Calculate the pH and concentration of all
species present at equilibrium in 0.1 M `H_(3)PO_(4)` solution.
`K_(a_(1))=7.5xx10^(-3), K_(a_(2))=6.2xx10^(-8), K_(a_(3))=4.2xx10^(-13)`

Answer

To calculate the pH and concentration of all species present at equilibrium in a 0.1 M `H₃PO₄` solution, we need to consider the dissociation of phosphoric acid, which is a triprotic acid. The dissociation can be represented in three steps:

1. **First dissociation**:
\[ H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^- \]
The equilibrium constant for this reaction is \( K_{a1} = 7.5 \times 10^{-3} \).

2. **Second dissociation**:
\[ H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-} \]
The equilibrium constant for this reaction is \( K_{a2} = 6.2 \times 10^{-8} \).

3. **Third dissociation**:
\[ HPO_4^{2-} \rightleftharpoons H^+ + PO_4^{3-} \]
The equilibrium constant for this reaction is \( K_{a3} = 4.2 \times 10^{-13} \).

### Step 1: Calculate the concentration of \( H^+ \) from the first dissociation

Assuming that the first dissociation is the most significant, we can set up an equilibrium expression for the first dissociation:

Let \( x \) be the concentration of \( H^+ \) produced from the first dissociation.

At equilibrium:
- \([H_3PO_4] = 0.1 - x\)
- \([H^+] = x\)
- \([H_2PO_4^-] = x\)

The equilibrium expression for the first dissociation is:
\[ K_{a1} = \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]} = \frac{x^2}{0.1 - x} \]

Assuming \( x \) is small compared to 0.1 M, we can approximate:
\[ K_{a1} \approx \frac{x^2}{0.1} \]
Substituting the value of \( K_{a1} \):
\[ 7.5 \times 10^{-3} = \frac{x^2}{0.1} \]
\[ x^2 = 7.5 \times 10^{-4} \]
\[ x = \sqrt{7.5 \times 10^{-4}} \approx 0.0274 \, \text{M} \]

### Step 2: Calculate the pH

The concentration of \( H^+ \) is approximately \( 0.0274 \, \text{M} \). Now we can calculate the pH:
\[ \text{pH} = -\log[H^+] = -\log(0.0274) \approx 1.56 \]

### Step 3: Calculate concentrations of species from the first dissociation

From the first dissociation:
- \([H_3PO_4] = 0.1 - x \approx 0.1 - 0.0274 \approx 0.0726 \, \text{M}\)
- \([H_2PO_4^-] = x \approx 0.0274 \, \text{M}\)

### Step 4: Consider the second dissociation

Now we consider the second dissociation. Let \( y \) be the concentration of \( H^+ \) produced from the second dissociation:
\[ H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-} \]
At equilibrium:
- \([H_2PO_4^-] = 0.0274 - y\)
- \([H^+] = 0.0274 + y\)
- \([HPO_4^{2-}] = y\)

The equilibrium expression for the second dissociation is:
\[ K_{a2} = \frac{[H^+][HPO_4^{2-}]}{[H_2PO_4^-]} = \frac{(0.0274 + y)y}{0.0274 - y} \]

Assuming \( y \) is small compared to 0.0274, we can approximate:
\[ K_{a2} \approx \frac{(0.0274)y}{0.0274} = y \]
Substituting the value of \( K_{a2} \):
\[ 6.2 \times 10^{-8} = y \]

### Step 5: Calculate concentrations of species from the second dissociation

From the second dissociation:
- \([HPO_4^{2-}] = y \approx 6.2 \times 10^{-8} \, \text{M}\)
- \([H^+] = 0.0274 + 6.2 \times 10^{-8} \approx 0.0274 \, \text{M}\) (since \( 6.2 \times 10^{-8} \) is negligible)
- \([H_2PO_4^-] = 0.0274 - 6.2 \times 10^{-8} \approx 0.0274 \, \text{M}\)

### Step 6: Consider the third dissociation

For the third dissociation, the concentration of \( H^+ \) produced is negligible due to the very small \( K_{a3} \). Thus, we can assume:
- \([HPO_4^{2-}] \approx 6.2 \times 10^{-8} \, \text{M}\)
- \([PO_4^{3-}] \approx 0 \, \text{M}\)

### Final Concentrations

- \([H_3PO_4] \approx 0.0726 \, \text{M}\)
- \([H_2PO_4^-] \approx 0.0274 \, \text{M}\)
- \([HPO_4^{2-}] \approx 6.2 \times 10^{-8} \, \text{M}\)
- \([PO_4^{3-}] \approx 0 \, \text{M}\)

### Final pH

The final pH of the solution is approximately \( 1.56 \).

### Summary of Results
- **pH**: 1.56
- **Concentrations**:
- \([H_3PO_4] \approx 0.0726 \, \text{M}\)
- \([H_2PO_4^-] \approx 0.0274 \, \text{M}\)
- \([HPO_4^{2-}] \approx 6.2 \times 10^{-8} \, \text{M}\)
- \([PO_4^{3-}] \approx 0 \, \text{M}\)