Question

A student used 36.02 ml of 0.544 M NaOH to react with 25.0 ml of Vinegar. If acetic acid present in the vinegar reacts with NaOH according to the following reaction, what is the molarity of the acetic acid?

CH3COOH + NAOH --->H2O + NaC2H3O2

Answer 1

`["H"_3"CCO"_2"H"]=0.7838*mol*L^-1`

Explanation:

We write the equation to remind us of the equivalence...

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

`"Moles of NaOH"-=36.02*mLxx10^-3*L*mL^-1xx0.544*mol*L^-1=1.96xx10^-2*mol`.

And so we can work out the concentration of the vinegar directly, as this molar quantity was present in the `25.0*mL` volume....

`[HOAc]=(1.96xx10^-2*mol)/(25xx10^-3*L)=?*mol*L^-1`.