Solid AgN O 3 is gradually added to a solution which is 0-01 M in Cl -and 0-01 M in CO 32- - K sp AgCl -1-8 - 10-10 molL -12 andK sp Ag 2 CO 3-4 - 10-12 molL -13 - The concentration of Cl -when Ag 2 CO 3 starts precipitating is-A- 1-8 - 10-9 MB- 1-8 - 10-5 MC- 1-2 - 10-4 MD- 9 - 10-6 M

The correct option is D 9 × 10^ 6 MGiven, K_sp of AgCl=1.8× 10^ 10 (molL^ 1)^2 K_spAgCl = s^2 ∴ Solubility of AgCl = √(1.8× 10^ 10) =1.34× 10^ 5 K_spAg_2CO_3= ...

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Byju's Answer

Standard XII

Chemistry

Lewis Theory

Solid AgN O 3...

Question

Solid $A g N O_{3}$ is gradually added to a solution which is $0.01 M$ in $C l^{-}$ and $0.01 M$ in $C O_{3}^{2 -}$ . $K_{s p} A g C l = 1.8 \times 10^{- 10} (m o l L^{- 1})^{2}$ and $K_{s p} A g_{2} C O_{3} = 4 \times 10^{- 12} (m o l L^{- 1})^{3}$ . The concentration of $C l^{-}$ when $A g_{2} C O_{3}$ starts precipitating is:

1.8 \times 10^{- 9} M

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1.8 \times 10^{- 5} M

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1.2 \times 10^{- 4} M

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9 \times 10^{- 6} M

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Solution

The correct option is D $9 \times 10^{- 6} M$
Given,
$K_{s p} o f A g C l = 1.8 \times 10^{- 10} (m o l L^{- 1})^{2}$
$K_{s p} A g C l = s^{2}$
$∴$ Solubility of $A g C l = \sqrt{1.8 \times 10^{- 10}} = 1.34 \times 10^{- 5}$

$K_{s p} A g_{2} C O_{3} = 4 \times 10^{- 12}$ .
$K_{s p} A g_{2} C O_{3} = 4 s^{3}$
So, solubility of $A g_{2} C O_{3}$ = $\frac{s^{\frac{1}{3}}}{4} = 10^{- 4}$

Hence $A g C l$ will precipitate first and concentration of $A g^{+}$ will keep on increasing slowly until it exceeds the $K_{s p} A g_{2} C O_{3}$ .
Concentration of $A g^{+}$ at the begining of the precipitation of $A g_{2} C O_{3}$

$[A g^{+}] = \sqrt{\frac{K_{s p} A g_{2} C O_{3}}{[C O_{3}]^{2 -}}}$
$= \sqrt{\frac{4 \times 10^{- 12}}{10^{- 2}}}$
$= 2 \times 10^{- 5} M$
Concentration of $C l^{-}$ at the begining of the precipitation of $A g_{2} C O_{3}$

$[C l^{-}] = \frac{[K_{s p} A g C l]}{[A g^{+}]}$
$= \frac{1.8 \times 10^{- 10}}{2 \times 10^{- 5}}$
$= 9 \times 10^{- 6} M$

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Similar questions

Q. A solution consists of a mixture of

0.01 M K I

and

0.1 M K C l

. If solid

A g N O_{3}

is added to the solution, what is the concentration of

I^{-}

when

A g C l

begins to precipitate?

[K_{s p (A g I)} = 1.5 \times 10^{- 16}; K_{s p (A g C l)} = 1.8 \times 10^{- 10}]

Q. A solution is made of

0.1 M

C l^{-}

and

10^{- 4} M C r O_{4}^{2 -}

. If solid

A g N O_{3}

is gradually added to this solution, then what will be the concentration of

C l^{-}

when

A g_{2} C r O_{4}

begins to precipitate?

[K_{s p} (A g C l) = 10^{- 10} M^{2}; K_{s p} (A g_{2} C r O_{4}) = 10^{- 12} M^{3}]

The concentrations of $C l^{-} & C r O_{4}^{2 -}$ in a solution are $0.1 M & 10^{- 4} M$ respectively. If solid $A g N O_{3}$ is gradually
added to this solution, what will be the concentration of $C l^{-}$ when $A g_{2} C r O_{4}$ begins to precipitate? $[K_{s p} (A g C l) = 10^{- 10} M^{2}; K_{s p} (A g_{2} C r O_{4}) = 10^{- 12} M^{3}]$