50 mL of 0.5 M oxalic acid is needed to neu | vedclass

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Question

(D) The neutralization reaction is: $(COOH)_2 + 2NaOH ightarrow (COONa)_2 + 2H_2O$. Using the equivalence concept: $n_{factor} 	imes M_1 	imes V_1

Vedclass · Accepted Answer

None of these

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(D) The neutralization reaction is: $(COOH)_2 + 2NaOH ightarrow (COONa)_2 + 2H_2O$. Using the equivalence concept: $n_{factor} 	imes M_1 	imes V_1 = n_{factor} 	imes M_2 	imes V_2$. For oxalic acid $(COOH)_2$,$n_{factor} = 2$. For $NaOH$,$n_{factor} = 1$. $2 	imes 0.5 	imes 50 = 1 	imes M_2 	imes 25$. $50 = 25 	imes M_2 \implies M_2 = 2 \ M$. Now,calculate the mass of $NaOH$ in $50 \ mL$ of $2 \ M$ solution: $Moles = Molarity 	imes Volume(L) = 2 	imes 0.050 = 0.1 \ mol$. $Mass = Moles 	imes Molar \ mass = 0.1 	imes 40 = 4 \ g$. Since $4 \ g$ is not among the options,the correct answer is $D$.

$50 \ mL$ of $0.5 \ M$ oxalic acid is needed to neutralize $25 \ mL$ of sodium hydroxide solution. The amount in grams of $NaOH$ in $50 \ mL$ of the given sodium hydroxide solution is:

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