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.
kPa1: 2.16
kPa2: 7.21
kPa3:12.32
[H3PO4] = ?
[H2PO4-] = ?
[HPO42-] = ?
[PO43-] = ?
[H+] = ?
[OH-] = ?
pH =?
http://www.jiskha.com/display.cgi?id=1362846408
Why did the acid go to therapy?
Because it had too many acidic protons and it needed to sort out its pH-ychological issues!
Now, let's estimate the concentrations of all the species in the 0.300 M phosphoric acid solution.
Since phosphoric acid is triprotic, it can donate three protons. Let's go step by step:
Step 1:
[H3PO4] = 0.300 M
This is the initial concentration of phosphoric acid in the solution.
Step 2:
[H2PO4-] = [H3PO4] × Ka1 / ([H3PO4] × Ka1 + Ka2)
[H2PO4-] = 0.300 M × 2.16 / (0.300 M × 2.16 + 7.21)
[H2PO4-] ≈ 0.123 M
This is the concentration of dihydrogen phosphate ion.
Step 3:
[HPO42-] = [H2PO4-] × Ka2 / ([H2PO4-] × Ka2 + Ka3)
[HPO42-] = 0.123 M × 7.21 / (0.123 M × 7.21 + 12.32)
[HPO42-] ≈ 0.063 M
This is the concentration of hydrogen phosphate ion.
Step 4:
[PO43-] = [HPO42-] × Ka3 / ([HPO42-] × Ka3 + Kw)
[PO43-] = 0.063 M × 12.32 / (0.063 M × 12.32 + Kw)
[PO43-] ≈ 0.051 M
This is the concentration of phosphate ion.
Step 5:
[H+] = [H3PO4] - [H2PO4-] + Kw / [H2PO4-]
[H+] = 0.300 M - 0.123 M + Kw / 0.123 M
[H+] ≈ 0.174 M
This is the concentration of hydrogen ions.
Step 6:
[OH-] = Kw / [H+]
[OH-] ≈ Kw / 0.174 M
[OH-] ≈ 1.15 × 10^-14 M
This is the concentration of hydroxide ions.
Step 7:
pH = -log[H+]
pH ≈ -log(0.174)
pH ≈ 0.76
So, in summary:
[H3PO4] ≈ 0.300 M
[H2PO4-] ≈ 0.123 M
[HPO42-] ≈ 0.063 M
[PO43-] ≈ 0.051 M
[H+] ≈ 0.174 M
[OH-] ≈ 1.15 × 10^-14 M
pH ≈ 0.76
I hope you enjoyed this pH-unny answer! If you have any more chemistry questions, feel free to ask!
To estimate the pH and concentrations of all the species in the phosphoric acid solution, we can use the dissociation constants (Ka values) provided.
Let's start by writing down the dissociation reactions for each proton:
1st dissociation: H3PO4 ⇌ H+ + H2PO4-
2nd dissociation: H2PO4- ⇌ H+ + HPO42-
3rd dissociation: HPO42- ⇌ H+ + PO43-
Considering the initial concentration of H3PO4 is 0.300 M, we can use an ICE (Initial, Change, Equilibrium) table to calculate the concentrations of each species based on these reactions.
[H3PO4] = Initial concentration = 0.300 M (given)
For the first dissociation:
[H3PO4] H+ H2PO4-
0.300 0 0
Let x be the concentration of [H+] and [H2PO4-].
[H3PO4] H+ H2PO4-
0.300-x x x
Using the relation for Ka1:
Ka1 = [H+][H2PO4-]/[H3PO4]
Substituting the values and rearranging:
2.16 = x*x/(0.300-x)
Since the value of x is expected to be small compared to 0.300, we can assume (0.300 - x) ≈ 0.300 and simplify the equation:
2.16 ≈ x^2/0.300
Multiplying both sides by 0.300:
0.648 ≈ x^2
Taking the square root of both sides, we get:
x ≈ 0.805 M
Now we can calculate the concentrations of [H2PO4-] and [HPO42-] because they are directly proportional to x.
[H2PO4-] ≈ x ≈ 0.805 M
[HPO42-] ≈ x ≈ 0.805 M
For the second dissociation:
[H2PO4-] H+ HPO42-
0.805 0 0
Let y be the concentration of [H+] and [HPO42-].
[H2PO4-] H+ HPO42-
0.805-y y y
Using the relation for Ka2:
Ka2 = [H+][HPO42-]/[H2PO4-]
Substituting the values and rearranging:
7.21 = y*y/(0.805-y)
Using the same approximation as before, we simplify the equation:
7.21 ≈ y^2/0.805
Multiplying both sides by 0.805:
5.80905 ≈ y^2
Taking the square root of both sides, we get:
y ≈ 2.41 M
Now we can calculate the concentrations of [HPO42-] and [PO43-] because they are directly proportional to y.
[HPO42-] ≈ y ≈ 2.41 M
[PO43-] ≈ y ≈ 2.41 M
For the third dissociation:
[HPO42-] H+ PO43-
2.41 0 0
Let z be the concentration of [H+] and [PO43-].
[HPO42-] H+ PO43-
2.41-z z z
Using the relation for Ka3:
Ka3 = [H+][PO43-]/[HPO42-]
Substituting the values and rearranging:
12.32 = z*z/(2.41-z)
Using the same approximation as before, we simplify the equation:
12.32 ≈ z^2/2.41
Multiplying both sides by 2.41:
29.6872 ≈ z^2
Taking the square root of both sides, we get:
z ≈ 5.44 M
Now we have calculated the concentrations of all the species:
[H3PO4] ≈ 0.300 M
[H2PO4-] ≈ 0.805 M
[HPO42-] ≈ 2.41 M
[PO43-] ≈ 2.41 M
[H+] ≈ 5.44 M
Finally, we can calculate the concentration of hydroxide ions [OH-] using the relation Kw = [H+][OH-], where Kw is the ion product of water. At room temperature, Kw is approximately 1.0 x 10^-14.
[H+][OH-] = 1.0 x 10^-14
[OH-] = 1.0 x 10^-14 / [H+]
[OH-] ≈ 1.0 x 10^-14 / 5.44
[OH-] ≈ 1.84 x 10^-15 M
To calculate the pH, we can use the equation pH = -log[H+]:
pH ≈ -log(5.44)
pH ≈ 0.264
Therefore, the estimated pH of the 0.300 M phosphoric acid solution is approximately 0.264.
Yes, I couldn't find it, sorry!
Thankssss!