Suppose you are titrating vinegar, which is an acetic acid solution of unknown strength, with a sodium hydroxide solution according to the equation H C 2 H 3 O 2 + N a O H ⟶ H 2 O + N a C 2 H 3 O 2 If you require 32.98 mL of 0.1776 M N a O H solution to titrate 10.0 mL of H C 2 H 3 O 2 solution, what is the concentration of acetic acid in the vinegar?

Here one mole of acetic acid is used to neutralize one mole of sodium hydroxide. According to the law of Volumetric analysis, we can write the equation as,

V1M1 = V2M2

V1 and M1 being the volume and molarity of NaOH

V2 and M2 being the volume and molarity of CH₃COOH

M2 = [tex]$\frac{V1M1}{V2}\\[/tex]

= [tex]$\frac{32.98 \times 0.1776}{10}[/tex]

= 0.5857 M

So the concentration of acetic acid in the vinegar is 0.5857 M.

Oseni Oseni · Answer 2 · 2021-09-27T18:56:21+02:00

Going by the reaction, the concentration of acetic acid in the vinegar would be 0.5857 M.

From the balanced equation of the reaction between acetic acid and sodium hydroxide:

H C 2 H 3 O 2 + N a O H ⟶ H 2 O + N a C 2 H 3 O 2

1 mole of acetic acid requires 1 mole of NaOH for complete reaction.

Using the equation:

CaVa/CbVb = Na/Nb where

Ca = concentration of acid,
Cb = concentration of base,
Va = volume of acid,
Vb = volume of base,
Na = number of moles of acid, and
Nb = number of moles of base

In this case, Ca = ?, Va = 10.0 mL, Cb = 0.1776 M, Vb = 32.98 mL, Na = 1, Nb = 1

Substitute into the equation

Ca = 0.1776 x 32.98/10

= 0.5857 M

Hence, the concentration of acetic acid in the vinegar is 0.5857 M.

More on titration can be found here: https://brainly.com/question/6389903