MyQuestionIcon

1

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

Basic Buffer Action

A solution is...

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. 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:

A

5 \times 10^{- 12} M

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

B

1.7 \times 10^{- 9} M

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

C

6 \times 10^{- 11} M

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

D

4 \times 10^{- 8} M

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

Open in App

Solution

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

Suggest Corrections

thumbs-up

0

Similar questions

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 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:

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 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

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}

A g B r

5.0 \times 10^{- 13}

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

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Buffer Solutions

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Basic Buffer Action

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon