# 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?

Jun 12, 2017

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

#### Explanation:

$\text{NaOH(aq)" + "H"_3"CCO"_2"H(l)" rarr "H"_3"CCO"_2^(-)"Na"^(+) + "H"_2"O(l)}$.

This is a $\text{1:1 reaction}$, and we have to use the density of acetic acid to find the molar equivalence.

This site reports that ${\rho}_{\text{HOAc}} = 1.049 \cdot g \cdot m {L}^{-} 1$.

And thus $\text{Moles of HOAc} = \frac{20 \cdot m L \times 1.049 \cdot g \cdot m {L}^{-} 1}{60.05 \cdot g \cdot m o {l}^{-} 1}$

$= 0.349 \cdot m o l$

And thus an equivalent quantity of $N a O H \equiv 0.349 \cdot m o l \times 40.0 \cdot g \cdot m o {l}^{-} 1 = 13.98 \cdot g$; of course we would put this mass into aqueous solution first.