20 , mL of a solution of a weak monobasic aci | vedclass

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(A) Let $N_1$ be the normality of the weak monobasic acid and $N_2$ be the normality of the $NaOH$ solution.For the acid-base neutralization: $N_1 	i

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$0.10$

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(A) Let $N_1$ be the normality of the weak monobasic acid and $N_2$ be the normality of the $NaOH$ solution.For the acid-base neutralization: $N_1 	imes 20 = N_2 	imes 22.18$  $(i)$For the $HCl$ and $NaOH$ neutralization: $N_{HCl} 	imes V_{HCl} = N_{NaOH} 	imes V_{NaOH}$$\frac{1}{10} 	imes 20 = N_2 	imes 21.5$$N_2 = \frac{2}{21.5} \approx 0.09302 \, N$ $(ii)$Substitute $N_2$ into equation $(i)$:$N_1 = \frac{N_2 	imes 22.18}{20} = \frac{2 	imes 22.18}{21.5 	imes 20} = \frac{22.18}{215} \approx 0.103 \, N$The normality of the acid is nearly $0.10 \, N$.

$20 \, mL$ of a solution of a weak monobasic acid neutralizes $22.18 \, mL$ of a solution of $NaOH$,and $20 \, mL$ of $N/10$ $HCl$ neutralizes $21.5 \, mL$ of the same $NaOH$ solution. The normality of the acid is nearly $.......$ $N$.

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