H3PO4 ==> H^+ + H2PO4^-
H2PO4^- ==> H^+ + HPO4^2-
HPO4^2- ==> H^+ + PO4^3-
Write the expressions for ka1, ka2, and ka3.
You solve k1 as if it were a monoprotic acid; ie.
......H3PO4 ==> H^+ + H2PO4^-
I......0.1......0......0
C......-x.......x......x
E.....0.1-x.....x......x
Since ka1 is so large (relatively) you will need to use the quadratic equation but that will give you (H^+), H2PO4^-, and the 0.1-x will be H3PO4.
Go to k2 expression. You know H^+, you know H2PO4^- and you solve for HPO4^2- (which by the way will be just k2 since (H^+) = (H2PO4^-.
Only one more to go. Use k3.
Ka3 = (H^+)(PO4^3-)/(HPO4^2-).
You know (H^+) from above in ka1, you know (HPO4^2-) from k2 so you can solve for (PO4^3-). Post your work if you get stuck.
Calculate the concentrations of all species present in a 0.100 M solution of weak polyprotic acid H3PO4. Ka1=7.5 x 10 ^-3, ka2 = 6.2 x 10 ^-8, Ka3= 3.6 x 10^-13
How the heck do I do this
5 answers
H2PO4 Undergoes 3 separate ionization steps. Each is treated as a monoprotic ionization. The concentrations at equilibrium for the products of the 1st ionization step are the beginning concentrations for the 2nd ionization and the products of the 2nd ionization step are the beginning concentrations for the 3rd ionization step. Once you see how these work, multiprotic ionizations can be quite easy. I'll try to illustrate...
1st Izn: H3PO4 <=> H + H2PO4
Ci: 0.100M 0 0
∆C -x +x +x
Ceq 0.10-x
≈0.10M
Ka1 = [H][H2PO4]/[H3PO4]
=(X)(X)/(0.1) = (X)^2/(0.10)
= 7.5E-3
Solving for x =>
x = [(7.5E-3)(0.10)]^0.5 = 0.027M for [H]1st izn = [H2PO4]1ST Izn
2nd Ionization:
H2PO4^- <=> H + HPO4^2-
Ceq:0.27M 0.027M x2
This is cool...
Ka2 = [H][HPO4]/[H2PO4]
Ka2 = ([H](0.027)/0.027) = [H] = 6.2E-8M
The concentration of [H]2 from second ionization step is simply = Ka2 = 6.2E-8M
The same is true for the 3rd ionization step. [H]3 = Ka3 = 3.6E-13M
In summary [H] from 1st izn stem can be determined from (Ka1[Acid])^1/2.
After the 1st ionization [H]2=Ka2 and [H]3 = Ka3
(This assumes no common ions or complex ions present. Only water)
1st Izn: H3PO4 <=> H + H2PO4
Ci: 0.100M 0 0
∆C -x +x +x
Ceq 0.10-x
≈0.10M
Ka1 = [H][H2PO4]/[H3PO4]
=(X)(X)/(0.1) = (X)^2/(0.10)
= 7.5E-3
Solving for x =>
x = [(7.5E-3)(0.10)]^0.5 = 0.027M for [H]1st izn = [H2PO4]1ST Izn
2nd Ionization:
H2PO4^- <=> H + HPO4^2-
Ceq:0.27M 0.027M x2
This is cool...
Ka2 = [H][HPO4]/[H2PO4]
Ka2 = ([H](0.027)/0.027) = [H] = 6.2E-8M
The concentration of [H]2 from second ionization step is simply = Ka2 = 6.2E-8M
The same is true for the 3rd ionization step. [H]3 = Ka3 = 3.6E-13M
In summary [H] from 1st izn stem can be determined from (Ka1[Acid])^1/2.
After the 1st ionization [H]2=Ka2 and [H]3 = Ka3
(This assumes no common ions or complex ions present. Only water)
The 'Drop x' simplification can be determined from this ... if the Conc/Ka > 100 the inherent error will be insignificant. For the 1st ionization I used just (0.10/x10^3) = 100. The concentrations for the 2nd and 3rd ionizations drop off so fast and make very little contribution to the 1st ionization. I punked the simplification rule and just went with the square root formula for Ka1 calculation.
Using the quadratic (H^+) from ka1 is 0.023 and not 0.027 which makes H3PO4 then 0.10-0.023 = 0.077 M
Yes, (H^+) from ka2 is = ka2 and that is insignificant.
Yes and No for part ka3. (H^+) is that from part 1; i.e., 0.023 and not = k3. Likewise, (HPO4^2-) is known from ka2 calculation. Therefore, (PO4^3-) is not = ka3.
Yes, (H^+) from ka2 is = ka2 and that is insignificant.
Yes and No for part ka3. (H^+) is that from part 1; i.e., 0.023 and not = k3. Likewise, (HPO4^2-) is known from ka2 calculation. Therefore, (PO4^3-) is not = ka3.
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1 answer