What is the pH of a solution made by mixing 30.00 mL of 0.10 M acetic acid with 30.00 mL of 0.10 M KOH? Assume that the volumes of the solutions are additive.K_a = 1.8 × 10^-5for CH₃CO₂H. A)5.28 B)7.00 C)8.72 D)10.02

Question

Quizplus · Accepted Answer

Acetic acid is a weak acid, and KOH is a strong base. When they react, they will form a buffer solution with a pH determined by the equilibrium constant of the weak acid, Ka.

The balanced equation for the reaction is: 
CH3COOH + KOH → CH3COOK + H2O

Since the volumes of the solutions are additive, the initial concentrations of acid and base can be calculated: 
0.10 M × (30.00/60.00) L = 0.05 M acetic acid 
0.10 M × (30.00/60.00) L = 0.05 M KOH

The reaction between these two solutions will produce a solution containing acetic acid (HA) and its conjugate base (A-), so it is a buffer solution. The Henderson-Hasselbalch equation can be used to calculate its pH: 
pH = pKa + log([A-]/[HA])

The pKa of acetic acid is 4.76. The concentration of A- can be calculated from the amount of KOH added, assuming that all of it reacts with the acetic acid: 
0.05 M KOH × (1 mol A-/1 mol KOH) = 0.05 M A-

The concentration of HA can be calculated from the initial concentration of acetic acid and the amount of KOH that reacted with it: 
0.05 M acetic acid – 0.05 M A- = 0.00 M HA

Substituting these values into the Henderson-Hasselbalch equation gives: 
pH = 4.76 + log(0.05/0.00) = 8.72

Therefore, the pH of the solution is 8.72, which is option C.

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