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 5.65 mL of the base is needed for the titration?

Answer 1

The concentration of acetic acid present in the vinegar is 0.113M

Step-by-step explanation:

1. As the titration occurs between an acid and a base is called neutralization. The balanced chemical reaction between the acetic acid and the NaOH is:

`NaOH+HC_(2)H_(3)O_(2)=NaC_(2)H_(3)O_(2)+H_(2)O`

2. Calculate the moles of NaOH used, taking in account the concentration of NaOH 1.00M:

`5.65mL*(1.00molesNaOH)/(1000mLNaOH)=0.00565molesNaOH`

3. Calculate the number of moles of acetic acid neutralized using stoichiometry:

`0.00565molesNaOH*(1molHC_(2)H_(3)O_(2))/(1molNaOH)=0.00565molesHC_(2)H_(3)O_(2)`

4. Calculate the concentration of acetic acid present in the vinegar:

`(0.00565molesHC_(2)H_(3)O_(2))/(50.0mLvinegar)*(1000mLvinegar)/(1Lvinegar)=0.113M`