173 quiz 6 - CHEM 173(07-08), QUIZ 6 Name: Section: 25...

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CHEM 173(07-08), QUIZ 6 Name: Section: 25 points 1. Answer the following appropriately ( filling in the blank, circling the correct answer etc. ) a. The equilibrium to be considered when calculating the solubility of Fe(OH) 2 (s) in Ba(OH) 2 is ans : Fe(OH) 2 * ) Fe 2+ (aq) + 2OH - (aq) b. The expression for the solubility product of PbF 2 (s) in terms of its solubility s in pure water is: K s = . ans : From the stoichiometry, if the solubility is s , [Fe 2+ ] eq = s , [OH - ] eq = 2 s , so K s = s × (2 s ) 2 = 4 s 3 . c. If K s = 4 . 1 × 10 - 8 the molar solubility of PbF 2 ( M = 245 . 2 g/mol) in pure water is and the solubility in g/L is ans : From the previous question, s 3 = K s / 4 . This gives the molar solubility as 2 . 17 × 10 - 3 mol/L. Use the molar mass to get the mass solubility: s × M = 0 . 533 g/L d. The solubility of PbF 2 in 0.200 M KF is ans : F - is a common ion: F - from KF is 0.200. Consider the solubility equilibrium of PbF 2 , with initial F - as 0.200; do not multiply this by 2 since it is not generated from the dissociation of PbF
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173 quiz 6 - CHEM 173(07-08), QUIZ 6 Name: Section: 25...

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