Which of the following happens when $$NH_{4}OH$$ is added gradually to the solution containing 1 M $$A^{2+}$$ and $$1MB^{3+}$$ ions? Given : $$K_{sp}[A(OH)_{2}]= 9 \times 10^{-10}$$ and $$K_{sp}[B(OH)_{3}]= 27 \times 10^{-18}$$ at 298 K .

To determine which precipitate forms first, we need to find the minimum $$[OH^-]$$ concentration required to start precipitation of each salt.

For $$A(OH)_2$$, the solubility product is given by $$K_{sp} = [A^{2+}][OH^-]^2$$. Substituting $$K_{sp} = 9 \times 10^{-10}$$ and assuming $$[A^{2+}] = 1$$ yields $$ 9 \times 10^{-10} = (1)[OH^-]^2 $$. Solving for $$[OH^-]$$ gives $$[OH^-] = \sqrt{9 \times 10^{-10}} = 3 \times 10^{-5} \text{ M}$$.

For $$B(OH)_3$$, the solubility product expression is $$K_{sp} = [B^{3+}][OH^-]^3$$. With $$K_{sp} = 27 \times 10^{-18}$$ and $$[B^{3+}] = 1$$, we have $$ 27 \times 10^{-18} = (1)[OH^-]^3 $$. This leads to $$[OH^-] = (27 \times 10^{-18})^{1/3} = 3 \times 10^{-6} \text{ M}$$.

Since $$B(OH)_3$$ requires a lower $$[OH^-]$$ concentration ($$3 \times 10^{-6}$$ M) to precipitate than $$A(OH)_2$$ ($$3 \times 10^{-5}$$ M), $$B(OH)_3$$ will precipitate first as $$NH_4OH$$ is gradually added.

The correct answer is Option 3: $$B(OH)_3$$ will precipitate before $$A(OH)_2$$.