Solution X contains Na_2CO_3 and NaHCO_3. 20 | vedclass

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(B) For titration of $Na_2CO_3$ and $NaHCO_3$ with $HCl$ using phenolphthalein,the reaction is: $Na_2CO_3 + HCl \longrightarrow NaHCO_3 + NaCl$. At th

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$0.1, 0.1$

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(B) For titration of $Na_2CO_3$ and $NaHCO_3$ with $HCl$ using phenolphthalein,the reaction is: $Na_2CO_3 + HCl \longrightarrow NaHCO_3 + NaCl$. At the phenolphthalein end point,$20 \ mL$ of $0.1 \ M$ $HCl$ is consumed,which corresponds to half the $Na_2CO_3$. $Moles \ of \ Na_2CO_3 = Moles \ of \ HCl = 20 \ mL 	imes 0.1 \ M = 2 \ mmol$. Concentration of $Na_2CO_3 = \frac{2 \ mmol}{20 \ mL} = 0.1 \ M$. Using methyl orange,both $Na_2CO_3$ and $NaHCO_3$ are neutralized. Total $HCl$ consumed = $60 \ mL 	imes 0.1 \ M = 6 \ mmol$. $HCl$ consumed by $Na_2CO_3$ (complete) = $2 	imes (2 \ mmol) = 4 \ mmol$. $HCl$ consumed by $NaHCO_3$ = $6 \ mmol - 4 \ mmol = 2 \ mmol$. Concentration of $NaHCO_3 = \frac{2 \ mmol}{20 \ mL} = 0.1 \ M$.

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