What is the pH of a solution made by mixing 10.00 mL of 0.10 M acetic acid with 10.00 mL of 0.10 M KOH? Assume that the volumes of the solutions are additive.K_a =1.8 × 10^-5 for CH₃CO₂H. A)5)28 B)7)00 C)8)72 D)10.02

Question

Quizplus · Accepted Answer

First, we need to determine the amount of each species present before any reaction occurs. Since acetic acid and KOH react in a 1:1 mole ratio, both solutions have the same number of moles of their respective solutes.

moles of acetic acid = (10.00 mL / 1000 mL/L) x 0.10 mol/L = 0.0010 mol 
moles of KOH = (10.00 mL / 1000 mL/L) x 0.10 mol/L = 0.0010 mol

When acetic acid reacts with KOH, they form water and the acetate ion:

CH3COOH + KOH --> CH3COO- + H2O

We can use the balanced chemical equation to determine the number of moles of each species present at equilibrium. Since there is excess KOH in this reaction, all of the acetic acid will be consumed in the reaction. Therefore, the number of moles of acetate ion present at equilibrium will be the same as the initial number of moles of acetic acid.

moles of acetate ion = 0.0010 mol

Now we need to use the definition of pKa to calculate the pH of the solution. The pKa of acetic acid is 4.76, which means that:

pH = pKa + log([acetate ion]/[acetic acid])
pH = 4.76 + log(0.0010/0.0010)
pH = 8.72

Therefore, the pH of the solution is 8.72, which means that the correct answer is C.

What Is the PH of a Solution Made by Mixing