If 25 mL of 0.20 M NaOH(aq)is added to 50 mL of 0.10 M aqueous HC₂H₃O₂(K_a = 1.8 × 10^-5),what is the pH? A)5.33 B)10.21 C)8.78 D)13.56 E)1.34

Question

Quizplus · Accepted Answer

First, let's determine the moles of HC₂H₃O₂ in the solution: moles = M x V = 0.10 mol/L x 0.050 L = 0.005 mol Next, let's determine the moles of NaOH added: moles = M x V = 0.20 mol/L x 0.025 L = 0.005 mol Since NaOH is a strong base and HC₂H₃O₂ is a weak acid, we can assume that all of the NaOH will react with the HC₂H₃O₂ to form its conjugate base, C₂H₃O-. The reaction is: NaOH + HC₂H₃O₂ --> C₂H₃O- + H2O Thus, the moles of HC₂H₃O₂ remaining in solution is: 0.005 mol (initial moles) - 0.005 mol (reacted with NaOH) = 0 mol This means that all of the HC₂H₃O₂ has been converted to its conjugate base, which makes the solution a buffer solution. The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([base]/[acid]) where pKa is the dissociation constant of the weak acid, [base] is the concentration of the conjugate base, and [acid] is the concentration of the weak acid. From the balanced equation, we know that the moles of HC₂H₃O₂ that reacted with NaOH is equal to the moles of C₂H₃O- formed, so the concentration of the conjugate base is: [base] = moles of C₂H₃O- / total volume of solution [base] = 0.005 mol / 0.075 L = 0.067 M The concentration of the weak acid is equal to the initial concentration, since all of it has been converted to its conjugate base: [acid] = 0.10 M The pKa of HC₂H₃O₂ is given as 1.8 × 10^-5 in the problem statement. plugging these values into the Henderson-Hasselbalch equation, we get: pH = 1.8 × 10^-5 + log(0.067/0.10) pH = 8.78 Therefore, the answer is choice C.

If 25 ML of 0