What mass of sodium hydroxide would be equivalent to a 20*mL20⋅mL volume of acetic acid, for which rho_"HOAc"=1.049*g*mL^-1ρHOAc=1.049⋅g⋅mL−1?

1 Answer
anor277
Jun 12, 2017

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

Explanation:

"NaOH(aq)" + "H"_3"CCO"_2"H(l)" rarr "H"_3"CCO"_2^(-)"Na"^(+) + "H"_2"O(l)"NaOH(aq)+H3CCO2H(l)→H3CCO−2Na++H2O(l).

This is a "1:1 reaction"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^-1ρHOAc=1.049⋅g⋅mL−1.

And thus "Moles of HOAc"=(20*mLxx1.049*g*mL^-1)/(60.05*g*mol^-1)Moles of HOAc=20⋅mL×1.049⋅g⋅mL−160.05⋅g⋅mol−1

=0.349*mol=0.349⋅mol

And thus an equivalent quantity of NaOH-=0.349*molxx40.0*g*mol^-1=13.98*gNaOH≡0.349⋅mol×40.0⋅g⋅mol−1=13.98⋅g; of course we would put this mass into aqueous solution first.

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