Solution X contains Na_2CO_3 and NaHCO_3. 20 | vedclass

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(B) For titration of a mixture of $Na_2CO_3$ and $NaHCO_3$ against $HCl$:$1$. With phenolphthalein indicator,the end point corresponds to the conversi

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$0.1, 0.1$

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(B) For titration of a mixture of $Na_2CO_3$ and $NaHCO_3$ against $HCl$:$1$. With phenolphthalein indicator,the end point corresponds to the conversion of $Na_2CO_3$ to $NaHCO_3$:$Na_2CO_3 + HCl ightarrow NaHCO_3 + NaCl$$Moles \ of \ HCl = Moles \ of \ Na_2CO_3 = 20 \ mL 	imes 0.1 \ M = 2 \ mmol$.Concentration of $Na_2CO_3 = \frac{2 \ mmol}{20 \ mL} = 0.1 \ M$.$2$. With methyl orange indicator,the end point corresponds to the complete neutralization of $Na_2CO_3$ and $NaHCO_3$:$Na_2CO_3 + 2HCl ightarrow 2NaCl + CO_2 + H_2O$$NaHCO_3 + HCl ightarrow NaCl + CO_2 + H_2O$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 $NaHCO_3 = \frac{2 \ mmol}{20 \ mL} = 0.1 \ M$.Thus,the concentrations are $0.1 \ M$ and $0.1 \ M$.

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