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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:

A

0.1 M, 0.1 M

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B

6 \times 10^{- 3} M, 0.1 M

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C

1.64 \times 10^{- 8} M, 0.1 M

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D

2.03 \times 10^{- 12} M, 1.64 \times 10^{- 8} M

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Solution

The correct option is A $0.1 M, 0.1 M$
$S O_{4}^{2 -}$ requires $0.1 M$ $A g^{+}$ for precipitation.
Hence, $1^{s t}$ and $3^{r d}$ ions will have $0.1 M$ concentration when $2^{n d}$ ion $(S O_{4}^{2 -})$ starts precipitating.

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Similar questions

Q. A solution is 0.1 M in each of

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

{C l}^{-}

{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}

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

.
The sequence of ions precipitation 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 first ion when the second ion starts precipitating is about:

Q. A solution of

0.1 M

{C l}^{-}

0.1 M

{B r}^{-}

0.1 M

I^{-}

A g {N O}_{3}

is gradually added to this solution. Assuming that the addition of

A g {N O}_{3}

does not change the volume.
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 I) = 8.5 \times 10^{- 17}

What will be the conc. of both ions when the third ion start precipitating.

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}

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:

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