Question

Assuming the density of 5 percent acetic acid by mass solution 1.0 g/mL, determine the volume of the acetic acid solution necessary to neutralize 25.0 mL of 0.10 M NaOH.

Answer 1

First, we can find the amount of NaOH in moles to be neutralized:

$n(NaOH)=c(NaOH) \cdot V(NaOH)=0.10\frac{mmol}{mL} \cdot 25.0 mL=2.5mmol$

According to the equation of neutralization

$AcOH + NaOH \rightarrow AcONa + H_2O$

the amount of AcOH in moles is the same as of NaOH:

$n(AcOH)=n(NaOH)=2.5mmol$

Mass of pure AcOH can be found:

$m(AcOH)=n(AcOH) \cdot M(AcOH) = 2.5mmol \cdot 60\frac{mg}{mmol}=150mg=0.15g$

Knowing the weight percentage of AcOH in solution, the mass of aqueous solution of AcOH can be found:

$m(AcOH_{(aq.)})=\frac{m(AcOH)}{\omega(AcOH)}=\frac{0.15g}{0.05}=3.00g$

Assuming the density of AcOH solution is 1.0 g/mL, the volume of aqueous acetic acid is 3.00 mL.