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Preparation of Haloalkanes from Alcohol

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Question

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

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Solution

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 there is no change in the volume of the solution given,
Given:
Ksp of $A g C l$ = $1.7 \times 10^{- 10}$
Ksp of $A g B r$ = $5.0 \times 10^{- 13}$
Ksp $(K_{2} C r O_{4}) = 1.9 \times 10^{- 12}$

$[A g]^{+}$ 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
$[B r^{-}] = \frac{K s p (A g B r)}{[A g^{+}]} = \frac{5 \times 10^{- 18}}{1.7 \times 10^{- 9}} = 3 \times 10^{- 4} M$

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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 first ion when the second ion starts precipitating is about:

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}

5.0 \times 10^{- 13}

(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

K C l, K B r

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}

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 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

K C l, K B r

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}

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

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