At 25°C, the solubility product of Mg(OH)2 is 1.0 x 10–11. At what pH, will Mg2+ ions start precipitating in the form of Mg(OH)2 from a solution of 0.001 M Mg2+ ions ?

KSP=1.0×10−11=Mg+2OH−21.0×10−11=(0.001)OH−2OH−=10−4pOH=4pH=14−4=10.

At 25°C, the solubility product of Mg(OH)2 is 1.0 x 10–11. At what pH, will Mg2+ ions start precipitating in the form of Mg(OH)2 from a solution of 0.001 M Mg2+ ions ?

KSP=1.0×10−11=Mg+2OH−21.0×10−11=(0.001)OH−2OH−=10−4pOH=4pH=14−4=10.

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At 25°C, the solubility product of Mg(OH)₂ is 1.0 x 10^–11. At what pH, will Mg²⁺ ions start precipitating in the form of Mg(OH)₂from a solution of 0.001 M Mg²⁺ ions ?

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$\begin{array}{l} K_{SP} = 1 .0 \times 10^{- 11} = [{Mg}^{+ 2}] {[{OH}^{-}]}^{2} \\ 1 .0 \times 10^{- 11} = (0 .001) {[{OH}^{-}]}^{2} \\ [{OH}^{-}] = 10^{- 4} \\ pOH = 4 \\ pH = 14 - 4 = 10 . \end{array}$

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At 25°C, the solubility product of Mg(OH)2 is 1.0 x 10–11. At what pH, will Mg2+ ions start precipitating in the form of Mg(OH)2 from a solution of 0.001 M Mg2+ ions ?

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At 25°C, the solubility product of Mg(OH)₂ is 1.0 x 10^–11. At what pH, will Mg²⁺ ions start precipitating in the form of Mg(OH)₂from a solution of 0.001 M Mg²⁺ ions ?