If 50 mL of 0.5 M oxalic acid is required t | vedclass

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(A) The reaction between oxalic acid $(H_2C_2O_4)$ and $NaOH$ is: $H_2C_2O_4 + 2NaOH \rightarrow Na_2C_2O_4 + 2H_2O$. Using the principle of equivalen

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$4$

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(A) The reaction between oxalic acid $(H_2C_2O_4)$ and $NaOH$ is: $H_2C_2O_4 + 2NaOH ightarrow Na_2C_2O_4 + 2H_2O$. Using the principle of equivalence: $n_{factor} 	imes M_1 	imes V_1 = n_{factor} 	imes M_2 	imes V_2$. For oxalic acid, $n_{factor} = 2$. For $NaOH$, $n_{factor} = 1$. $2 	imes 0.5 	imes 50 = 1 	imes M_{NaOH} 	imes 25$. $50 = 25 	imes M_{NaOH} \Rightarrow M_{NaOH} = 2 \ M$. Now, calculate the mass of $NaOH$ in $50 \ mL$ of this solution: $Mass = Molarity 	imes Molar \ mass 	imes Volume (in \ L) = 2 	imes 40 	imes (50 	imes 10^{-3}) = 4 \ g$.

If $50 \ mL$ of $0.5 \ M$ oxalic acid is required to neutralize $25 \ mL$ of $NaOH$ solution, the amount of $NaOH$ in $50 \ mL$ of the given $NaOH$ solution is . . . . . . $g$.

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