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The `K_(sp)` of `Mg(OH)_(2)` is `1 xx 10^(-12), 0.01 M Mg(OH)_(2)` will precipitate at the limiting pHA. 3B. 9C. 5D. 8 |
Answer» Correct Answer - B `Mg(OH)_(2) Mg^(2+) + 2OH^(-)` `K_(sp) = [Mg^(2+)] [OH^(-)]^(2)` `1 xx 10^(-12) = 0.01 [OH^(-)]^(2)` `[OH^(-)]^(2) = 1 xx 10^(-10) rArr [OH^(-)] = 10^(-5)` `[H^(+) = 10^(-14)//10^(-5) = 10^(-9)` `pH = -log[H^(+)] = -log[10^(-9)] = 9`. |
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