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.
pKa1: 2.16
pKa2: 7.21
pKa3: 12.32
Hint: Convert the pKa values to Ka values.
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-]
To find [OH]
pOH =14 - pH
[OH-] = 10^ - pOH
mol ratio is not required
Would it be
pOH = 14 - pH
Then, [OH-] = 10^-pOH
After finding [OH-] find the mol ratio?
substitute your M for the M found here (0.350)
How do you find the concentration of[OH-]? :C
To estimate the pH and concentrations of all species in a 0.300 M phosphoric acid solution, we can use the relationship between the concentrations of the acidic species and their respective Ka values.
First, let's convert the pKa values to Ka values using the formula Ka = 10^(-pKa).
For pKa1: Ka1 = 10^(-2.16) = 6.31 x 10^(-3)
For pKa2: Ka2 = 10^(-7.21) = 1.29 x 10^(-8)
For pKa3: Ka3 = 10^(-12.32) = 3.98 x 10^(-13)
Now, let's look at the dissociation reactions for phosphoric acid:
H3PO4 ⇌ H+ + H2PO4-
H2PO4- ⇌ H+ + HPO4^2-
HPO4^2- ⇌ H+ + PO4^3-
Since phosphoric acid is triprotic, it will undergo three successive dissociations.
1. First Dissociation:
Let's assume that x Moles/L of H3PO4 dissociate into H+ and H2PO4-. From the balanced equation, we can see that x Moles/L of H3PO4 will release x Moles/L of H+ and x Moles/L of H2PO4-. Since the initial concentration of H3PO4 is 0.300 M, the concentration of H+ and H2PO4- will be x.
[H3PO4] = 0.300 - x M
[H+] = x M
[H2PO4-] = x M
Using the Ka1 value, we can set up the equilibrium expression and solve for x:
Ka1 = [H+][H2PO4-] / [H3PO4]
6.31 x 10^(-3) = x * x / (0.300 - x)
Simplifying the equation gives us a quadratic equation, which we can solve to find the value of x.
2. Second Dissociation:
Now, suppose y Moles/L of H2PO4- dissociate into H+ and HPO4^2-. The concentration of H2PO4- will be (x - y) M, and thus the concentration of H+ and HPO4^2- will also be (x - y) M.
Using the Ka2 value, we can set up the equilibrium expression and solve for y:
Ka2 = [H+][HPO4^2-] / [H2PO4-]
1.29 x 10^(-8) = (x - y) * (x - y) / x
Simplifying the equation gives us another quadratic equation, which we can solve to find the value of y.
3. Third Dissociation:
Finally, suppose z Moles/L of HPO4^2- dissociate into H+ and PO4^3-. The concentration of HPO4^2- will be ((x - y) - z) M, and thus the concentration of H+ and PO4^3- will also be ((x - y) - z) M.
Using the Ka3 value, we can set up the equilibrium expression and solve for z:
Ka3 = [H+][PO4^3-] / [HPO4^2-]
3.98 x 10^(-13) = ((x - y) - z) * ((x - y) - z) / (x - y)
Simplifying the equation gives us another quadratic equation, which we can solve to find the value of z.
Once we have the values of x, y, and z, we can calculate the concentrations of all species in the phosphoric acid solution. The pH can be determined using the expression pH = -log[H+].
.......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-).