Question

If 40. mL of .10M NaOH is used to titrate 25 mL of acetic acid, how many grams of acetic acid were present?

Thanks in advance

Answer 1

`"Mass of acetic acid = 240 mg"`

Explanation:

As always we write the stoichiometric equation to inform our reasoning..

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

And thus there is 1:1 equivalence between acetic acid and sodium hydroxide...

`n_"NaOH"=40.0*mLxx10^-3*L*mL^-1xx0.10*mol*L^-1=4.0*mmol`

And thus there were an equivalent molar quantity of acetic acid in the `25*mL` volume...

`[HOAc]=(4.0xx10^-3*mol)/(25xx10^-3*L)=0.160*mol*L^-1`.

And ......................

`"mass"_"HOAc"=25xx10^-3*Lxx0.160*mol*L^-1xx60.05*g*mol^-1=0.240*g`