To 0.360 L of 0.140 M NH3 is added 0.150 L of 0.100 M MgOH2. How many grams of (NH4)2SO4 should be present to prevent precipitation of Mg(OH)2 (s)?


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  1. Just to clarify the question, I wonder if that 0.150L of 0.100 M MgOH2 was supposed to be 0.150 L of 0.100 M MgCl2?

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  2. @DrBob222 - No, it is suppose to be Mg(OH)2
    The Ksp to be used for this problem is 1.8*10^-11

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  3. Oh wait, nevermind. Sorry! It is suppose to be 0.150L of 0.100M MgCl2 :)

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