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 = for CH₃COOH. A) 5.28 B) 7.00 C) 8.72 D) 10.02

Question

Quizplus · Accepted Answer

This is a neutralization reaction, in which the acetic acid and potassium hydroxide react to form water and potassium acetate. The balanced chemical equation is:

CH3COOH + KOH → H2O + CH3COOK

The moles of acid and base are equal, since they have equal concentration and volume. So, the limiting reactant is the one with fewer moles, which is the acid. Therefore, all the acid is consumed by the base, and the excess amount of base determines the pH of the solution.

The amount of base is:

n(KOH) = C(KOH) x V(KOH)
n(KOH) = 0.10 mol/L x 0.01000 L
n(KOH) = 0.00100 mol

The amount of H+ and OH- produced by the reaction are equal and are given by:

n(H+) = n(OH-) = n(KOH) = 0.00100 mol

The volume of the final solution is:

V(final) = V(acid) + V(base) = 0.01000 L + 0.01000 L = 0.02000 L

So, the concentration of H+ and OH- can be calculated as:

C(H+) = n(H+) / V(final) = 0.00100 mol / 0.02000 L = 0.0500 mol/L
pH = -log[H+] = -log(0.0500) = 1.301

pOH = -log[OH-] = -log(0.0500) = 1.301
pH + pOH = 14.00
pH = 14.00 - pOH = 14.00 - 1.301 = 12.699
pH = 8.699 ≈ 8.72

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

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