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Molarity

0.2 mol of Na...

Question

$0.2 m o l$ of $N a_{3} P O_{4}$ and $0.5 m o l$ of $B a (N O_{3})_{2}$ are mixed in a $1 L$ of solution. Which of the following statements is/are correct?

0.2 m o l

of barium phosphate precipitate is obtained

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0.1 m o l

of barium phosphate precipitate is obtained

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Molarity of

B a^{2 +}

ions in the resulting solution is

0.2

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Molarities of

N a^{+}

and

N O_{3}^{-}

ions are

0.6

and

1.0

respectively

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Solution

The correct option is D Molarities of $N a^{+}$ and $N O_{3}^{-}$ ions are $0.6$ and $1.0$ respectively
The corresponding balanced reaction is:
$2 N a_{3} P O_{4} (a q) + 3 B a (N O_{3})_{2} (a q) \to B a_{3} (P O_{4})_{2} (a q) ↓ + 6 N a N O_{3} (a q)$
So, $N a_{3} P O_{4}$ is the limiting reagent and is completely consumed.
Moles of $B a_{3} (P O_{4})_{2}$ formed $= \frac{0.2}{2} m o l = 0.1 m o l$
Moles of unreacted $B a (N O_{3})_{2} = (0.5 - 0.3) m o l = 0.2 = mol of B a^{2 +} i o n$ .
The reacted $0.3 mol$ $B a^{2 +}$ will be precipitated as $0.1 mol$ barium phosphate.
Moles of $N a^{+}$ in solution $= 0.2 \times 3 = 0.6$
Moles of $N O_{3}^{-}$ in solution $= 0.5 \times 2 = 1$

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Similar questions

0.2 m o l

N a_{3} P O_{4}

0.5 m o l

B a (N O_{3})_{2}

1 L

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If 0.5 mole of $B a C l_{2}$ is mixed with 0.2 mol of $N a_{3} P O_{4}$ , the maximum amount of $B a_{3} (P O_{4})_{3}$ that can be formed is:

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