Question

A 35.0-mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 MNaOHsolution. Calculate the pH after the following volumes of base have been added.

35.5 mL

Express your answer using two decimal places.

Answer 1

To solve this problem we must determine the number of moles of NaOH and present after the neutralization
reaction.

The product of the volume and concentration of each solution gives the number of moles of each reactant present before the neutralization:

(0,150 M CH₃COOH )*(0,035 L) = 0,00525 moles CH₃COOH

(0,150 M NaOH)*(0,0355 L) = 0,00532 moles NaOH

This amount is greater then the moles of acid that is present. The 0,00532 moles OH^- neutralizes the 0,00525 moles CH₃COOH. To find how much OH^- will be in excess we subtract the amount of acid and hydroxide:

moles of hydroxide in excess: 0,00532 moles - 0,00525 moles= 7,00 x 10^-5 moles OH^-

To determine the OH- concentration we need to find the total volume in the solution.

The total volumein the solution is the 35,0 ml of the acid and the 35,5 ml of the base, so the total volume will be 70,5 ml or 0,0705 L.

The concentration of OH- will be:

[OH-] = 7,50 x 10^-5 moles OH^- / 0,0705 L = 1,06 x 10^-3 M

The pOH value will be pOH=-log(1,06 x 10^-3) = 2,97, the pH value is 11,03.