Question

What mass of sodium hydroxide would be equivalent to a `20*mL` volume of acetic acid, for which `rho_"HOAc"=1.049*g*mL^-1`?

Answer 1

We need a stoichiometric equation.........

Explanation:

`"NaOH(aq)" + "H"_3"CCO"_2"H(l)" rarr "H"_3"CCO"_2^(-)"Na"^(+) + "H"_2"O(l)"`.

This is a `"1:1 reaction"`, and we have to use the density of acetic acid to find the molar equivalence.

This site reports that `rho_"HOAc"=1.049*g*mL^-1`.

And thus `"Moles of HOAc"=(20*mLxx1.049*g*mL^-1)/(60.05*g*mol^-1)`

`=0.349*mol`

And thus an equivalent quantity of `NaOH-=0.349*molxx40.0*g*mol^-1=13.98*g`; of course we would put this mass into aqueous solution first.