Question

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 0.200 M phosphoric acid solution.
pKa1 = 2.16
pKa2 = 7.21
pKa3 = 12.32
Ok I haven't been able to solve problems like this. I'm scared to start it because I desperately need the points and I don't get this concept. I asked my teacher and I've look up other college websites and I still don't know where to begin.
H3PO4 =
H2PO4- =
HPO4^2- =
PO4^3- =
H^+ =
OH^- =
pH =

Answer 1

A complete calculation can be obtained at https://www.google.com/search?q=calculate+M+ions+in+H3PO4&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a&channel=sb and click on the Chapter 7 pdf file document. They use 5.0M H3PO4 and not 0.2 but the process is the same. Below I've gone through the first ka1.
You begin with the first K; i.e., ka1 and you work them just like you would a 0.2M solution of acetic acid or NH3. Start with an ICE chart.
.......H3PO4 --> H^+ + H2PO4^2-
I.......0.2M.....0........0
C.......-x.......x........x
E......0.2-x.....x........x
If pKa1 = 2.16 then Ka1 = 6.92E-3
Ka1 = (H^+)(H2PO4^2-)/(H3PO4)
6.92E-3 = (x)(x)/(0.2-x)
and solve for x = (H^+), then convert to pH.

For k2 you know (H^+) = (H2PO4^-) which makes (HPO4^2-) = k2.

For the last one you have
k3 = (H^+)(PO4^3-)/(HPO4^2-)
Plug in (H^+) from your calculation in k1. Plug in (HPO4^2-) from your calculation in k2. Solve for PO4^3-