A solution that is mixed to be 0.010M MgCl2 andalso 0.10 NH3. Ksp (Mg(OH)2) = 18

Question

A solution that is mixed to be 0.010M MgCl2 andalso 0.10 NH3. Ksp (Mg(OH)2) = 18. x 10^-11 , Kb (NH3) = 1.8 x 10^-5 i, The concentration of OH- available from solution is 1.3 x 10^-3M How do you know whether it is talking about the concentration of the OH in MgOH2 or NH3, cuz they give different answers and apparently they are using NH3 not mgoh2

Explanation / Answer

The question is whether Mg(OH)2 will precipitate or not. We have two things to start with, 0.010M MgCl2 which dissociates completely to form MG2+ and Cl- ions and 0.10 NH3 which dissociates partially into NH3+ and OH- ions So, as you can see it is the NH3 that is producing the OH- ions. (as u have calculated it gives [OH-] = 1.3 x 10^-3 M) Now, we have to find if the concentration of OH- (produced by NH3) is high enough to cause the precipitation of Mg(OH)2. This is done by comparing The ion product Qsp with the solubility product Ksp Qsp = [mg2+][OH-] = 1.3 x 10^-5 > Ksp => precipitation occurs Now, about your doubt that why we havent used Mg(OH)2 for [OH-] As it is NH3 that is producing OH- , initial concentration of OH- will be determined by NH3. It is this concenteration that causes the formation of Mg(OH)2 not Mg(OH)2 that makes OH-

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