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Solubility Product

A solution is...

Question

A solution is 0.1 M in each of ${P O}_{4}^{3 -}, {S O}_{4}^{- 3}$ and ${C l}^{-}$ ions. Solid ${A g N O}_{3}$ is added gradually to this solution which causes no change in volume of the solution. Given $K_{a p} (A g C l) = 1.8 \times 10^{- 10}$ , $K_{a p} ({A g}_{2} {S O}_{4}) = 1.21 \times 10^{- 3}$ and $K_{a p} ({A g}_{3} {S O}_{4}) = 2.7 \times 10^{- 18}$ .
The sequence of ions precipitation is:

{C l}^{-}, {P O}_{4}^{3 -}, {S O}_{4}^{3 -}

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{C l}^{-}, {S O}_{4}^{2 -}, {P O}_{4}^{3 -}

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{P O}_{4}^{3 -}, {C l}^{-}, {S O}_{4}^{2 -}

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{P O}_{4}^{3 -}, {S O}_{4}^{2 -}, {C l}^{-}

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Solution

The correct option is A ${C l}^{-}, {P O}_{4}^{3 -}, {S O}_{4}^{3 -}$
$[P O_{4}^{3 -}] = [C l^{-}] = [S O_{4}^{2 -}] = 10^{- 1} M$
For $A g C l$ , $K_{s p} = [A g^{+}] [C l^{-}]$
$\Rightarrow 1.8 \times 10^{- 10} = [A g^{+}] \times 10^{- 1}$
$∴ [A g^{+}] = 1.8 \times 10^{- 9} M . . . . (1)$
For $A g_{2} S O_{4}$ , $K_{s p} = [A g^{+}]^{2} [S O_{4}^{2 -}]$
$\Rightarrow [A g^{+}]^{2} = 1.21 \times 10^{- 2}$
$∴ [A g^{+}] = 0.11 M . . . . (2)$
For $A g_{3} P O_{4}$ , $K_{s p} = [A g^{+}]^{3} [P O_{4}^{3 -}]$
$\Rightarrow [A g^{+}]^{2} = 2.7 \times 10^{- 17}$
$∴ [A g^{+}] = 3 \times 10^{- 6} M . . . . (3)$
Here, $A g C l$ requires least concentration of $A g^{+}$ for precipitation. This is followed by $A g_{3} P O_{4}$ . $A g_{2} S O_{4}$ requires more $A g^{+}$ concentration for precipitation. Hence, the order of ions precipitation is:
(i) $C l^{-}$ , (ii) $P O_{4}^{3 -}$ , (iii) $S O_{4}^{2 -}$

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Similar questions

Q. A solution is 0.1 M in each of

{P O}_{4}^{3 -}, {S O}_{4}^{- 3}

and

{C l}^{-}

ions. Solid

{A g N O}_{3}

is added gradually to this solution which causes no change in volume of the solution. Given

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

K_{a p} ({A g}_{2} {S O}_{4}) = 1.21 \times 10^{- 3}

and

K_{a p} ({A g}_{3} {S O}_{4}) = 2.7 \times 10^{- 18}

.
The concentrations of first and third ions when the second ion starts precipitating, respectively, are:

Question 13
Write the molecular formulae of all the compounds that can be formed by the combination of following ions
$C u^{2 +}, N a^{+}, F e^{3 +}, C l^{-}, S O_{4}^{2}, P O_{4}^{3 -}$

Q. A solution contains

0.1 M

{C l}^{-}

and

0.001 M

C r O_{4}^{2 -}

. If solid

A g {N O}_{3}

is gradually added to this solution, which will precipitate first,

A g C l

{A g}_{2} C r O_{4}

? Assume that the addition causes no change in volume. Given

K_{s p} A g C l = 1.7 \times 10^{- 10} M^{2}

K_{s p} {A g}_{2} C r O_{4} = 1.79 \times 10^{- 12} M^{3}

What percent of

{C l}^{-}

remains in the solution when

C r O_{4}^{2 -}

starts precipitating?

Q. A solution contains

{C l}^{-}, I^{-}

and

S O_{4}^{2 -}

ions in it. Which of the following ion is capable to precipitate all of above when added in this solution?

Q. A solution is made of

0.1 M

C l^{-}

and

10^{- 4} M C r O_{4}^{2 -}

. If solid

A g N O_{3}

is gradually added to this solution, then 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}]

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A solution is 0.1 M in each of PO3−4,SO−34 and Cl− ions. Solid AgNO3 is added gradually to this solution which causes no change in volume of the solution. Given Kap(AgCl)=1.8×10−10, Kap(Ag2SO4)=1.21×10−3 and Kap(Ag3SO4)=2.7×10−18.The sequence of ions precipitation is: