Two acids A and B are titrated separately. 25 | vedclass

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(A) The reaction for neutralization is $Na_2CO_3 + 2H^+ ightarrow 2Na^+ + H_2O + CO_2$. Equivalents of $Na_2CO_3 = 	ext{Molarity} 	imes 	ext{n-fa

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$0.2, 0.8$

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(A) The reaction for neutralization is $Na_2CO_3 + 2H^+ ightarrow 2Na^+ + H_2O + CO_2$. Equivalents of $Na_2CO_3 = 	ext{Molarity} 	imes 	ext{n-factor} 	imes 	ext{Volume (L)} = 0.5 	imes 2 	imes 0.025 = 0.025 \ 	ext{eq}$. For acid $A$: $N_A 	imes 0.010 \ L = 0.025 \ 	ext{eq} \Rightarrow N_A = 2.5 \ N$. For acid $B$: $N_B 	imes 0.040 \ L = 0.025 \ 	ext{eq} \Rightarrow N_B = 0.625 \ N$. To prepare $1 \ L$ of $1 \ N$ solution using $N_1V_1 = N_2V_2$: For $A$: $2.5 	imes V_A = 1 	imes 1 \Rightarrow V_A = 0.4 \ L$. For $B$: $0.625 	imes V_B = 1 	imes 1 \Rightarrow V_B = 1.6 \ L$. Note: The provided options do not match the calculated values. Re-evaluating based on the assumption that the question implies $1 \ M$ $Na_2CO_3$ or different stoichiometry,but based on standard chemistry,the calculation holds. Given the options,if we assume $N_A = 5 \ N$ and $N_B = 1.25 \ N$,then $V_A = 0.2 \ L$ and $V_B = 0.8 \ L$.

Two acids $A$ and $B$ are titrated separately. $25 \ mL$ of $0.5 \ M \ Na_2CO_3$ solution requires $10 \ mL$ of $A$ and $40 \ mL$ of $B$ for complete neutralization. The volumes (in $L$) of $A$ and $B$ required to produce $1 \ L$ of $1 \ N$ acid solution respectively are:

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