PS7 solutions

# PS7 solutions - PROBLEMS 1. 5. (a) (a) (b) (c) 9. (a) (b)...

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141 PROBLEMS 1. (a) 3+ 2 2 3 sp [Co ] [S ] K (c) 2+ 2 4 sp 2 7 [Zn ] [P O ] K (d) 5. (a) 14 28 6 2.5 10 [OH ] 1.67 10 1.5 10 −− × == × × ; 4 [OH ] 1.3 10 M (b) 36 2+ 3 28 42 7.6 10 [Cu ] 1.3 10 (2.4 10 ) x × × ; 2+ 10 [Cu ] 5.1 10 M (c) 8 2+ 6 3 2.7 10 [Zn ] 3.1 10 8.8 10 x xM × 9. (a) 7 sp 2 2+ 4 22 BaF 1.8 10 [Ba ] 2.9 10 [F ] (0.025) K M × = × (b) 7 25 1.8 10 [F ] = 4.0 10 0.0045 × ; 3 [F ] 6.3 10 M 3 2 6.3 10 % F left 100 25% 2.5 10 × = × ; % F precipitated = 75% 13. (a) 2+ 4 2 13.00 [Hg ] 0.0021 mol/L 7.2 10 38.0 M = × 25.0 [Cl ] 0.015 0.10 38.0 M M = 6 1 8 (7.2 10 ) (1.0 10 ) 7.2 10 1 10 Q ×× = × > × ; yes (b) 2+ 2 5 Hg 0.0021 mol/L 0.01300L 2.7 10 mol n = × 4 Cl 0.015 mol/L 0.025 L 3.75 10 mol n = × 2+ 2 Hg is limiting; 5 5.4 10 mol Cl used × ; 4 3.2 10 mol Cl left × 43 [Cl ] 3.2 10 mol/0.038 L 8.4 10 mol/L 2+ 18 3 2 14 2 [Hg ] (1 10 ) (8.4 10 ) 1 10 mol/L × 53 3 [NO ] 2(2.7 10 mol)/0.035 L=1.4 10 M × 17. 10 sp 1.8 10 K (a) 21 0 1.8 10 s ; 143.4 g mol 1.34 10 1.9 10 g/L L1 m o l s × = × (b) 10 97 143.4 g 1.8 10 mol 3.6 10 5.2 10 g/L 0.050 L 1 mol s × × × = × (c) 10 143.4 g 1.8 10 mol 1.1 10 1.6 10 g/L 0.17 L 1 mol s × × × = ×

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142 CHAPTER 16 23. (a) 3+ 65 [Al ] 0.40 0.14 190 M M = ; 2+ 125 [Fe ] 0.17 0.11 190 M M = to precipitate 3 Al(OH) :
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## This homework help was uploaded on 04/08/2008 for the course CHEM 128 taught by Professor All during the Spring '08 term at UConn.

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PS7 solutions - PROBLEMS 1. 5. (a) (a) (b) (c) 9. (a) (b)...

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