• # question_answer A solution contains a mixture of $N{{a}_{2}}C{{O}_{3}}$ and NaOH. Using phenolphthalein as indicator, 25 mL of mixture required 19.5 mL of 0.995 N HCI for the end point. Whith methyl organge, 25 mL of solution required 24 mL of the same HCI for the end point. The grams per litre of $N{{a}_{2}}C{{O}_{3}}$ in the mixture is A)  23.2                       B)  18.5     C)  19.9                       D)  12.8

Let the moles of $N{{a}_{2}}C{{O}_{3}}$ and NaOH in 25 mL mixture be x and y respectively. Case -1: When HPh is used as indicator $\underset{x}{\mathop{NaOH}}\,+\underset{x}{\mathop{HCl\to }}\,NaCl+{{H}_{2}}O$ $\underset{y}{\mathop{N{{a}_{2}}C{{O}_{3}}}}\,+\underset{y}{\mathop{HCl\to }}\,NaHC{{O}_{3}}+NaCl$ So$,$$x+y=19.5\times {{10}^{-3}}\times (0.995\times 1)$        ?.1 Case - 2 : When MeOH is used as indicator. $\underset{x}{\mathop{NaOH}}\,+\underset{x}{\mathop{HCl\to }}\,NaCl+{{H}_{2}}O$ $\underset{y}{\mathop{N{{a}_{2}}C{{O}_{3}}}}\,+\underset{2y}{\mathop{2HCl\to 2}}\,NaCl+{{H}_{2}}O+C{{O}_{2}}$ .....(ii)  On solving equation (i) and (ii), we get, $x=13.93\times {{10}^{-3}}mol$ and $y=5.4725\times {{10}^{-3}}mol$ Now, weight of NaOH in 25 mL mixture $=13.93\times {{10}^{-3}}\times 40=557.2\times {{10}^{-3}}g$ Weight of $N{{a}_{2}}C{{O}_{3}}$ in 25 mL mixture $=5.4725\times {{10}^{-3}}\times 106=580.085\times {{10}^{-3}}g$ $\therefore$ Weight of NaOH per litre $=\frac{557.2\times {{10}^{-3}}\times 1000}{25}=23.2034g/L$ weight of $N{{a}_{2}}C{{O}_{3}}$per litre $=\frac{580.085\times {{10}^{-3}}\times 1000}{25}=23.2034g/L$