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Common Ion Effect

Ksp of CaSO...

Question

$K_{s p}$ of $C a S O_{4}$ is $4 \times 10^{- 12} . C a S O_{4}$ is precipitated on mixing equal volumes of the following solutions.

A

3 \times 10^{- 10} M C a C l_{2}

and

3 \times 10^{- 6} M (N H_{4})_{2} S O_{4}

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B

4 \times 10^{- 6} M C a C l_{2}

and

3 \times 10^{- 6} M (N H_{4})_{2} S O_{4}

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C

6 \times 10^{- 6} M C a C l_{2}

and

3 \times 10^{- 6} M (N H_{4})_{2} S O_{4}

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D

All of the above

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Solution

The correct option is D $6 \times 10^{- 6} M C a C l_{2}$ and $3 \times 10^{- 6} M (N H_{4})_{2} S O_{4}$
A precipitate will form when two solutions are mixed if the ion product is greater than the solubility product. This solution is called super saturated solution
$[C a^{2 +}] [S O_{4}^{2 -}] > K_{s p}$ .

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Similar questions

Q. Equal volumes of

0.02 M

C a C l_{2}

0.04 M

N a_{2} S O_{4}

are mixed. Will a precipitate form?

(K_{s p}

C a S O_{4} = 2.4 \times 10^{- 5})

Q. Equal volume of the following

{C a}^{2 +}

F^{-}

solutions are mixed. In which solutions will the precipitation occur?

K_{s p}

C a F_{2} = 1.7 \times 10^{- 10}

10^{- 2} M {C a}^{2 +} + 10^{- 5} M F^{-}

10^{- 3} M {C a}^{2 +} + 10^{- 3} M F^{-}

10^{- 4} M {C a}^{2 +} + 10^{- 2} M F^{-}

10^{- 2} M {C a}^{2 +} + 10^{- 3} M F^{-}

Select the correct answer using the codes given below:

A solution prepared by mixing $200 m l$ of $0.025 M$ $C a C l_{2}$ and $400 m l$ of $0.15 M$ $N a_{2} S O_{4}$ ? Given $K_{s p}$ of $C a S O_{4}$ = $2.4 \times 10^{- 5}$ Then

The $K_{S P}$ of Agl is $1.5 \times 10^{- 16}$ . On mixing equal volumes of the following

solutions, precipitation will occur only with

(C a S O_{4})

K_{s p} = 4.9 \times 10^{- 5}

is found to have

4 \times 10^{- 7} M

C a^{2 +}

2.5 \times 10^{- 8} M

S O_{4}^{2 -}

.
The equilibrium state of this solution with respect to

(C a S O_{4})

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