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Ideal Gas Equation

The ksp of ...

Question

The $k_{s p}$ of $A g_{2} C r O_{4}$ is $1.1 \times 10^{- 12}$ at $298 k$ . The solubility (in mol/l) of $A g_{2} C r O_{4}$ in a 0.1M $A g N O_{3}$ solution is :

1.1 \times 10^{-} 10 S

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3 \times 10^{- 8} S

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2 \times 10^{- 8} S

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2.5 \times 10^{- 8} S

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Solution

The correct option is A $1.1 \times 10^{-} 10 S$
$A g_{2} C r O_{4} \to 2 A g^{+} + C r O_{4} 2 -$
Ag $N O_{3} \to A g^{+} + N O_{3}^{-}$
$K_{s p} o f A g_{2} C r O_{4} = 1.1 \times 10^{- 12}$
$\Rightarrow (2 S + 0.1)^{2} (S) = 1.1 \times 10^{- 12}$ (Common ion effect)
Since, S <<< 0.1 , 2S can be neglected.
$\Rightarrow (0.01) S = 1.1 \times 10^{- 12}$
$\Rightarrow S = \frac{1.1 \times 10^{- 12}}{0.01} = 1.1 \times 10^{- 10}$
Hence, the correct option is A.

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Similar questions

The $K_{s p}$ of $A g_{2} C r O_{4}$ is $1.1 \times 10^{- 12}$ at 298K. The solubility (in mol/L) of $A g_{2} C r O_{4}$ in 0.1M $A g N O_{3}$ solution is:

Q. The

K_{s p} o f A g_{2} C r O_{4} i s 1.1 \times 10^{- 12} a t 298 K .

The solubility (in

m o l / L

) of

A g_{2} C r O_{4}

in a

0.1 M A g N O_{3}

solution is

Q. The

K_{s p}

A g_{2} C r O_{4}

1.1 \times 10^{- 12}

at 298 K. The solubility (in mole / L) of

A g_{2} C r O_{4}

in a

0.1 M A g N O_{3}

solution is

Q. The

K_{s p}

{A g}_{2} {C r O}_{4}

1.1 \times 10^{- 12}

298 K

. The solubility (in mol/L) of

{A g}_{2} {C r O}_{4}

0.1 M

A g {N O}_{3}

solution is:

Q. The

K_{s p}

A g_{2} C r O_{4}

1.1 \times 10^{- 10}

, what is the value of solubility,

S

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The ksp of Ag2CrO4 is 1.1×10−12 at 298k. The solubility (in mol/l) of Ag2CrO4 in a 0.1M AgNO3 solution is :

The correct option is A 1.1×10−10SAg2CrO4→2Ag++CrO42−AgNO3→Ag++NO−3KspofAg2CrO4=1.1×10−12⇒(2S+0.1)2(S)=1.1×10−12 (Common ion effect)Since, S <<< 0.1 , 2S can be neglected. ⇒(0.01)S=1.1×10−12⇒S=1.1×10−120.01=1.1×10−10Hence, the correct option is A.

The $k_{s p}$ of $A g_{2} C r O_{4}$ is $1.1 \times 10^{- 12}$ at $298 k$ . The solubility (in mol/l) of $A g_{2} C r O_{4}$ in a 0.1M $A g N O_{3}$ solution is :