Question

zA 25.0 mL sample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 12.5 mL of the base is added. The concentration of acetic acid in the sample was ________ M. 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 12.5 mL of the base is added. The concentration of acetic acid in the sample was ________ M. 0.365 0.119 0.0875 0.175 none of the above

Answer 1

0.0875M of Acetic Acid

Step-by-step explanation:

Based on the reaction of acetic acid (HX) with sodium hydroxide (NaOH):

HX + NaOH → NaX + H₂O

1 mole of acetic acid reacts with 1 mole of NaOH

That means the moles of sodium hydroxide added = Moles the acetic acid at the equivalence point.

To solve this question we have to find the moles of NaOH = Moles HX. With the moles and the volume (25.0mL = 0.0250L) we can find the molarity of the acetic acid:

Moles NaOH = Moles Acetic acid:

12.5mL = 0.0125L * (0.175mol / L) = 0.00219 moles Acetic Acid

Molarity acetic acid:

0.00219 moles Acetic Acid / 0.0250L =

0.0875M of Acetic Acid