06_Equilibrium_Calculations-page9

# 06_Equilibrium_Calculations-page9 - Our mass balance for...

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Unformatted text preview: Our mass balance for ammonia species is 3.79x10-3 +] = [NH4 + [NH3] [NH +] Substituting for ammonia using: 4 = [NH3] 0.11 Gives us [NH4+] + +] [NH4 0.11 = 3.79x10-3 [NH4+] = 3.8x10-4 M Two down, one to go! We can now solve for phosphate. [HPO42-] = 3.79 x 10-3 M [PO43-] = 0.012 [HPO42-] = 4.6x 10-5M KSP MgNH4PO4 = [Mg2+][NH4][PO4] = (4.7x10-4)(3.8x10-4)(4.6x10-5) = 8.2x10-12 For phosphate, we need to determine which species exist at significant levels at pH 10.2 [H2PO4-] [H3PO4] KA1 [H3O+] = KA2 [H3O+] 4 = [H PO -] 2 4 KA3 [H3O+] 4 = [HPO 2-] 4 [HPO [PO 2-] 3-] = 1.88x108 = 1.51x103 = 0.012 It appears that most of our phosphate exists as HPO42- ...
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## This note was uploaded on 01/07/2012 for the course CHEM 290 taught by Professor Harvey during the Fall '08 term at SUNY Stony Brook.

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