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Common Ion Effect

0.01 mol of ...

Question

$0.01$ mol of $A g N O_{3}$ is added to $1$ L of a solution which is $0.1 M$ $N a_{2} C r O_{4}$ and $0.005 M$ in $N a I O_{3}$ . Calculate $[A g^{\oplus}], [I O_{3}^{⊝}]$ and $[C r O_{4}^{2 -}]$ .
$K_{s p} A g_{2} C r O_{4}$ and $A g I O_{3}$ are $10^{- 8}$ and $10^{- 13}$ respectively.

A

[A g^{\oplus}] = 0.316 \times 10^{- 4} M, [I O_{3}^{⊝}] = 0.31 \times 10^{- 10} M, [C r O_{2}^{2 -}] = 0.0975 M

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B

[A g^{\oplus}] = 3.16 \times 10^{- 4} M, [I O_{3}^{⊝}] = 3.1 \times 10^{- 10} M, [C r O_{2}^{2 -}] = 0.975 M

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C

[A g^{\oplus}] = 31.6 \times 10^{- 4} M, [I O_{3}^{⊝}] = 31 \times 10^{- 10} M, [C r O_{2}^{2 -}] = 9.75 M

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D

None of these

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Solution

The correct option is A $[A g^{\oplus}] = 0.316 \times 10^{- 4} M, [I O_{3}^{⊝}] = 0.31 \times 10^{- 10} M, [C r O_{2}^{2 -}] = 0.0975 M$
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:
$[A g^{\oplus}] [I O_{3}^{⊝}] = 10^{- 13}$
$[A g^{\oplus}]_{n e e d e d} = \frac{10^{- 13}}{[0.005]} = 2 \times 10^{- 11}$
$[A g^{\oplus}]^{2} [C r C_{4}^{2 -}] = 10^{- 8}$
$[A g^{\oplus}]_{n e e d e d} = \sqrt{\frac{10^{- 8}}{0.1}} = 3.16 \times 10^{- 4}$
Thus, $A g I O_{3}$ will be precipitated first.
Now, in order to precipitate $A g I O_{3}$ , we get,

$A g N O_{3} + N a I O_{3} ⟶ A g I O_{3} + N a N O_{3}$
0.01 0.005 0 0
0.005 0 0.005 0.005
The left mole of $A g N O_{3}$ are now used to precipitate $A g_{2} C r O_{4}$
$2 A g N O_{3} + N a_{2} C r O_{4} ⟶ A g_{2} C r O_{4} + 2 N a N O_{3}$
0.005 0.1 0 0
0 0.0975 0.0025 0.005
Thus, $[C r O_{4}^{2 -}]$ left in solution = 0.0975
Now, solution has $A h I O_{3} (s) + A g_{2} C r O_{4} (s) + C r O_{4}^{2 -}$ ions
0.005 0.0025 0.0975
$[A g^{\oplus}]_{l e f t} = \frac{K_{s p} A g_{2} C r O_{4}}{[C r C_{4}^{2}]} = \sqrt{\frac{10^{- 8}}{0.0975}} = 3.2 \times 10^{- 4} M$
$[I O_{3}^{⊝}]_{l e f t} = \frac{K_{s p} A g I O_{3}}{[A g^{\oplus}]} = \frac{10^{- 13}}{3.2 \times 10^{- 4}} = 3.1 \times 10^{- 4} M$

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Similar questions

0.01

A g {N O}_{3}

is added to one litre of a solution which is

0.1 M

{N a}_{2} C r O_{4}

0.005 M

N a I O_{3}

. The concentration of precipitate formed at equilibrium and the concentrations of

{A g}^{+}, {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}

respectively) :

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

Q. In a saturated solution of the sparingly soluble strong electrolyte

A g I O_{3}

(Molecular mass = 283) the equilibrium which sets in is

A g I O_{3} (s) ⇌ A g^{+} (a q) + I O_{3}^{-} (a q)

If the solubility product constant

K_{s p}

A g I O_{3}

at a given temperature is

1.0 \times 10^{- 8}

, what is the mass of

A g I O_{3}

contained in 100 ml of its saturated solution ?

K_{s p}

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

A g l

are respectively,

1.1 \times 10^{- 12}, 1.8 \times 10^{- 10}, 5.0 \times 10^{- 13},

. Which of these will precepitate last if

A g N O_{3}

solution is added to a solution containing equal moles of NaCl, NaBr, NaI and

N a_{2} C r O_{4}

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}

]

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