To find the pH and the concentrations of all species in a 0.250 M phosphoric acid ([tex]$H_3PO_4$[/tex]) solution, let's proceed step-by-step.
### Step 1: Ionization Constants
Given:
- [tex]\( pK_{a1} = 2.16 \)[/tex]
- [tex]\( pK_{a2} = 7.21 \)[/tex]
- [tex]\( pK_{a3} = 12.32 \)[/tex]
The dissociation reactions are:
1. [tex]\( H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^- \)[/tex] with [tex]\( K_{a1} = 10^{-2.16} \)[/tex]
2. [tex]\( H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-} \)[/tex] with [tex]\( K_{a2} = 10^{-7.21} \)[/tex]
3. [tex]\( HPO_4^{2-} \rightleftharpoons H^+ + PO_4^{3-} \)[/tex] with [tex]\( K_{a3} = 10^{-12.32} \)[/tex]
### Step 2: Approximations for Triprotic Acids
Considering [tex]\( K_{a1} \)[/tex] is the largest, we can approximate that the majority of the [tex]\( H^+ \)[/tex] ions come from the first dissociation step.
[tex]\[ H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^- \][/tex]
### Step 3: First Dissociation Equilibrium
Let [tex]\( x \)[/tex] be the concentration of [tex]\( H^+ \)[/tex] (and [tex]\( H_2PO_4^- \)[/tex]) at equilibrium, we start with:
Initial concentrations:
[tex]\[ [H_3PO_4] = 0.250 \, \text{M}, \quad [H^+] = 0, \quad [H_2PO_4^-] = 0 \][/tex]
At equilibrium:
[tex]\[ [H_3PO_4] = 0.250 - x \][/tex]
[tex]\[ [H^+] = x \][/tex]
[tex]\[ [H_2PO_4^-] = x \][/tex]
Using the expression for [tex]\( K_{a1} \)[/tex]:
[tex]\[ K_{a1} = \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]} = \frac{x \cdot x}{0.250 - x} \approx \frac{x^2}{0.250} \][/tex]
Given that [tex]\( K_{a1} = 10^{-2.16} \)[/tex]:
[tex]\[ 10^{-2.16} = \frac{x^2}{0.250} \][/tex]
[tex]\[ x^2 = 0.250 \times 10^{-2.16} \][/tex]
[tex]\[ x^2 \approx 0.250 \times 0.00692 \][/tex]
[tex]\[ x^2 \approx 0.00173 \][/tex]
[tex]\[ x \approx \sqrt{0.00173} \][/tex]
[tex]\[ x \approx 0.0416 \, \text{M} \][/tex]
Thus, at equilibrium:
[tex]\[ [H^+] = 0.0416 \, \text{M} \][/tex]
### Step 4: Check Concentrations of All Species
- [tex]\( [H_3PO_4] = 0.250 - 0.0416 = 0.2084 \, \text{M} \)[/tex]
- [tex]\( [H_2PO_4^-] = 0.0416 \, \text{M} \)[/tex]
The further dissociations are minimal due to the much smaller [tex]\( K_{a2} \)[/tex] and [tex]\( K_{a3} \)[/tex] values.
- [tex]\( [HPO_4^{2-}] \approx 0 \)[/tex]
- [tex]\( [PO_4^{3-}] \approx 0 \)[/tex]
### Step 5: Calculate pH
[tex]\[ pH = -\log[H^+] \][/tex]
[tex]\[ pH = -\log(0.0416) \][/tex]
[tex]\[ pH \approx 1.38 \][/tex]
### Summary of Concentrations and pH
[tex]\[ [H_3PO_4] = 0.2084 \, \text{M} \][/tex]
[tex]\[ [H_2PO_4^-] = 0.0416 \, \text{M} \][/tex]
[tex]\[ [HPO_4^{2-}] \approx 0 \, \text{M} \][/tex]
[tex]\[ [PO_4^{3-}] \approx 0 \, \text{M} \][/tex]
[tex]\[ [H^+] = 0.0416 \, \text{M} \][/tex]
[tex]\[ pH = 1.38 \][/tex]
This completes the detailed, step-by-step solution for estimating the pH and the concentrations of all species in a 0.250 M phosphoric acid solution.
