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# hw11sol - H omework#11 solution outlines SOL-4(a Ksp...

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3.091 Fall Term 2004 Homework #11 solution outlines SOL-4. (a) K sp = [Ba 2+ ] [CrO 4 2– ] (b) K sp = [Ca 2+ ] [CO 3 2– ] (c) K sp = [Pb 2+ ] [F ] 2 (d) K sp = [Ag + ] 2 [S 2– ] SOL-8. If [Hg 2 2+ ] = C s then [Cl ] = 2 C s K sp = [Hg 2 2+ ] [Cl ] 2 = [C s ] [2 C s ] 2 = 4 C s 3 Answer (c) SOL-10. Let C s = [Pb 2+ ] for PbSO 4 1.6 × 10 –8 = C s 2 C s = 1.3 × 10 –4 mol L for PbI 2 7.1 × 10 –9 = 4 C s 3 C s = 1.2 × 10 –3 mol L Lead (II) iodide is more soluble. SOL-16. the lithium carbonate dissolves according to Li 2 CO 3 = 2 Li + + CO 3 2– which by stoichiometry gives [Li+] = 2 [CO 3 2– ] = 2 c s , where cs means the concentration of solute or solubility of lithium carbonate for this reaction K sp = [Li + ] 2 [CO 3 2– ] = 4 c s 3 2 3 2 3 s 2 3 1.36 g Li CO 1 mol Li CO 1000 0.184 M = c 100 mL water 73.89 g Li CO 1 L mL × = = K sp = 2.49 × 10 –2 NOTE: By definition, molarity means moles of solute per liter of solution. In this instance we have taken 100 mL water to be 100 mL of solution. Stricly speaking, this is not correct, but the difference between the two is small enough to allow the approximation. SOL-26. (b). The presence of fluoride ion from NaF represses the solubility of MgF 2 . This is an example of the common ion effect in action.

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