CameraIcon

SearchIcon

MyQuestionIcon

35

You visited us 35 times! Enjoying our articles? Unlock Full Access!

Common Ion Effect

The concentra...

Question

The concentration of a solution is $10^{- 4} M$ for $C l^{-}, 10^{- 5} M$ for $B r^{-}$ and $10^{- 3} M$ for $I^{-} . A g N O_{3} (s)$ is added slowly to the solution. If the minimum concentration of $A g^{+}$ required to start precipitation of all three ions is $10^{- x}$ , then the value of x is

[Given, $K_{s p (A g C l)} = 10^{- 10}, K_{s p (A g B r)} = 10^{- 13}, K_{s p (A g I)} = 10^{- 17}$ ]

A

8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

10

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

6

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is C 6
Since AgCl has highest solubility. So , AgCl will precipitate at last.
For precipitation of AgCl,
$[A g^{+}] [C l^{-}] = 10^{- 10}$
As the concentration of $C l^{-} = 10^{- 4} M$ ,

So, the minimum concentration of $A g^{+}$ required is $10^{- 6} M$ .

The value of x is 6.

Suggest Corrections

thumbs-up

0

Similar questions

Q. The concentration of a solution is

10^{- 4} M

C l^{-}, 10^{- 5} M

B r^{-}

10^{- 3} M

I^{-} .

A g N O_{3} (s)

is added slowly to the solution. The minimum concentration of

A g^{+}

required to start precipitation of all three ions is:
[Given,

K_{s p (A g C l)} = 10^{- 10}, K_{s p (A g B r)} = 10^{- 13}, K_{s p (A g I)} = 10^{- 17}

]

Q. A solution is

10^{- 4} M i n C l^{-}, 10^{- 5} M i n B r^{-}, 10^{- 3} M i n I^{-} . A g N O_{3} (s)

is added slowly to the solution. The minimum concentration of

A g^{+}

required to start precipitation of all three ions is

10^{- x}

then the value of x is:
[Given ,

K_{s p (A g C l)} = 10^{- 10}, K_{s p (A g B r)} = 10^{- 13}, K_{s p (A g I)} = 10^{- 17}]

Q. A solution of

0.1 M

{C l}^{-}

0.1 M

{B r}^{-}

0.1 M

I^{-}

A g {N O}_{3}

is gradually added to this solution. Assuming that the addition of

A g {N O}_{3}

does not change the volume.
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 I) = 8.5 \times 10^{- 17}

What will be the conc. of both ions when the third ion start precipitating.

A g N O_{3}

is slowly added to a solution that is

0.001

N a C l, N a B r

N a I

(A) The concentration of

A g^{+}

required to initiate precipitation of each of

A g I, A g B r

A g C l

1.5 \times 10^{- 13}

3.3 \times 10^{- 10}

1.6 \times 10^{- 7}

(B) The percentage of

I^{-}

precipitated before

A g B r

precipitates is

99.95

(C) The percentage of

I^{-}

precipitated before

C l^{-}

precipitates is

99.999917

(K_{s p} (A g I) = 1.5 \times 10^{- 16},

K_{s p} (A g B r) = 3.3 \times 10^{- 13},

K_{s p} (A g C l) = 1.8 \times 10^{- 10})

Which of the above-given option(s) is/are correct?

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

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Degree of Dissociation

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Common Ion Effect

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon