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.

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-

Answer 2

I understand that this concept might seem intimidating, but I'll guide you through the process step-by-step so that you can solve this problem with confidence.

To find the pH and concentrations of all species in a 0.200 M phosphoric acid (H3PO4) solution, we'll use the given pKa values. Let's begin:

Step 1: Identify the relevant equilibrium reactions
In this case, we have three acidic protons in phosphoric acid. Each proton can ionize to form different species. The equilibrium reactions are as follows:

H3PO4 ⇌ H+ + H2PO4- (Equation 1)
H2PO4- ⇌ H+ + HPO4^2- (Equation 2)
HPO4^2- ⇌ H+ + PO4^3- (Equation 3)

Step 2: Determine the ionization constants
The given pKa values correspond to the negative logarithm of the ionization constants for each equilibrium reaction. To obtain the ionization constants, we need to take the antilog of the given pKa values:

Ka1 = 10^(-pKa1) = 10^(-2.16)
Ka2 = 10^(-pKa2) = 10^(-7.21)
Ka3 = 10^(-pKa3) = 10^(-12.32)

Step 3: Calculate the concentrations of all species
Since the initial concentration of phosphoric acid (H3PO4) is 0.200 M, we'll start by assuming that all of it is in the form of H3PO4. Let x be the concentration of H+ for each equilibrium. Now, let's calculate the concentrations of all species in each equilibrium step:

Equation 1:
H3PO4 ⇌ H+ + H2PO4-
Initial: 0.200 M 0 0
Change: -x +x +x
Equilibrium: 0.200 - x x x

Equation 2:
H2PO4- ⇌ H+ + HPO4^2-
Initial: x 0 0
Change: -x +x +x
Equilibrium: x x x

Equation 3:
HPO4^2- ⇌ H+ + PO4^3-
Initial: x 0 0
Change: -x +x +x
Equilibrium: x-x x x

Step 4: Determine x and concentrations at equilibrium
Since phosphoric acid (H3PO4) is a triprotic acid, we'll have three equilibrium steps. However, the third ionization constant (Ka3) is significantly smaller than the other two, indicating that the third equilibrium will have a negligible effect on the solution.

Therefore, we can focus on Equations 1 and 2. To find x, we'll solve for x by setting up an expression for the total concentration of H+:

[H+] = [H2PO4-] + [HPO4^2-]
[H+] = x + x
[H+] = 2x

Now, using the expression for x and the given initial concentration of H3PO4:

[H3PO4] = 0.200 M
[H3PO4] = [H+ in Equation 1]
0.200 = 0.200 - x
x = 0.200 M

So, the concentration of H+ at equilibrium is 0.200 M.

Now, using the equilibrium concentrations for each species, we can complete the table:

H3PO4: 0.200 - x = 0.200 - 0.200 = 0 M
H2PO4-: x = 0.200 M
HPO4^2-: x = 0.200 M
PO4^3-: 0 M

Step 5: Calculate pH
Since the concentration of H+ is equal to x, the pH can be calculated using the formula:

pH = -log [H+]
pH = -log (0.200)
pH ≈ 0.70

So, the estimated pH of a 0.200 M phosphoric acid solution is approximately 0.70.

To summarize:
H3PO4 = 0 M
H2PO4- = 0.200 M
HPO4^2- = 0.200 M
PO4^3- = 0 M
H+ = 0.200 M
OH- = We didn't calculate this, but you can determine it by using the equation Kw = [H+][OH-] = 1.0 x 10^-14 at room temperature.
pH ≈ 0.70