CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Common Ion Effect

A mixture of ...

Question

A mixture of $H_{2} O$ and solid $A g C l$ shaken to obtain saturated solution. The solid is filtered and to $100 m L$ filtrate, $100 m L$ of $0.03 M$ $N a B r$ is added, will a precipitate formed?
$K_{s p} A g C l = 10^{- 10}, K_{s p} A g B r = 5 \times 10^{- 13}$

Open in App

Solution

Let solubility of $A g C l$ is ' $S$ '
$∴$ $K_{s p} A g C l = S^{2}$
$S = \sqrt{K_{s p}} = 10^{- 5}$
When $100 m L$ of both $A g C l$ and $N a B r$ are mixed concentration of ${A g}^{+}$ and ${B r}^{-}$ will be:
$[{A g}^{+}] = \frac{100 \times 10^{- 5}}{200} = 0.5 \times 10^{- 5} M$
$[{B r}^{-}] = \frac{100 \times 0.03}{200} = 0.015 M$
Ionic product of $A g B r = 0.5 \times 10^{- 5} \times 0.015 = 7.5 \times 10^{- 8}$
Since, ionic product of $A g B r$ is greater than its solublity product, hence $A g B r$ will be precipitated.

Suggest Corrections

thumbs-up

0

Similar questions

Q. 100 mL of a clear saturated solution of

A g_{2} S O_{4}

is added to 250 mL of a clear saturated solution of

P b C r O_{4}

. What is the precipitate formed? Given,

K_{s p}

A g_{2} S O_{4}, A g_{2} C r O_{4}, P b C r O_{4}

P b S O_{4}

1.4 \times 10^{- 5}, 2.4 \times 10^{- 12}, 2.8 \times 10^{- 13}

1.6 \times 10^{- 5}

100 m L

of a saturated solution of

A g_{2} S O_{4}

250 m L

of a saturated solution of

P b C r O_{4}

. Will any precipitate form and if so what?

K_{s p}

A g_{2} S O_{4}, A g_{2} C r O_{4}, P b C r O_{4}

P b S O_{4}

1.4 \times 10^{- 5},

2.4 \times 10^{- 12},

2.8 \times 10^{- 13}

1.6 \times 10^{- 8}

Q. A solution consists of a mixture of

0.01 M K I

0.1 M K C l

A g N O_{3}

is added to the solution, what is the concentration of

I^{-}

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 a mixture of

0.01

K I

0.1

K C l

A g N O_{3}

is added to the solution, what is the

[I^{-}]

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}]

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?

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