Which of the following happens when NH_{4}OH | vedclass

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(A) The condition for precipitation is $Q_{ip} > K_{sp}$. For $A(OH)_{2}$: $[A^{2+}][OH^{-}]^2 > 9 	imes 10^{-10}$ Given $[A^{2+}] = 1 \ M$,we have $

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

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(A) The condition for precipitation is $Q_{ip} > K_{sp}$. For $A(OH)_{2}$: $[A^{2+}][OH^{-}]^2 > 9 	imes 10^{-10}$ Given $[A^{2+}] = 1 \ M$,we have $[OH^{-}]^2 > 9 	imes 10^{-10}$,which implies $[OH^{-}] > 3 	imes 10^{-5} \ M$. For $B(OH)_{3}$: $[B^{3+}][OH^{-}]^3 > 27 	imes 10^{-18}$ Given $[B^{3+}] = 1 \ M$,we have $[OH^{-}]^3 > 27 	imes 10^{-18}$,which implies $[OH^{-}] > 3 	imes 10^{-6} \ M$. Since the concentration of $OH^{-}$ required for the precipitation of $B(OH)_{3}$ $(3 	imes 10^{-6} \ M)$ is lower than that required for $A(OH)_{2}$ $(3 	imes 10^{-5} \ M)$,$B(OH)_{3}$ will precipitate first.

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

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