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. Ka = 1.8 × 10^-5 for CH3CO2H.

• A. 8.74
• B. 4.74
• C. 3.00
• D. 7.00

Answer: (D)

Verified step by step guidance

1. Step 1: Begin by identifying the chemical reaction that occurs when acetic acid (CH3CO2H) is mixed with potassium hydroxide (KOH). The reaction is a neutralization reaction: CH3CO2H + KOH → CH3CO2K + H2O.
2. Step 2: Calculate the initial moles of acetic acid and KOH. Use the formula: moles = concentration × volume. For acetic acid: moles = 0.10 M × 0.030 L = 0.003 moles. For KOH: moles = 0.10 M × 0.030 L = 0.003 moles.
3. Step 3: Determine the limiting reactant. Since both acetic acid and KOH have the same number of moles (0.003 moles), they will completely react with each other, resulting in the formation of 0.003 moles of CH3CO2K.
4. Step 4: After the reaction, calculate the concentration of the acetate ion (CH3CO2-) in the solution. The total volume of the solution is 60.00 mL or 0.060 L. The concentration of CH3CO2- is: concentration = moles / volume = 0.003 moles / 0.060 L = 0.05 M.
5. Step 5: Use the Henderson-Hasselbalch equation to find the pH of the solution. The equation is: pH = pKa + log([A-]/[HA]). Since all acetic acid is converted to acetate, [HA] = 0 and [A-] = 0.05 M. The pKa is calculated from Ka: pKa = -log(1.8 × 10^-5). Substitute these values into the equation to find the pH.