The solubility product constants of M(OH)_3 a | vedclass

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(B) For $M(OH)_3$,$K_{sp} = [M^{3+}][OH^-]^3 = 10^{-23}$.For $M(OH)_2$,$K_{sp} = [M^{2+}][OH^-]^2 = 10^{-14}$.Precipitation occurs when the ionic prod

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$M^{3+}$

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(B) For $M(OH)_3$,$K_{sp} = [M^{3+}][OH^-]^3 = 10^{-23}$.For $M(OH)_2$,$K_{sp} = [M^{2+}][OH^-]^2 = 10^{-14}$.Precipitation occurs when the ionic product exceeds the solubility product $(K_{sp})$.For $M^{3+}$,$[OH^-]^3 = 10^{-23} / [M^{3+}]$.For $M^{2+}$,$[OH^-]^2 = 10^{-14} / [M^{2+}]$.Since $M(OH)_3$ has a much lower $K_{sp}$ value,it requires a much lower concentration of $OH^-$ ions to exceed its solubility product compared to $M(OH)_2$.Therefore,$M^{3+}$ will precipitate first.

The solubility product constants of $M(OH)_3$ and $M(OH)_2$ are $10^{-23}$ and $10^{-14}$ respectively. If both ions are present in a solution,which one will precipitate first upon the addition of $NH_4OH$?

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