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 0.100 M phosphoric acid solution.

pKa1 = 2.16
pKa2 = 7.21
pKa3 = 12.32

[H3PO4] = ?
[H2PO4-] = ?
[HPO42-] = ?
[PO43-] = ?
[H+] = ?
[OH-] = ?
pH =?

To estimate the pH and concentrations of all the species in a 0.100 M phosphoric acid solution, we need to consider the ionization reactions of phosphoric acid. Here are the ionization reactions and their corresponding equilibrium constants (Ka values):

1. H3PO4 ⇌ H+ + H2PO4-
Ka1 = 10^-pKa1 = 10^-(2.16)

2. H2PO4- ⇌ H+ + HPO42-
Ka2 = 10^-pKa2 = 10^-(7.21)

3. HPO42- ⇌ H+ + PO43-
Ka3 = 10^-pKa3 = 10^-(12.32)

Let's start estimating the concentrations.

1. [H3PO4]:
Since the initial concentration of phosphoric acid is given as 0.100 M, the concentration of H3PO4 remains the same.

[H3PO4] = 0.100 M

2. [H2PO4-]:
We can use the Ka1 value to calculate [H+], assuming the concentration of H3PO4 fully dissociates in the first ionization step.

Ka1 = [H+][H2PO4-]/[H3PO4]
[H+][H2PO4-] = Ka1[H3PO4]
[H2PO4-] = [H+][H3PO4]/Ka1

Since [H3PO4] = 0.100 M, we need to solve for [H+] first.

[H+] = 10^-pH

We'll use the Henderson-Hasselbalch equation to approximate [H+] and simplify the calculations.
pH ≈ pKa1 + log([H2PO4-]/[H3PO4])

Rearranging the equation and solving for [H+], we have:
[H+] ≈ 10^-(pH - pKa1)

Now we can substitute [H+] back into the equation for [H2PO4-]:
[H2PO4-] ≈ [H+][H3PO4]/Ka1
[H2PO4-] ≈ (10^-(pH - pKa1))(0.100)/10^-pKa1
[H2PO4-] ≈ 10^(pKa1 - pH) * 0.100 (Equation 1)

3. [HPO42-]:
The concentration of [HPO42-] can be estimated in a similar manner using Equation 2:
[HPO42-] ≈ 10^(pKa1 - pH) * [H2PO4-] / 10^-pKa2 (Equation 2)

4. [PO43-]:
Similarly, we can calculate [PO43-] using Equation 3:
[PO43-] ≈ 10^(pKa2 - pH) * [HPO42-] / 10^-pKa3 (Equation 3)

5. [OH-]:
The concentration of hydroxide ions can be found using the equation:
[OH-] = 10^-pOH

6. pH:
To calculate the pH, we need to consider the concentrations of H+ and OH- in the solution. Using the equation for water dissociation:
kw = [H+][OH-] = 10^-14
Taking the negative logarithm of both sides:
-log(kw) = -log([H+][OH-]) = 14
pH + pOH = 14
rearranging the equation:
pH = 14 - pOH

Now, we can use Equations 1, 2, 3, and the pH equation to estimate the concentrations and pH value.