25 mL of the given HCl solution requires 30 | vedclass

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(A) Step $1$: Determine the normality of $HCl$ using the reaction with $Na_2CO_3$.$Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$.Using the law of e

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$25$

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(A) Step $1$: Determine the normality of $HCl$ using the reaction with $Na_2CO_3$.$Na_2CO_3 + 2HCl ightarrow 2NaCl + H_2O + CO_2$.Using the law of equivalence: $N_1V_1 = N_2V_2$.For $Na_2CO_3$,$N = M 	imes 	ext{valency factor} = 0.1 	imes 2 = 0.2 \ N$.$N_{HCl} 	imes 25 \ mL = 0.2 \ N 	imes 30 \ mL$.$N_{HCl} = \frac{6}{25} = 0.24 \ N$.Step $2$: Determine the volume of $HCl$ required for $NaOH$.$HCl + NaOH ightarrow NaCl + H_2O$.Using $N_1V_1 = N_2V_2$ for $HCl$ and $NaOH$.$0.24 \ N 	imes V_{HCl} = 0.2 \ M 	imes 1 	imes 30 \ mL$.$0.24 	imes V_{HCl} = 6$.$V_{HCl} = \frac{6}{0.24} = 25 \ mL$.

$25 \ mL$ of the given $HCl$ solution requires $30 \ mL$ of $0.1 \ M$ sodium carbonate solution. What is the volume of this $HCl$ solution required to titrate $30 \ mL$ of $0.2 \ M$ aqueous $NaOH$ solution? .............. $mL$

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