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

Step-by-step 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}}`

=
`(12 g)/(60 g/mol)`

= 0.2 mol

Molarity of acetic acid is calculated as follows.

Density =
`(mass)/(volume)`

1 g/ml =
`(100 g)/(volume)`

volume = 100 ml

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

=
`(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 * 1 M = 10 mL * 2 M` (as 1 L = 1000 ml)

x = 20 L

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