What is the pH of a buffer that results when 0.50 mole of H₃PO₄ is mixed with 0.75 mole of NaOH and diluted with water to 1.00 L? (The acid dissociation constants of phosphoric acid are K_a1 = 7.5 × 10^-3,K_a2 = 6.2 × 10^-8,and K_a3 = 3.6 × 10^-13) A) 1.82 B) 2.12 C) 6.91 D) 7.21 E) 12.44

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To solve this problem, we need to use the Henderson-Hasselbalch equation: pH = pKa + log([base]/[acid]) First, we need to determine which species in the solution are the acid and the base. From the given information, we know that we have "H₃PO₄" and NaOH. The NaOH is a strong base and will completely dissociate into Na+ and OH-. The H₃PO₄ is the conjugate acid of H2PO4-, which can function as either an acid or a base depending on the solution conditions. In this case, since we are adding NaOH (a strong base), the H₃PO₄ will act as an acid to neutralize the base. Next, we need to determine the initial concentrations of the acid and base before they react. We know that we have 0.50 moles of H₃PO₄ and 0.75 moles of NaOH, so we can use the volume (1.00 L) to calculate their concentrations: [H₃PO₄] = 0.50 M [OH-] = 0.75 M [H2PO4-] = [H₃PO₄] (because 1 mole of NaOH reacts with 1 mole of H2PO4- to form H₃PO₄) Now we can use the Henderson-Hasselbalch equation to calculate the pH: pH = pKa + log([base]/[acid]) pH = pKa + log([OH-]/[H2PO4-]) pH = pKa + log(0.75/0.50) pH = pKa + 0.176 pH = 7.21 We can use the pKa values given for the three dissociation steps of H3PO4 to determine which one corresponds to the equilibrium equation we used: K_a1 = 7.5 × 10^-3 (corresponds to H3PO4 ⇌ H2PO4- + H+) K_a2 = 6.2 × 10^-8 (corresponds to H2PO4- ⇌ HPO42- + H+) K_a3 = 3.6 × 10^-13 (corresponds to HPO42- ⇌ PO43- + H+) Since we used the equilibrium equation for the first dissociation step (H3PO4 ⇌ H2PO4- + H+), the corresponding pKa value is 7.5 × 10^-3. We can see that this pKa value is closest to a pH of 7.21, which confirms that we used the correct pKa. Therefore, the buffer resulting from this mixture has a pH of 7.21, so the answer is (D).

What Is the PH of a Buffer That Results When