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Degree of Dissociation

If a 25 mL 5....

Question

If a $25 m L$ $5.0 \times 10^{- 4} M C a^{2 +}$ solution and $25 m L$ $2.5 \times 10^{- 4} M C_{2} O_{4}^{2 -}$ solution are mixed, then which of the following option is correct ?
Given: $K_{s p} (C a C_{2} O_{4}) = 2.3 \times 10^{- 9}$ .

A

Precipitate is formed

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B

No precipitate is formed

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C

Can't be determined.

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D

None of the above

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Solution

The correct option is A Precipitate is formed
After mixing,
Total volume= $25 + 25 = 50 m l .$

To calculate $[C a^{2 +}]$ using the formula:
$M_{1} V_{1} = M_{2} V_{2}$
$(5.0 \times 10^{- 4} M) \times (25 m L) = M_{2} \times (50 m L)$
$[C a^{2 +}] = 2.5 \times 10^{- 4} M$
Calculate $[C_{2} O_{4}^{2 -}]$ using the formula:
$M_{1} V_{1} = M_{2} V_{2}$
$(2.5 \times 10^{- 4} M) (25 m L) = M_{2} (50 m L)$ $[C_{2} O_{4}^{2 -}] = 1.25 \times 10^{- 4} M$

$Ionic Product = [C a^{2 +}] [C_{2} O_{4}^{2 -}] = [2.5 \times 10^{- 4}] [1.25 \times 10^{- 4}] = 3.125 \times 10^{- 8}$

Since $Ionic Product > K_{s p}$
$\Rightarrow$ Precipitate is formed.

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Degree of Dissociation

Standard XII Chemistry

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