What is the pH of a 6.00 M H3PO4 solution, given that Ka1 = 7.5 × 10⁻³, Ka2 = 6.2 × 10⁻⁸, and Ka3 = 4.2 × 10⁻¹³?

• A. 3.50
• B. 0.78
• C. 2.15
• D. 1.25

Answer: (B)

Verified step by step guidance

1. Identify that H3PO4 is a triprotic acid, meaning it can donate three protons (H+ ions) in solution. The dissociation occurs in three steps, each with its own equilibrium constant (Ka1, Ka2, Ka3).
2. Since the concentration of H3PO4 is relatively high (6.00 M), the first dissociation step will contribute the most to the [H+] concentration. Therefore, we will focus on the first dissociation: H3PO4 ⇌ H+ + H2PO4⁻ with Ka1 = 7.5 × 10⁻³.
3. Set up the expression for the equilibrium constant Ka1: \( K_{a1} = \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]} \). Assume that the initial concentration of H3PO4 is 6.00 M and that the change in concentration due to dissociation is x, where x is the concentration of H+ produced.
4. Substitute the equilibrium concentrations into the Ka1 expression: \( 7.5 \times 10^{-3} = \frac{x^2}{6.00 - x} \). Since x is small compared to 6.00, you can approximate the denominator as 6.00, simplifying the equation to \( 7.5 \times 10^{-3} = \frac{x^2}{6.00} \).
5. Solve for x, which represents the [H+] concentration. Once x is found, calculate the pH using the formula \( pH = -\log[H^+] \).