Unformatted text preview: 3 (PO 4 ) 2 x moles Ca 3 (PO 4 ) 2 3 moles Ca 2+ [Ca 2+ ] =  X  = 3x M 1 liter sol’n 1 mole Ca 3 (PO 4 ) 2 x moles Ca 3 (PO 4 ) 2 2 moles PO 4 3[PO 4 3] =  X  = 2x M 1 Liter sol’n 1 mole Ca 3 (PO 4 ) 2 K sp = [Ca 2+ ] 3 [PO 4 3] 2 = (3x) 3 (2x) 2 = (27x 3 )(4x 2 ) = 108x 5 = 2.0x1029 x 5 = 1.85x1031 x = molar solubility = 7.1x107...
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This note was uploaded on 04/03/2008 for the course CHEM 114 taught by Professor Martin during the Spring '08 term at Moravian.
 Spring '08
 Martin
 Solubility

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