HW3_19

# HW3_19 - 3(PO 4 2 x moles Ca 3(PO 4 2 3 moles Ca 2[Ca 2 = X...

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Homework due 3/19 Calculate the molar solubility of Ca 3 (PO 4 ) 2 . (K sp Ca 3 (PO 4 ) 2 = 2.0X10 -29 ) The molar solubility is the concentration of the dissolve salt as if it was completely unionized. Since the salt is really 100 % ionized in solution, the molar solubility, which is assigned the variable x, can be used to determine the concentration of the ions present. Then this expression can be placed in the Ksp expression and the value of x calculated. Ca 3 (PO 4 ) 2 (s) “Ca 3 (PO 4 ) 2 (aq)” 3 Ca 2+ + 2 PO 4 3- K sp = [Ca 2+ ] 3 [PO 4 3- ] 2 let x = molar solubility of Ca
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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 = 2.0x10-29 x 5 = 1.85x10-31 x = molar solubility = 7.1x10-7...
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