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:
[tex]NaOH+HC_{2}H_{3}O_{2}=NaC_{2}H_{3}O_{2}+H_{2}O[/tex]
2. Calculate the moles of NaOH used, taking in account the concentration of NaOH 1.00M:
[tex]5.65mL*\frac{1.00molesNaOH}{1000mLNaOH}=0.00565molesNaOH[/tex]
3. Calculate the number of moles of acetic acid neutralized using stoichiometry:
[tex]0.00565molesNaOH*\frac{1molHC_{2}H_{3}O_{2}}{1molNaOH}=0.00565molesHC_{2}H_{3}O_{2}[/tex]
4. Calculate the concentration of acetic acid present in the vinegar:
[tex]\frac{0.00565molesHC_{2}H_{3}O_{2}}{50.0mLvinegar}*\frac{1000mLvinegar}{1Lvinegar}=0.113M[/tex]
