Question

A `5.00*mL` volume of acetic acid required a `45*mL` volume of `0.100*mol*L^-1` `NaOH` for equivalence. What is the concentration of the acetic acid solution?

Answer 1

Approx. `1*mol*L^-1`, with respect to acetic acid.

Explanation:

`"Concentration"` `=` `"Moles of solute"/"Volume of solution"`, or `C=n/V`, and clearly `"concentration"` has units of `mol*L^-1`.

We need a reaction:

`H_3C-CO_2H(aq) + NaOH(aq) rarr H_3C-CO_2^(-)""^(+)Na(aq) + H_2O(aq)`

And thus, since
`"moles of sodium hydroxide "-=" moles of acetic acid"`,

there were `45.0*mLxx10^-3L*mL^-1xx0.100*mol*L^-1=` `4.50xx10^-3*mol` of acetic acid in that original `5*mL` volume.

Again, `"Concentration"` `=` `"Moles of solute"/"Volume of solution"`

`=` `(4.50xx10^-3*mol)/(5xx10^-3L)`