Speciation of acid doesn't depend on the base (other than through pH). But just like there are four forms of the phosphoric acid present in the solution, there are several forms of each base present in the solution - and you should calculate concentration of each form using Kb values.
Yes, I see - this I've done before (in my practice, it seems that the equations needed are analogous to the speciation calculations for acids, with Ca replaced by Cb, Ka by Kb and [H3O+] by [OH-]).
Calculate pH first, then calculate speciation using the same approach we are talking about all the time.
Let me analyse what I think I know and then ask my doubts:
The salt Al2(HPO4)3 is assumed to dissolve entirely into solution (in the absence of a Ksp value for it). This leaves 2 Al3+
ions and 3 HPO42-
ions for every mole of the salt dissolved. The ions for which I need concentration at equilibrium are therefore Al3+
(let us assume Kb1 refers to the loss of OH- from Al(OH)3 to Al(OH)2+
, Kb2 from Al(OH)2+
, and Kb3 from Al(OH)2+
, and H3
(let us assume Ka1 refers to the loss of H+ from H3PO4 to H2PO4-
, Ka2 refers to the loss of H+ from H2PO4-
and Ka3 to the loss of H+ from HPO42-
, just like in a solution of the normal acid H3
I assume, when you say that the same approach fits all, you mean the speciation equation will still solve this. Here is my interpretation of that. Please point out the mistakes where they are there, or tell me if this is correct.
Let the concentration of Al2(HPO4)3 be Cs
. Use the equation as you would for a normal base (with Al(OH)3 the form which has all hydroxides, Al(OH)2+
the form which has lost 1, Al(OH)2+
the form which has lost 1, and Al(OH)2+
the form which has lost 2, so forth - i.e. just like we would calculate Al(OH)3 for a base, we use the same one of the speciation equations to find the concentration of Al(OH)3 for this solution), but for Cb
, use Cb
(since the coefficient of Al in the formula of this salt is 2). Plug that in along with the original Kb values in the correct order from the base Al(OH)3
, and with the calculated [OH-], and you can find the concentration of each species from Al3+
Next, use the equation as you would for a normal acid (i.e. to calculate [H3
] in this solution we want the same form of the speciation equation as used to calculate H3
from a solution of that acid), but replacing Ca
(since the coefficient on the source of the anion PO43-
is 3). Plug that in along with the original Ka values in the correct order from the acid H3PO4, and with the calculated [H3O+], and you can find the concentration of each species from PO43-
, to H3
Does this method work to find the ionic concentrations?