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

Answer:

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

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:

$5.65mL*\frac{1.00molesNaOH}{1000mLNaOH}=0.00565molesNaOH$

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

$0.00565molesNaOH*\frac{1molHC_{2}H_{3}O_{2}}{1molNaOH}=0.00565molesHC_{2}H_{3}O_{2}$

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

$\frac{0.00565molesHC_{2}H_{3}O_{2}}{50.0mLvinegar}*\frac{1000mLvinegar}{1Lvinegar}=0.113M$