A solution has 0-1 M in each KCl - KBr and K 2 CrO 4 and to this solution solid AgNO 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 - Ag -required to start the precipitation of AgBr is-A- 5 - 10-12 MB- 1-7 - 10-9 M- 6 - 10-11 MD- 4 - 10-8 M

The correct option is A 5× 10^ 12 M[Ag^+] to start the ppt. Of AgBr =Ksp(AgBr)/[Br^ ]=5× 10^ 13/0.1=5× 10^ 12 M ...

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Standard XII

Chemistry

Concentration Terms

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 $[A g^{+}]$ required to start the precipitation of AgBr is:

5 \times 10^{- 12} M

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1.7 \times 10^{- 9} M

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6 \times 10^{- 11} M

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4 \times 10^{- 8} M

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Solution

The correct option is A $5 \times 10^{- 12} M$
$[A g^{+}]$ to start the ppt. Of AgBr
$= \frac{K s p (A g B r)}{[B r^{-}]} = \frac{5 \times 10^{- 13}}{0.1} = 5 \times 10^{- 12} M$

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Similar questions

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 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

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:

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. 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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The correct option is A 5×10−12 M[Ag+] to start the ppt. Of AgBr =Ksp(AgBr)[Br−]=5×10−130.1=5×10−12 M

The correct option is A $5 \times 10^{- 12} M$
$[A g^{+}]$ to start the ppt. Of AgBr
$= \frac{K s p (A g B r)}{[B r^{-}]} = \frac{5 \times 10^{- 13}}{0.1} = 5 \times 10^{- 12} M$