25 cm^3 of oxalic acid completely neutralize | vedclass

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Question

(B) The balanced chemical equation for the reaction is: $(COOH)_2 + 2NaOH \rightarrow (COONa)_2 + 2H_2O$. From the stoichiometry,$1 \ mole$ of oxalic

Vedclass · Accepted Answer

$0.032$

Vedclass · Answer

(B) The balanced chemical equation for the reaction is: $(COOH)_2 + 2NaOH ightarrow (COONa)_2 + 2H_2O$. From the stoichiometry,$1 \ mole$ of oxalic acid reacts with $2 \ moles$ of $NaOH$. Number of moles of $NaOH = \frac{	ext{mass}}{	ext{molar mass}} = \frac{0.064 \ g}{40 \ g/mol} = 0.0016 \ mol$. Since $2 \ moles$ of $NaOH$ react with $1 \ mole$ of oxalic acid,the moles of oxalic acid $= \frac{0.0016}{2} = 0.0008 \ mol$. Volume of oxalic acid $= 25 \ cm^3 = 0.025 \ L$. Molarity $(M) = \frac{	ext{moles of solute}}{	ext{volume of solution in } L} = \frac{0.0008 \ mol}{0.025 \ L} = 0.032 \ M$.

$25 \ cm^3$ of oxalic acid completely neutralized $0.064 \ g$ of sodium hydroxide. The molarity of the oxalic acid solution is: (in $M$)

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