0.4 g mixture of NaOH, Na_{2}CO_{3} and some | vedclass

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(B) Let $x$ be the milliequivalents of $NaOH$ and $y$ be the milliequivalents of $Na_{2}CO_{3}$.At the first end point (phenolphthalein indicator),$Na

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$2$

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(B) Let $x$ be the milliequivalents of $NaOH$ and $y$ be the milliequivalents of $Na_{2}CO_{3}$.At the first end point (phenolphthalein indicator),$NaOH$ is neutralized and $Na_{2}CO_{3}$ is converted to $NaHCO_{3}$:$x + y = \frac{1}{10} 	imes 17.5 = 1.75 	ext{ meq} \dots (1)$At the second end point (methyl orange indicator),$NaHCO_{3}$ is neutralized:$y = \frac{1}{10} 	imes 1.5 = 0.15 	ext{ meq} \dots (2)$Mass of $Na_{2}CO_{3} = 	ext{milliequivalents} 	imes 	ext{equivalent weight} 	imes 10^{-3}$$= 0.15 	imes 53 	imes 10^{-3} = 0.00795 \ g$Weight percentage of $Na_{2}CO_{3} = \frac{0.00795}{0.4} 	imes 100 = 1.9875 \%$Rounding off to the nearest integer,we get $2 \%$.

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