A 25.0 mL sample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 26.6 mL of the base is added. The concentration of acetic acid in the sample was __________ M.
A)0.365
B)0.175
C)0.119
D)0.186
E)none of the above

Question

Quizplus · Accepted Answer

The moles of NaOH used at the equivalence point are equal to the moles of acetic acid initially present. Moles of NaOH = 0.175 M * 0.0266 L = 0.004655 mol. Since the reaction is 1:1, moles of acetic acid = 0.004655 mol. Concentration of acetic acid = 0.004655 mol / 0.025 L = 0.186 M.

A 250 ML Sample of an Acetic Acid Solution Is Titrated