A solution has 0-1 M in each KCl - KBr and K 2 CrO 4 and to this solution solid AgN O 3 is gradually added- Assume there is no change in the volume of the solution given-Ksp AgCl -1-7 - 10-10Ksp AgBr -5-0 - 10-13 and Ksp K 2 CrO 4-1-9 - 10-12The concentration of first ion when the second ion starts precipitating is about-A- 0-03 MB- 0-003 MC- 0-0003 MD- 0-1 M

The correct option is C 0.0003M[Ag^+] to start the ppt. Of Cl^ =Ksp(AgCl)/[Cl^ ]=1.7× 10^ 10/0.1=1.7× 10^ 9 M When Cl^ starts precipitating [Br^ ] =Ksp(AgBr) ...

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Byju's Answer

Standard XII

Chemistry

Preparation of Haloalkanes from Alcohol

A solution ha...

Question

A solution has 0.1 M in each KCl, KBr and $K_{2} C r O_{4}$ and to this solution solid $A g N O_{3}$ is gradually added. Assume there is no change in the volume of the solution given:
$K s p (A g C l) = 1.7 \times 10^{- 10}$
$K s p (A g B r) = 5.0 \times 10^{- 13} a n d K s p (K_{2} C r O_{4}) = 1.9 \times 10^{- 12}$
The concentration of first ion when the second ion starts precipitating is about:

0.1M

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0.03M

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0.003M

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0.0003M

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Solution

The correct option is D 0.0003M
$[A g^{+}]$ to start the ppt. Of $C l^{-}$
$= \frac{K s p (A g C l)}{[C l^{-}]} = \frac{1.7 \times 10^{- 10}}{0.1} = 1.7 \times 10^{- 9} M$
When $C l^{-}$ starts precipitating
$[B r^{-}] = \frac{K s p (A g B r)}{[A g^{+}]} = \frac{5 \times 10^{- 13}}{1.7 \times 10^{- 9}} = 3 \times 10^{- 4} M$

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Similar questions

Q. A solution is 0.1 M in each

K C l

K B r

and

K ­_{2} C r O_{4}

to this solution, solid

A g N O_{3}

is gradually added. Assuming that there is no change in the volume of the solution given, The concentration (in M) of first ion when the second ion starts precipitating is about .

Given:
Ksp of

A g C l

1.7 \times 10^{- 10}

Ksp of

A g B r

5.0 \times 10^{- 13}

Ksp

(K_{2} C r O_{4}) = 1.9 \times 10^{- 12}

Q. A solution has 0.1 M in each KCl, KBr and

K_{2} C r O_{4}

and to this solution solid

A g N O_{3}

is gradually added. Assume there is no change in the volume of the solution given:

K s p (A g C l) = 1.7 \times 10^{- 10}

K s p (A g B r) = 5.0 \times 10^{- 13} a n d K s p (K_{2} C r O_{4}) = 1.9 \times 10^{- 12}

The concentration of

[A g^{+}]

required to start the precipitation of AgBr is:

Q. A solution is 0.1 M in each KCl, KBr and

K ­_{2} C r O_{4}

to this solution solid

A g N O_{3}

is gradually added. Assume there is no change in the volume of the solution given:
Ksp (AgCl) =

1.7 \times 10^{- 10}

Ksp (AgBr) =

5.0 \times 10^{- 13}

and Ksp

(K_{2} C r O_{4}) = 1.9 \times 10^{- 12}

The concentration of

[A g^{+}]

required to start the precipitation of AgBr is:

Q. A solution is

0.1 M

in each

K C l, K B r

and

K_{2} C r O_{4}

. To this solution solid

A g N O_{3}

is gradually added. Assume no change in the volume of the solution.
Given:

K_{s p} (A g C l) = 1.7 \times 10^{- 10}

K_{s p} (A g B r) = 5 \times 10^{- 13}

and

K_{s p} (A g_{2} C r O_{4}) = 1.9 \times 10^{- 12}

The concentration of first ion when the second ion starts precipitating is:

Q. A solution is

0.1 M

in each

K C l, K B r

and

K_{2} C r O_{4}

. To this solution solid

A g N O_{3}

is gradually added. Assume no change in the volume of the solution.
Given:

K_{s p} (A g C l) = 1.7 \times 10^{- 10}

K_{s p} (A g B r) = 5 \times 10^{- 13}

and

K_{s p} (A g_{2} C r O_{4}) = 1.9 \times 10^{- 12}

The concentration of [

A g^{+}

] required to start the precipitation of

A g B r

is:

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The correct option is D 0.0003M[Ag+] to start the ppt. Of Cl− =Ksp(AgCl)[Cl−]=1.7×10−100.1=1.7×10−9 M When Cl− starts precipitating [Br−]=Ksp(AgBr)[Ag+]=5×10−131.7×10−9=3×10−4 M