A `5.00 g` sample of vinegar is titrated with `0.108 M NaOH`. If the vinger requires `39.1 mL` of the `NaOH` for complete reaction, the mass percentage of acetic acid `(CH_(3)CO_(2)H)` in the vinegar is

Options

A. 0.0605
B. 0.0407
C. 0.0786
D. 0.0506

Answer

4

Explanation

The reaction is
`CH_(3)CO_(2)H(aq.)+NaOH(aq.) rarr CH_(3)CObar(O)Na^(+)(aq.)+H_(2)O(l)`
Convert the volume of NaOH to moles of NaOH using the molarity formula. Then we convert moles NaOH to moles `CH_(3)CO_(2)H` using the balanced chemical equation. Finally we covert moles `CH_(3)CO_(2)H` to grams `CH_(3)CO_(2)H`.
Mass percentage of acetic acid in the vinegar
`=("Mass of acetic acid")/("Mass of vinegar")xx100`
`n_(NaOH)=V_(NaOH "soln")xxM_(NaOH "soln")`
`=(39.1xx10^(-3)L)(0.108 mol L^(-1))`
`=4.22xx10^(-3) mol`
According to equation, moles of `CH_(3)CO_(2)H` and `NaOH` reacting are equal. Thus
`n_(CH_(3)COOH)=n_(NaOH)=4.22xx10^(-3) mol`
`mass_(CH_(3)CO_(2)H)=n_(CH_(3)CO_(2)H)xx "molar mass"_(CH(3)CO_(2)H)`
`=(4.22xx10^(-3) mol)(60 g mol^(-1))`
`=0.253 g`
Thus mass% `=(0.253 g)/(5.00 g)xx100%=5.06%`