The K sp of Ag 2 CrO 4 is 1-1- 10 -12 at 298K- The solubility in mol-L of Ag 2 CrO 4 in 0-1M Ag NO 3 solution is-

The correct option is D 1.1× 10 ^ 10 Ag_2CrO_4⇌ 2Ag^++CrO_4^2 #160; #160; #160; #160; #160; #160; #160; #160; #160; # ...

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Byju's Answer

Standard XII

Chemistry

Solubility Product

The K sp 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 $0.1 M$ $A g {N O}_{3}$ solution is:

1.1 \times 10^{- 11}

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1.1 \times 10^{- 10}

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1.1 \times 10^{- 12}

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1.1 \times 10^{- 9}

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Solution

The correct option is D $1.1 \times 10^{- 10}$
$A g_{2} C r O_{4} ⇌ 2 A g^{+} + C r O_{4}^{2 -}$
$2 S$ $S$

$K_{s p} = [A g^{+}]^{2} [C r O_{4}^{2 -}] = (2 S + 0.1)^{2} (S)$

$(2 S + 0.1) = 0.1$ as $S << 1$

So $K_{s p} = 0.01 (S) = 1.1 \times 10^{- 12}$

$S = 1.1 \times 10^{- 10}$

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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

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}

in a 0.1M

A g N O_{3}

solution is :

Q. The solubility product constant of

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

and

A g B r

1.1 \times 10^{- 12}

and

5.0 \times 10^{- 13}

respectively. Calculate the ratio of the molarities of their saturated solutions.

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The Ksp of Ag2CrO4 is 1.1×10−12 at 298K. The solubility (in mol/L) of Ag2CrO4 in 0.1M AgNO3 solution is:

The correct option is D 1.1×10−10Ag2CrO4⇌2Ag++CrO2−4 2S SKsp=[Ag+]2[CrO2−4]=(2S+0.1)2(S)(2S+0.1)=0.1 as S<<1So Ksp=0.01(S)=1.1×10−12S=1.1×10−10

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 $0.1 M$ $A g {N O}_{3}$ solution is:

The correct option is D $1.1 \times 10^{- 10}$
$A g_{2} C r O_{4} ⇌ 2 A g^{+} + C r O_{4}^{2 -}$
$2 S$ $S$

$K_{s p} = [A g^{+}]^{2} [C r O_{4}^{2 -}] = (2 S + 0.1)^{2} (S)$

$(2 S + 0.1) = 0.1$ as $S << 1$

So $K_{s p} = 0.01 (S) = 1.1 \times 10^{- 12}$

$S = 1.1 \times 10^{- 10}$