3PO4 <-> H2PO4- and H+
because the second (H2PO4- <-> HPO4-- and H+) and the third (HPO4-- <-> PO4--- and H+) do not practically happen in this low pH. The next step is to use the formula that shoe provided, by substituting 2.5 for the pH and 2.15 for the pKa. We then solve the equation for the logarithmic ratio:
2.5 = 2.15 + log([H2PO4-]/[H3PO4]) ->
0.35 = log([H2PO4-]/[H3PO4]) ->
(we solve for the logarithm) [H2PO4-]/[H3PO4] = 10^0.35 [[["^" means "to the power"]]]
[H2PO4-]/[H3PO4] = 2.24.
The ration therefore of dissociated to non-dissociated acid is 2.24. This cannot be solved further since we know neither how much [H2PO4-] nor [H3PO4] is. We can however find [H2PO4-].
The concentration of the dissociated [H2PO4-] is exactly the same as the [H+], as shown in the dissociation equation H3PO4 <-> H2PO4- and H+. According the the stoichiometry of the reaction, there are same amount of species [H+] as there are species of [H2PO4-]. At a pH of 2.5 there are -log[H+] of pH, or:
2.5 = -log[H+] ->
-2.5 = log[H+] ->
[H+] = 10^(-2.5) ->
[H+] = 0.0031 moles/litre
or, in other words, [H+] = [H2PO4-] = 0.0031.
We now go back to the [H2PO4-]/[H3PO4] ratio and by substituting we get:
[H2PO4-]/[H3PO4] = 2.24 ->
0.0031 / [H3PO4] = 2.24 ->
[H3PO4] = 0.0013.
So in the dissociation equation we have:
H3PO4 <-> H2PO4- and H+
0.0013 and 0.0031 and 0.0031 or if we convert these into grams /litre taking into account that the molecular weight of H3PO4 is 98, the molecular weight of H2PO4- is 97 and the molecular weight of H+ is 1:
H3PO4 <-> H2PO4- and H+
(0.1274 and 0.3007 and 0.0031) grams per litre,
or a total of 0.4312 grams of phosphoric acid in 1 litre of water. And finally, if one has a 25% solution of phosphoric acid, that means 25 grams phosphoric acid/100ml water therefore if one gets in order for one to get (100*0.4312)/25 = 1.725 ml of a 25% phosphoric acid in a total of 1 litre solution to get a pH of 2.5
I understand that may be far more complicated than the half-assed Henderson-Hasselbach equation given by shoe, so please feel free to ask back if something is obscure or does not make much sense!
