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Mole Concept and Avogadro's Number

When 10 g m...

Question

When $10 g$ mixture of $N a_{2} C O_{3}$ and $N a O H$ was titrated against $H C l$ by using phenolphthalein followed by adding methyl orange after first equivalence point, $7.7818 g$ of $H C l$ was consumed in total. Find the mass of $N a_{2} C O_{3}$ in the mixture.

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Solution

$10 g$ of ${N a}_{2} {C O}_{3}$ and $N a O H$ , total $H C l$ moles equal to twice ${N a}_{2} {C O}_{3}$ moles $+$ $N a O H$ moles
$7.78189$ of $H C l = 0.2132$ moles
Let $x$ $g$ of ${N a}_{2} {C O}_{3} = \frac{x}{106}$ moles
moles of $N a O H = \frac{10 - x}{40}$ moles
$0.2132 = 2 (\frac{x}{106}) + \frac{10 - x}{40}$
$0.2132 = \frac{1}{4} + x (\frac{1}{53} \frac{- 1}{40})$
$x \approx 6.0012$ g

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Q. 200 mL of a solution mixture

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