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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]
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
= 8.2x10-12 For phosphate, we need to determine which
species exist at significant levels at pH 10.2
[H3O+] = KA2
= [H PO -]
= [HPO 2-]
[PO 2-] 3-] = 1.88x108
= 0.012 It appears that most of our phosphate exists as HPO42- ...
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