CameraIcon

SearchIcon

MyQuestionIcon

1

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

Common Ion Effect

0.01 mole of ...

Question

$0.01$ mole of $A g {N O}_{3}$ is added to one litre of a solution which is $0.1 M$ in ${N a}_{2} C r O_{4}$ and $0.005 M$ in $N a I O_{3}$ . The concentration of precipitate formed at equilibrium and the concentrations of ${A g}^{+}, {I O}_{3}^{-}$ and $C r O_{4}^{2 -}$ is ( $K_{s p}$ values of ${A g}_{2} C r O_{4}$ and $A g I O_{3}$ are $10^{- 8}$ and $10^{- 13}$ respectively) :

A

2.6 \times 10^{- 10} M

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

B

2.8 \times 10^{- 10} M

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

C

3.1 \times 10^{- 10} M

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

D

none of these

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

Open in App

Solution

The correct option is C $3.1 \times 10^{- 10} M$
The $K_{s p}$ values of ${A g}_{2} {C r O}_{4}$ and $A g I O_{3}$ reveals that $C r O_{4}^{2 -}$ and $I O_{3}^{-}$
will be precipitated on addition of $A g {N O}_{3}$ as:
$[{A g}^{+}] [I O_{3}^{-}] = 10^{- 13}$
${[{A g}^{+}]}_{n e e d e d} = \frac{10^{- 13}}{(0.005)} = 2 \times 10^{- 11} M$
${[{A g}^{+}]}^{2} [C r O_{4}^{2 -}] = 10^{- 8}$
$∴$ ${[{A g}^{+}]}_{n e e d e d} = \sqrt{\frac{10^{- 8}}{0.1}} = 3.16 \times 10^{- 4} M$
Thus $A g {I O}_{3}$ will be precipiated first.
Now in order to precipitate $A g {I O}_{3}$ , one can slow.
$A g {N O}_{3} 0.01 0.005 + N a I O_{3} 0.005 0 ⟶ A g I O_{3} 0 0.005 + N a {N O}_{3} 0 0.005$
The left mole of $A g {N O}_{3}$ are now used to precipitate ${A g}_{2} {C r O}_{4}$
$2 A g {N O}_{3} 0.005 0 + {N a}_{2} {C r O}_{4} 0.1 0.0975 ⟶ A g I O_{3} 0 0.0025 + 2 N a {N O}_{3} 0 0.005$
Thus $[C r O_{4}^{2 -}]$ left in solution $= 0.0975$
Now solution has precipitates of $A g {I O}_{3} (s)$ $(0.005 m o l e) + {A g}_{2} {C r O}_{4} (s)$ ( $0.0025 m o l e$ ) and $C r O_{4}^{2 -}$ ions ( $0.0975$ )
Thus, ${[{A g}^{+}]}_{l e f t} = \sqrt{\frac{K_{{s p}_{{A g}_{2} {C r O}_{4}}}}{[C r O_{4}^{2 -}]}} = \sqrt{\frac{10^{- 8}}{0.0975}} = 3.2 \times 10^{- 4} M$
${[I O_{3}^{-}]}_{l e f t} = \frac{K_{{s p}_{A g I O_{3}}}}{[{A g}^{+}]} = \frac{10^{- 13}}{3.2 \times 10^{- 4}} = 3.1 \times 10^{- 10} M$

Suggest Corrections

thumbs-up

0

Similar questions

0.01

A g N O_{3}

1

L of a solution which is

0.1 M

N a_{2} C r O_{4}

0.005 M

N a I O_{3}

[A g^{\oplus}], [I O_{3}^{⊝}]

[C r O_{4}^{2 -}]

K_{s p} A g_{2} C r O_{4}

A g I O_{3}

10^{- 8}

10^{- 13}

A g N O_{3}

is added to a solution which is

0.1 M

C l^{-}

0.1 M

C r O_{4}^{2 -}

K_{s p}

A g C l

A g_{2} C r O_{4}

1.7 \times 10^{- 10}

1.9 \times 10^{- 12}

respectively. Find the minimum concentration of

C l^{-}

A g_{2} C r O_{4}

starts precipitating.

A g N O_{3}

is added to a solution which is

0.1 M

C l^{-}

0.1 M

C r O_{4}^{2 -}

K_{s p}

A g C l

A g_{2} C r O_{4}

1.7 \times 10^{- 10}

1.9 \times 10^{- 12}

respectively. Find the minimum concentration of

C l^{-}

A g_{2} C r O_{4}

starts precipitating.

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

K_{s p}

A g_{2} C r O_{4}, A g C l, A g B r

A g I

are respectively,

1.1 \times 10^{- 12}

1.8 \times 10^{- 10}

5.0 \times 10^{- 13}

8.3 \times 10^{- 17}

. Which of the following salts will precipitate last if

A g N O_{3}

solution is added to the solution containing equal moles of

N a C l, N a B r, N a I

N a_{2} C r O_{4}

?

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