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 = 7.1 × 10^-3,K_a2 = 6.2 × 10^-8,and K_a3 = 4.5 × 10^-13)

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To solve this problem, we need to use the Henderson-Hasselbalch equation which is given as: pH = pKa + log([A-]/[HA]) where: pKa = -log(Ka) = acid dissociation constant [A-] = concentration of conjugate base [HA] = concentration of weak acid First, we need to identify the weak acid and its conjugate base in the problem. From the given information, we can see that H₃PO₄ is a polyprotic acid with three dissociation constants. The acid dissociation constants given are in decreasing order, which suggests that H₃PO₄ is a stronger acid than its conjugate base _-1. Therefore, we can consider H₃PO₄ as the weak acid and _-1 as its conjugate base. Next, we need to determine the concentrations of the weak acid and its conjugate base after they react with NaOH. Since NaOH is a strong base, it will react completely with the weak acid to form its conjugate base: H₃PO₄ + OH- → _-1 + H2O We can use stoichiometry to determine the moles of _-1 formed: 0.75 mol NaOH × 1 mol _-1/mol NaOH = 0.75 mol _-1 Now, we need to determine the relative concentrations of the weak acid and its conjugate base in the buffer solution. To do this, we can use the mole balance equation: moles of weak acid = initial moles of weak acid - moles of strong base added moles of conjugate base = initial moles of conjugate base + moles of strong base added Substituting the values obtained above, we get: moles of H₃PO₄ = 0.50 mol - 0 mol = 0.50 mol moles of _-1 = 0 mol + 0.75 mol = 0.75 mol Now, we can calculate the concentrations of the weak acid and its conjugate base: [HA] = moles of weak acid/volume of solution = 0.50 M [A-] = moles of conjugate base/volume of solution = 0.75 M Finally, we can substitute these values into the Henderson-Hasselbalch equation to obtain the pH of the buffer: pH = pKa + log([A-]/[HA]) pH = pKa + log(0.75/0.50) pH = -log(4.5 × 10^-13) + log(1.50) pH = 7.21 Therefore, the pH of the buffer is 7.21. The correct answer is D.

What Is the PH of a Buffer That Results When