The acid-dissociation constants of phosphoric acid (H3PO4) are Ka1= 7.5×10-3, Ka2 =6.2×10-8, and Ka3 =4.2×10-13 at 25.0 degrees celcius. What is the pH of a 3.0M aqueous solution of phosphoric acid?
The Correct Answer and Explanation is :
To calculate the pH of a 3.0 M aqueous solution of phosphoric acid (H₃PO₄), we must consider the acid dissociation constants (Ka1, Ka2, and Ka3) and recognize that phosphoric acid undergoes stepwise dissociation in water. The dissociation reactions and the given values for Ka are:
- First dissociation:
[
H₃PO₄ (aq) \rightleftharpoons H⁺ (aq) + H₂PO₄⁻ (aq)
]
Ka1 = ( 7.5 \times 10^{-3} ) - Second dissociation:
[
H₂PO₄⁻ (aq) \rightleftharpoons H⁺ (aq) + HPO₄²⁻ (aq)
]
Ka2 = ( 6.2 \times 10^{-8} ) - Third dissociation:
[
HPO₄²⁻ (aq) \rightleftharpoons H⁺ (aq) + PO₄³⁻ (aq)
]
Ka3 = ( 4.2 \times 10^{-13} )
Since the solution is initially 3.0 M in H₃PO₄, the first dissociation is the most significant, as Ka1 is much larger than Ka2 and Ka3. This means the concentration of H⁺ from the first dissociation will dominate the pH.
Step 1: Set up an ICE table for the first dissociation
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| H₃PO₄ | 3.0 | -x | 3.0 – x |
| H⁺ | 0 | +x | x |
| H₂PO₄⁻ | 0 | +x | x |
Using the expression for Ka1:
[
Ka1 = \frac{[H⁺][H₂PO₄⁻]}{[H₃PO₄]} = 7.5 \times 10^{-3}
]
Substituting the equilibrium concentrations:
[
7.5 \times 10^{-3} = \frac{x^2}{3.0 – x}
]
Since Ka1 is relatively large compared to Ka2 and Ka3, we can assume that the change in concentration (x) is small relative to 3.0 M, so (3.0 – x \approx 3.0). This simplifies the equation to:
[
7.5 \times 10^{-3} = \frac{x^2}{3.0}
]
[
x^2 = 7.5 \times 10^{-3} \times 3.0 = 2.25 \times 10^{-2}
]
[
x = \sqrt{2.25 \times 10^{-2}} = 0.15 \, \text{M}
]
Step 2: Calculate pH
The concentration of H⁺ ions is approximately 0.15 M. The pH is calculated as:
[
pH = -\log[H⁺] = -\log(0.15) \approx 0.82
]
Thus, the pH of the 3.0 M solution of phosphoric acid is approximately 0.82.
Explanation:
Phosphoric acid dissociates in steps, with the first dissociation being the most important for determining the pH. Since Ka1 is significantly larger than Ka2 and Ka3, the contribution of H⁺ ions from the first dissociation far outweighs those from the second and third dissociation steps. This results in a relatively low pH for a concentrated phosphoric acid solution, typical of strong acids.
Let me also generate an image to visually represent this concept.
Here is the image showing the dissociation of phosphoric acid (H₃PO₄) in water, with the first dissociation step and the resulting pH calculation. This visual helps illustrate the equilibrium process and the pH level of the 3.0 M solution of phosphoric acid. Let me know if you need any further details!
