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(D) The thermal decomposition of sodium bicarbonate is given by: $2NaHCO_{3(s)} \xrightarrow{\Delta} Na_2CO_{3(s)} + H_2O_{(g)} + CO_{2(g)}$When pheno

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$84$

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(D) The thermal decomposition of sodium bicarbonate is given by: $2NaHCO_{3(s)} \xrightarrow{\Delta} Na_2CO_{3(s)} + H_2O_{(g)} + CO_{2(g)}$When phenolphthalein is used as an indicator,the titration of $Na_2CO_3$ with $HCl$ proceeds only up to the formation of $NaHCO_3$:$Na_2CO_3 + HCl \xrightarrow{Ph} NaHCO_3 + NaCl$Moles of $HCl$ used $= 0.1 \ M 	imes 0.5 \ L = 0.05 \ mol$From the stoichiometry of the titration reaction,$1 \ mol$ of $Na_2CO_3$ reacts with $1 \ mol$ of $HCl$. Thus,moles of $Na_2CO_3$ produced $= 0.05 \ mol$.From the decomposition reaction,$2 \ mol$ of $NaHCO_3$ produce $1 \ mol$ of $Na_2CO_3$. Therefore,initial moles of $NaHCO_3 = 2 	imes 0.05 \ mol = 0.1 \ mol$.Mass of $NaHCO_3 = 0.1 \ mol 	imes 84 \ g/mol = 8.4 \ g$.$PERCENT$ purity $= (8.4 \ g / 10 \ g) 	imes 100 = 84 \%$.

$10 \ g$ $NaHCO_{3(s)}$ sample is heated in open to constant weight,the residue is titrated against $0.1 \ M \ HCl$ in presence of phenolphthalein as an indicator,$500 \ mL$ of acid was required to reach end point hence $PERCENT$ purity of sample is $............... \%$.

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