What is the pH of a buffer that results when 0.50 mole of H₃PO₄ is mixed with 0.25 mole of NaOH and diluted with water to 1.00 L? (The acid dissociation constants of phosphoric acid are K_a1 = 7.5 $\times$ 10^-3,K_a2 = 6.2 $\times$ 10^-8,and K_a3 = 3.6 $\times$ 10^-13) A) 1.82 B) 2.12 C) 6.91 D) 7.21 E) 12.44

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First, we need to determine the relevant equation for the buffer system, which is: H₃PO₄ + OH- → ₃PO₄- + H2O The initial moles of H₃PO₄ and NaOH are 0.50 and 0.25, respectively. When they react, they will form the conjugate pair H₃PO₄- and H2O. Since the number of moles of the reactants are not equal, one reactant will be completely consumed and the other will be partially consumed. In this case, NaOH will be completely consumed since it is in excess. The amount of NaOH used can be determined from its concentration: 0.25 mole / 1 L = 0.25 M Since the reaction consumes the same number of moles of NaOH and H₃PO₄, the concentration of H₃PO₄- can also be determined: 0.25 M / (0.50 mole / 1 L) = 0.50 M Now, we can use the Henderson-Hasselbalch equation to find the pH of the buffer: pH = pKa + log([A-]/[HA]) Since we have three dissociation constants for phosphoric acid, we need to determine which one to use. The pH of the buffer will be determined by the concentration of the conjugate pair that is present in the highest concentration. Using the concentration of H₃PO₄- that we calculated earlier, we can determine the relevant dissociation constant: pH = pKa2 + log([A-]/[HA]) = 6.2 + log(0.50/0.50) = 6.2 Therefore, the answer is B) 2.12.

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