Question
An aqueous solution of $$6.3g$$ oxalic acid dihydrate is made up to $$250\,{\text{ml}}{\text{.}}$$ The volume of $$0.1\,\,N\,\,NaOH$$ required to completely neutralize $${\text{10}}\,{\text{ml}}$$ of this solution is
A.
$$40\,ml$$
B.
$$20\,ml$$
C.
$$10\,ml$$
D.
$$4\,ml$$
Answer :
$$40\,ml$$
Solution :
TIPS/Formulae :
Equivalents of
$$\eqalign{
& {H_2}{C_2}{O_4}.2{H_2}O = {\text{Equivalents}}\,\,{\text{of}}\,\,NaOH \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{At}}\,{\text{equivalence}}\,{\text{point}}} \right) \cr
& {\text{Strength}}\,\,{\text{of}}\,\,{H_2}{C_2}{O_4}.2{H_2}O\left( {{\text{in}}\,g/L} \right){\text{ = }}\frac{{6.3}}{{\frac{{250}}{{1000}}}} \cr
& {\text{ = 25}}{\text{.2}}\,g/L\, \cr
& {\text{Normality}}\,{\text{of}}\,\,{H_2}{C_2}{O_4}.2{H_2}O = \frac{{{\text{strength}}}}{{{\text{Eq}}{\text{.}}\,{\text{wt}}}} \cr
& = \frac{{25.2}}{{63}} = 0.4N \cr
& \left\{ {{\text{Eq}}{\text{.}}\,{\text{wt}}{\text{.}}\,{\text{of}}\,{\text{oxalic}}\,{\text{acid}} = \frac{{{\text{Mol}}{\text{.}}\,{\text{wt}}}}{2} = \frac{{126}}{2} = 63} \right\} \cr
& {\text{Using}}\,{\text{normality}}\,{\text{equation}}: \cr
& {N_1}{V_1} = {N_2}{V_2} \cr
& \left( {{H_2}{C_2}{O_4}.2{H_2}O} \right)\,\,\,\left( {NaOH} \right) \cr
& 0.4 \times 10 = 0.1 \times {V_2}\,or\,{V_2} = \frac{{0.4 \times 10}}{{0.1}} = 40ml. \cr} $$