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Solubility Product

The Ksp of Ag...

Question

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

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 B $1.1 \times 10^{- 10}$
PLAN In presence common ion (in this case $A g^{+}$ ion) solubiility of sparingly soluble salt is decreased.

Let solubility of $A g_{2} C r O_{4}$ in presence of 0.1 M
$A g N O_{3} = x$
$A g_{2} C r O_{4} ⇌ 2 A g^{+} + C r O_{4}^{2 -}$
$A g N O_{3} ⇌ A g^{+} 0.1 + N O_{3}^{-} 0.1$
Total $[A g^{+}]$ = (2x + 0.1) M $\approx$ 0.1 M
as x <<< 0.1 M
$[C r O_{4}^{2 -}] = x M$
Thus, $[A g^{+}]^{2} [C r O_{4}^{2 -}] = K_{s p}$
$(0.1)^{2} (x) = 1.1 \times 10^{12}$
$∵ x = 1.1 \times 10^{- 10} M$

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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. Find the solubility product of a saturated solution of

A g_{2} C r O_{4}

in water at

298 K

if the emf of the cell

A g | A g^{+}

(satd.

A g_{2} C r O_{4} s o l n .) | | A g^{+} (0.1 M) | A g

0.164 V

298 K

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

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