What is the pH of a solution made by mixing 10.00 mL of 0.10 mol L^-1 acetic acid with 10.00 mL of 0.10 mol L^-1 KOH? Assume that the volumes of the solutions are additive. K_a = 1.8 × 10^-5for CH₃COOH. A)5.28 B)7.00 C)8.72 D)10.02

Question

Quizplus · Accepted Answer

First, we need to find the moles of acetic acid and KOH:
moles of CH3COOH = 0.1 mol/L x 0.01 L = 0.001 mol
moles of KOH = 0.1 mol/L x 0.01 L = 0.001 mol

Since KOH is a strong base, it will completely dissociate into its ions:
KOH → K+ + OH-

The OH- ions will react with the CH3COOH to form water and the acetate ion (CH3COO-):
CH3COOH + OH- → CH3COO- + H2O

The initial moles of CH3COOH (0.001 mol) will be reduced by the same amount as the moles of OH- added (0.001 mol). Therefore, the remaining moles of CH3COOH will be:
0.001 mol - 0.001 mol = 0 mol

The moles of CH3COO- formed will be the same as the moles of OH- added (0.001 mol). Therefore, the concentration (in mol/L) of the acetate ion will be:
0.001 mol / 0.02 L = 0.05 mol/L

Now we can use the Henderson-Hasselbalch equation to find the pH:
pH = pKa + log([CH3COO-]/[CH3COOH])

The pKa for acetic acid is 4.76. Substituting in the values, we get:
pH = 4.76 + log(0.05/0.1) = 8.72

Therefore, the pH of the solution is 8.72, which corresponds to answer choice C.

What Is the PH of a Solution Made by Mixing