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 0.300 M phosphoric acid solution.

Question

pKa1: 2.16
pKa2: 7.21
pKa3: 12.32

Hint: Convert the pKa values to Ka values.

Answer 1

.......H3PO4 ==> H^++ H2PO4^-
I........0.3......0......0
C........-x.......x.......x
E......0.3-x......x.......x

k1 = (H^)(H2PO4^-)/(H3PO4)
Substitute the E line into Ks expression and solve for x = (H^+), convert to pH.

........H2PO4^- ==> H^+ + HPO4^2-
From k1 ionization you see(H+) = (H2PO4^-)
k2 = (H^+)(HPO4^2-)/(H2PO4-)
and since (H^+) is in the numerator and H2PO4^- is is the denominator, they cancel and k2 = (HPO4^2-)

.......HPO4^2- ==> H^+ + PO4^3-

k3 = (H^+)(PO4^3-)/(HPO4^2-)
I would substitute (H^+) from the first calculation you did, (HPO4^-) from the second calculation, then solve for (PO4^3-).

Answer 2

How do you find the concentration of[OH-]? :C

Answer 3

Would it be
pOH = 14 - pH
Then, [OH-] = 10^-pOH
After finding [OH-] find the mol ratio?

Answer 4

To find [OH]
pOH =14 - pH
[OH-] = 10^ - pOH

mol ratio is not required

Answer 5

KA1= 6.9e-3
KA2= 6.2e-8
KA3= 4.8e-13
KA2 and KA3 are so small, only use KA1 for [H+]

H3PO4 -> H+ H2PO4
I .350 0 0
C -x +x +x
E .35-x x x
KA1= x^2/.35-x 6.9e-3
x= .046 = [H+]
pH= -log(.046)=1.34
[OH-]=1e-14/.046
=2.2e-13

[H3PO4]=(.35-.046)M
=.304M

H2PO4- ->H+HPO42-
I .046 .046 0
C -y +y +y
E .046-y .046-y y

KA2= y(.046+y)/(.046-y)
=6.2e-8
y=6.2e-8

[HPO4^2-]=6.2e-8M
[H2PO4^-]=.046-6.2e-8=.046M

HPO4^2- >H+PO4^3-
I 6.2e-8 .046 0
C -z +z +z
E 6.2e-8-z .046+z z

KA3= z(.0460/
(6.2e-8)-z=4.8e-13

z=6.5e-19M=
[PO4^3-]

Answer 6

substitute your M for the M found here (0.350)