Which of the following happens when is added gradually to the solution containing 1M and 1M ions? [2025]

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Solution

Q.
Which of the following happens when $N H_{4} O H$ is added gradually to the solution containing 1M $A^{2 +}$ and 1M $B^{3 +}$ ions

$Given: K_{s p} [A(OH)_{2}] = 9 \times 10^{- 10} and K_{s p} [B(OH)_{3}] = 27 \times 10^{- 18} at 298 K.$ [2025]

1 Both $A {(OH)}_{2}$ and $B {(OH)}_{3}$ do not show precipitation with ${NH}_{4} OH$ .

2 $B {(OH)}_{3}$ will precipitate before $A {(OH)}_{2}$ .

3 $A {(OH)}_{2}$ will precipitate before $B {(OH)}_{3}$ .

4 $A {(OH)}_{2}$ and $B {(OH)}_{3}$ will precipitate together.

Ans.
(2)

$B {(O H)}_{3} ⇌ B^{3 +} + 3 O H^{-}$

$K_{i p} (B {(O H)}_{3}) = [B^{3 +}] {[O H^{-}]}^{3}$

For precipitation of $B {(O H)}_{3}$ to occur,

$K_{i p} (B {(O H)}_{_{3}}) > K_{s p} (B {(O H)}_{3})$

$[B^{3 +}] {[O H^{-}]}^{3} > 27 \times 10^{- 18}$

$1 \times {[O H^{-}]}^{3} > 27 \times 10^{- 18}$

$[O H^{-}] > 3 \times 10^{- 6} M$

$A {(O H)}_{2} ⇌ A^{2 +} + 2 O H^{-}$

$K_{i p} (A {(O H)}_{2}) = [A^{2 +}] {[O H^{-}]}^{2}$

For precipitation of $A {(O H)}_{2}$ to occur,

$K_{i p} (A {(O H)}_{2}) > K_{s p} (A {(O H)}_{2})$

$[A^{2 +}] {[O H^{-}]}^{2} > 9 \times 10^{- 10}$

$1 \times {[O H^{-}]}^{2} > 9 \times 10^{- 10}$

$[O H^{-}] > 3 \times 10^{- 5} M$

So precipitation of $B {(O H)}_{3}$ occurs before that of $A {(O H)}_{2}$ .

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