06_Equilibrium_Calculations-page8

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

Ask a homework question - tutors are online