06_Equilibrium_Calculations-page8

06_Equilibrium_Calculations-page8 - MgNH4PO4 As bad as...

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Unformatted text preview: MgNH4PO4 As bad as things look, we still can take the same approach to solve this littel jewel. KSP Mg2+ + OHKSP Mg(OH)2 + NH4+ + OH- + PO43- KA3 HPO42- KB KA2 NH3 + H2 O H2PO4- The H3O+ have been omitted from phosphate equilibria for clarity. KA1 H3PO4 There’s obviously more to this than we first thought! KSP Mg(OH)2 =1.2x10-11 = [Mg2+][OH-]2 KB NH4+ = 1.78x10-5 [OH-][NH +] = [NH ] 4 3 For phosphate KA1 = 7.5x10-3 = [H3O+][H2PO4-] [H3PO4] KA2 = 6.0x10-8 = [H3O+][HPO42-] [H2PO4-] KA3 = 4.8x10-13 = [H3O+][PO43-] [HPO42-] For Mg2+ we can see if we’ve exceeded the KSP for Mg(OH)2. KSP =1.2 x 10-11 = [Mg2+][OH-]2 [Mg2+] = KSP/[OH-]2 = 4.7x10-4 We put 3.79x10-3 into the solution so some must have precipitated. The [Mg2+] is then based on the KSP for Mg(OH)2. One down, two to go! To determine the KSP, we’ll need to calculate the actual concentrations using other equilibrium expressions. These are all [H3O+] or [OH-] based. Fortunately, we know the pH. What do we already know? [H3O+] [OH-] = 3.98x10-11 = 2.51x10-4 We also have several mass balances 3.79x10-3 = [NH4+] + [NH3] = [H3PO4]+[H2PO4-]+[HPO42-]+[PO43-] This actually works out to be three problems in one. We’ll start with Mg2+. For ammonium, nothing will precipitate out of solution. However, some of it can be expected to be converted to ammonia. KB NH4+ = 1.78x10-5 = KB [NH4+] -] = [NH ] [OH 3 [OH-][NH4+] [NH3] = 0.11 ...
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