Phosphoric acid, H3PO4(aq), is a triprotic acid, meaning that one molecule of the acid has three acidic protons. Estimate the pH, and the concentrations of all species in a 0.250 M phosphoric acid solution.

[H3PO4] = ?
[H2PO4^-] =?
[HPO4^2-]=?
[PO4^3-]=?
[H^+]=?
[OH^-]=?
pH=?

Tell me what you don't understand about this.

i converted from pKa to Ka for all three values. i don't know where to go from there.

What about looking at the three Ka values seprately? You notice that they are about 10^5 from each other. That means that the (H^+) is determined primarily by the first ionization. The second and third ionization produce little H^+. So do the first calculation as you would a monoprotic acid and (at least for the time being) ignore the second and third. Do that and see if you can see where to go from there.

Ka1 is relatively large; probably you will need to solve the quadratic for Ka1.

To estimate the pH and concentrations of species in a phosphoric acid solution, we need to consider the equilibrium reactions that occur when phosphoric acid (H3PO4) dissociates.

Phosphoric acid can dissociate in three stages, as follows:

1. H3PO4 ⇌ H+ + H2PO4^-
2. H2PO4^- ⇌ H+ + HPO4^2-
3. HPO4^2- ⇌ H+ + PO4^3-

First, let's calculate the concentration of H3PO4 in the solution. Given that the initial concentration of phosphoric acid is 0.250 M, we can assume that it completely dissociates in the solution, meaning that the concentration of [H3PO4] is still 0.250 M.

Next, let's determine the concentrations of the different species:

For the first dissociation reaction:
[H3PO4] = 0.250 M
[H+] = 0.250 M
[H2PO4^-] is formed from dissociation, so its concentration is also 0.250 M.

For the second dissociation reaction:
[H2PO4^-] = 0.250 M
[H+] is formed from dissociation, so its concentration will increase. It will be 0.250 M + [H+] (we need to solve for [H+]).

For the third dissociation reaction:
[HPO4^2-] is formed from dissociation, so its concentration is [HPO4^2-] = [H+] (as it's a 1:1 ratio).

Lastly, we can calculate the [PO4^3-] concentration by using the fact that the total concentration of phosphoric acid is 0.250 M:
[HPO4^2-] + [PO4^3-] = 0.250 M
[PO4^3-] = 0.250 M - [HPO4^2-] = 0.250 M - [H+] (since [HPO4^2-] = [H+] as stated before).

To find the pH of the solution, we need to determine the concentration of H+ ions. Since the first dissociation reaction produces H+ and H2PO4^-, we need to consider both [H+] and [H2PO4^-]:
[H+] = 0.250 M + [H+]

To solve for [H+], we can use the equilibrium expression for the second dissociation reaction, which is:

Ka2 = ([H+][HPO4^2-])/[H2PO4^-]

Given that the Ka2 value for phosphoric acid is 6.3 × 10^-8, we can substitute the concentrations in the equation and solve for [H+].

Once we have the [H+], we can calculate [OH^-] using the fact that water at 25°C is neutral, meaning [H+] = [OH^-]. So, [OH^-] will have the same concentration as [H+].

Lastly, we can calculate the pH using the equation: pH = -log[H+].

Please note that the calculations involve solving equations and may require a scientific calculator or software.