The acid-dissociation constants of phosphoric acid (H3PO4) at 25 degrees Celsius are Ka1 = 7.510-3, Ka2 =6.2 10-8, and Ka3 =4.210-13. What is the PH of a 2.5M aqueous solution of phosphoric acid?
The Correct Answer and Explanation is :
To calculate the pH of a 2.5 M aqueous solution of phosphoric acid (H₃PO₄), we need to consider its stepwise dissociation. Phosphoric acid is a triprotic acid, meaning it can donate three protons (H⁺) in a stepwise manner. The dissociation constants are given as:
- Ka₁ = 7.5 × 10⁻³ (for the dissociation of H₃PO₄ to H₂PO₄⁻)
- Ka₂ = 6.2 × 10⁻⁸ (for the dissociation of H₂PO₄⁻ to HPO₄²⁻)
- Ka₃ = 4.2 × 10⁻¹³ (for the dissociation of HPO₄²⁻ to PO₄³⁻)
At high concentrations, the first dissociation is the most significant. So, we will first focus on the first dissociation:
H₃PO₄ ⇌ H⁺ + H₂PO₄⁻
Step 1: Solve for the concentration of H⁺ from the first dissociation
We use the expression for Ka₁:
[
Ka₁ = \frac{[H⁺][H₂PO₄⁻]}{[H₃PO₄]}
]
Let x be the concentration of H⁺ produced. Initially, [H₃PO₄] is 2.5 M, and we assume that most of the dissociation comes from the first step. Therefore:
[
Ka₁ = \frac{x^2}{2.5 – x}
]
Given Ka₁ = 7.5 × 10⁻³, we substitute:
[
7.5 \times 10^{-3} = \frac{x^2}{2.5 – x}
]
Assuming x is small relative to 2.5 M, we can approximate 2.5 – x ≈ 2.5. This simplifies to:
[
7.5 \times 10^{-3} = \frac{x^2}{2.5}
]
Solving for x:
[
x^2 = (7.5 \times 10^{-3}) \times 2.5
]
[
x^2 = 1.875 \times 10^{-2}
]
[
x = 1.37 \times 10^{-1} \, \text{M}
]
Thus, the concentration of H⁺ from the first dissociation is 0.137 M.
Step 2: Calculate the pH
The pH is defined as:
[
\text{pH} = -\log[H⁺]
]
Substituting [H⁺] = 0.137 M:
[
\text{pH} = -\log(0.137) \approx 0.86
]
Thus, the pH of a 2.5 M aqueous solution of phosphoric acid is approximately 0.86.
The second and third dissociations are negligible at this concentration, so they do not significantly affect the pH. The major contributor is the first dissociation.
Now, I will generate an image of this calculation.
Here is the image showing the detailed mathematical calculation for the pH of a 2.5M aqueous solution of phosphoric acid. It includes the steps for solving the first dissociation and calculating the pH. Let me know if you need further clarification!
