A buffer solution is made such that the initial concentrations of lactic acid (HC₃H₅O₃)and the lactate ion (C₃H₅O₃^-)are 0.600 M and 0.620 M,respectively.What is the resulting pH if 100.0 mL of 0.100 M potassium hydroxide is added to 0.500 L of the buffer solution? (The K_a of HC₃H₅O₃ is 1.4 $\times$10^-4.)

Question

Quizplus · Accepted Answer

Step 1: Calculate the initial pH of the buffer solution using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])
where pKa = 3.86 (from the given Ka value), [A-] = concentration of lactate ion = 0.620 M, [HA] = concentration of lactic acid = 0.600 M

pH = 3.86 + log(0.620/0.600) = 3.88

Thus, the initial pH of the buffer solution is 3.88.

Step 2: Determine the moles of KOH added and the resulting concentrations of lactate and lactic acid:

Moles of KOH = Molarity x Volume = 0.100 M x 0.100 L = 0.010 mol
Volume of buffer solution = 0.500 L + 0.100 L = 0.600 L (after addition of KOH)
Moles of lactate ion after addition of KOH = initial moles + moles from KOH = (0.620 M x 0.600 L) + 0.010 mol = 0.374 mol
Concentration of lactate ion = moles/volume = 0.374 mol/0.600 L = 0.623 M
Moles of lactic acid after addition of KOH = initial moles - moles reacting with KOH = (0.600 M x 0.600 L) - 0.010 mol = 0.350 mol
Concentration of lactic acid = moles/volume = 0.350 mol/0.600 L = 0.583 M

Step 3: Calculate the new pH of the solution using the Henderson-Hasselbalch equation with the new [A-] and [HA] values:

pH = pKa + log([A-]/[HA])
where pKa = 3.86, [A-] = 0.623 M, [HA] = 0.583 M

pH = 3.86 + log(0.623/0.583) = 3.90

Thus, the resulting pH of the solution after addition of KOH is 3.90, and the best choice is B.

A Buffer Solution Is Made Such That the Initial Concentrations ×\times×

A Buffer Solution Is Made Such That the Initial Concentrations $\times$