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Standard XII

Chemistry

Common Ion Effect

A solution co...

Question

A solution contains $A g^{+}$ ion of concentration $2 \times 10^{- 18} M .$ What is the concentration of $N a_{2} S$ solution to be used in order to just start precipitation? $(K_{s p} (A g_{2} S) = 4 \times 10^{- 48})$

1 \times 10^{- 30} M

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1 \times 10^{- 24} M

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1 \times 10^{- 18} M

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1 \times 10^{- 12} M

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Solution

The correct option is C $1 \times 10^{- 12} M$
$K_{s p} (A g_{2} S) = [A g^{+}]^{2} [S^{2 -}]$

$4 \times 10^{- 48} = [2 \times 10^{- 18}]^{2} [S^{2 -}]$

$[S^{2 -}] = \frac{4 \times 10^{- 48}}{(2 \times 10^{- 18})^{2}} = 1 \times 10^{- 12} M$ .

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Similar questions

Q. The solubility product of

A g C l

25^{\circ} C

1 \times 10^{- 10}

. A solution of

A g^{+}

at a concentration of

4 \times 10^{- 3} M

just fails to yield a precipitate of

A g C l

with a concentration of

1 \times 10^{- 3} M

C l^{-}

when the concentration of

N H_{3}

in the solution is

2 \times 10^{- 2} M

. Calculate the equilibrium constant for

[A g (N H_{3})_{2}]^{+} ⇌ A g^{+} + 2 N H_{3}

Q. A solution is a mixture of

0.05 M N a C l

and

0.05 M N a I

. The concentration of iodide ion in the solution when

A g C l

just starts precipitating is equal to:

K_{s p} (A g C l) = 1 \times 10^{- 10} M^{2}; K_{s p} (A g I) = 4 \times 10^{- 16} M^{2}

Q. What is the minimum concentration of

B a^{+ 2}

ions required in order to initiate the precipitation of

B a S O_{4}

from a solution containing

0.002 m o l L^{- 1}

S O_{4}^{- 2}

ions?
(Given

K_{s p}

for

B a S O_{4} = 1.4 \times 10^{- 10}

)

Q. The solubility product of

A g C l

25^{\circ} C

1 \times 10^{- 10}

. A solution of

A g^{+}

at a concentration of

4 \times 10^{- 3} M

just fails to yield a precipitate of

A g C l

with a concentration of

1 \times 10^{- 3} M

C l^{-}

when the concentration of

N H_{3}

in the solution is

2 \times 10^{- 2} M

. Calculate the equilibrium constant for

[A g (N H_{3})_{2}]^{+} ⇌ A g^{+} + 2 N H_{3}

Q. What is the minimum concentration of

S O_{4}^{2 -}

required to precipitate

B a S O_{4}

in a solution containing

1 \times 10^{- 4}

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for

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A solution contains Ag+ ion of concentration 2×10−18M. What is the concentration of Na2S solution to be used in order to just start precipitation? (Ksp(Ag2S)=4×10−48)

The correct option is C 1×10−12MKsp(Ag2S)=[Ag+]2[S2−]4×10−48=[2×10−18]2[S2−][S2−]=4×10−48(2×10−18)2=1×10−12M.

A solution contains $A g^{+}$ ion of concentration $2 \times 10^{- 18} M .$ What is the concentration of $N a_{2} S$ solution to be used in order to just start precipitation? $(K_{s p} (A g_{2} S) = 4 \times 10^{- 48})$

The correct option is C $1 \times 10^{- 12} M$
$K_{s p} (A g_{2} S) = [A g^{+}]^{2} [S^{2 -}]$

$4 \times 10^{- 48} = [2 \times 10^{- 18}]^{2} [S^{2 -}]$

$[S^{2 -}] = \frac{4 \times 10^{- 48}}{(2 \times 10^{- 18})^{2}} = 1 \times 10^{- 12} M$ .