Question

Assuming that an acetic acid solution is 12% by mass and that the density of the solution is 1.00 g/mL, what volume of 1 M NaOH is needed to fully neutralize a 10 mL aliquot of the acetic acid solution

Answer 1

Explanation:

Let us assume that total mass of the solution is 100 g. And, as it is given that acetic acid solution is 12% by mass which means that mass of acetic acid is 12 g and 88 g is the water.

Now, calculate the number of moles of acetic acid as its molar mass is 60 g/mol.

    No. of moles = $\frac{mass}{\text{molar mass}}$

                           = $\frac{12 g}{60 g/mol}$

                           = 0.2 mol

Molarity of acetic acid is calculated as follows.

              Density = $\frac{mass}{volume}$

                 1 g/ml = $\frac{100 g}{volume}$

                    volume = 100 ml

Hence, molarity = $\frac{\text{no. of moles}}{volume}$

                           = $\frac{0.2 mol}{0.1 L}$

                           = 2 mol/l

As reaction equation for the given reaction is as follows.

     $NaOH + CH_{3}COOH \rightarrow CH_{3}COONa + H_{2}O$

So,          moles of NaOH = moles of acetic acid

Let us suppose that moles of NaOH are "x".

          $x \times 1 M = 10 mL \times 2 M$     (as 1 L = 1000 ml)

                        x = 20 L

Thus, we can conclude that volume of NaOH required is 20 ml.