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Conductivity MP BP and Electronegativity (Group 2)

A 0.141 g s...

Question

A $0.141$ g sample of phosphorus containing compound was digested in a mixture of $H N O_{3}$ and $H_{2} S O_{4}$ , which resulted in formation of $C O_{2}, H_{2} O$ and $H_{3} P O_{4}$ . Addition of ammonium molybdate yielded a solid having the composition $(N H_{4})_{3} P O_{4} .12 M o O_{3}$ . The precipitate was filtered, washed and dissolved in $50.0$ mL of $0.20 M N a O H$ .
$(N H_{4})_{3} P O_{4} .12 M o O_{3} (s) + 26 O H^{-} ⟶ H P O_{4}^{2 -} + 12 M o O_{4}^{2 -} + 14 H_{2} O$
$+ 3 N H_{3} (g)$
After boiling the solution to remove the $N H_{3}$ , the excess of $N a O H$ was titrated with $14.1$ mL of $0.174 M$ $H C l$ . Calculate the percentage of phosphorus in the sample.

8.9 %

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12.7 %

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6.4 %

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3.2 %

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Solution

The correct option is D $6.4 %$
$P O x i d i z e d - ---- \to H_{3} P O_{4}$
$P^{0} ⟶ P^{5 +} + 5 e^{-}$
Thus, $1$ eq. of $P$ is converted to $1$ eq. of $H_{3} P O_{4}$ .
or, $1$ millimole of $P$ is converted to $1$ mole of $H_{3} P O_{4}$ .
Millimoles of $N a O H$ added $= 50 \times 0.2 = 10$
Millimoles of $N a O H$ used by $H C l = 14.1 \times 0.174 = 2.453$ (both are monovalent)
$∴$ Millimoles of $N a O H$ used for sample contents after digestion and precipitation $= 10 - 2.453 = 7.547$
$∵ 26$ mmol of $N a O H$ is used for $1$ mmol $P$ .
$∴$ Millimoles of $P$ present $= \frac{7.547}{26} = 0.290$
$∴ \frac{w_{P}}{31} \times 1000 = 0.290$
$∴ w_{P} = 9.00 \times 10^{- 3}$
$% P = \frac{9.00 \times 10^{- 3}}{0.141} \times 100 = 6.38 %$

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