Question

Acetic acid (HC2H3O2) is an important ingredient of vinegar. A sample of 50.0 mL of a commercial vinegar is titrated against a 1.00 M NaOH solution. What is the concentration (in M) of acetic acid present in the vinegar if 5.75 mL of the base is needed for the titration?

Answer 1

Step-by-step explanation:

The given data is as follows.

`M_(1)` = 1.00 M,
`V_(1)` = 50.0 ml

`M_(2)` = ?,
`V_(2)` = 5.75 ml

Therefore, calculate the concentration (in M) of acetic acid as follows.

`M_(1)V_(1) = M_(2)V_(2)`

`1.00 M * 50.0 ml = M_(2) * 5.75 ml`

`M_(2)` =
`(1.00 M *s 50.0 ml)/(5.75 ml)`

= 8.69 M

Thus, we can conclude that the concentration (in M) of acetic acid present in the given vinegar is 8.69 M.

Answer 2

The answer to your question is: C₂ = 0.115 M

Step-by-step explanation:

Data

C₂H₄O₂ = 50 ml concentration = ?

NaOH = 1 M; 5.75 ml

Formula

C₁V₁ = C₂V₂

C₂ = C₁V₁/V₂

C₂ = (1)(5.75) / 50

C₂ = 0.115 M