230_Sp10_L34_post

# 230_Sp10_L34_post - CHEM 230 Sp10 Lecture 34 Review for...

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CHEM 230 Sp10 Lecture 34 Review for Exam 3

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What is the solubility of Fe(OH) 3 ? K sp = 2.06x10 -39 Fe(OH) 3(s) Fe 3+ (aq) + 3OH 1- (aq) K sp =2.06x10 -39 =[Fe 3+ ][OH 1- ] 3 0 0 =s(3s) 3 +s +3s =27s 4 s 3s 9.35x10 -11 = s What is the pH of this solution? [OH - ]=3s=2.8x10 -10 pOH=-log[2.8x10 -10 ]=9.55…pH = 4.47 [OH - ] v. small; consider auto ionziation of water. Consider also Fe +3 [Fe(H 2 O) 6 ] 3+ + H 2 O [Fe(OH - )(H 2 O) 5 ] 2+ + H 3 O + K a = 3.5x10 -3 creates H 3 O + which affects pH Fe +3 + 7H 2 O [Fe(H 2 O) 5 (OH - )] 2+ + H 3 O + K a = 3.5x10 -3 Fe(OH) 3(s) Fe 3+ (aq) + 3OH 1- (aq) K sp =2.06x10 -39 __________________________________________________ 7H 2 O + Fe(OH) 3(s) [Fe (OH - )(H 2 O) 5 ] 2+ + 2H 2 O + 2OH 1- (aq) 5H 2 O + Fe(OH) 3(s) [Fe (OH - )(H 2 O) 5 ] 2+ + 2OH 1- (aq) K= 3.5x10 -3 * 2.06x10 -39 =7.21x10 -42
5H 2 O + Fe(OH) 3(s) [Fe (OH - ) (H 2 O) 5 ] 2+ + 2OH 1- (aq) K =7.21x10 -42 I 0 0 C +x +2x E x 2x K =7.21x10 -42 =[[Fe (OH - ) (H 2 O) 5 ] 2+ ][OH - ] 2 = x(2x) 2 = 4x 3 1.217x10 -14 = x = [[Fe (OH - ) (H 2 O) 5 ] 2+ [H 3 O + ][OH - ]=1x10 -14 [OH - ]=[H 3 O + ] + 1/2 [[Fe(H 2 O) 5 (OH - )] 2+ ] 1/2 is due to charge of 2+ Solve for [H 3 O + ] = 9.999999696x10 -8 pH = 7.000000013

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A 25.0 mL of unknown concentration of FeSO
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## This note was uploaded on 01/24/2012 for the course PCHEM 230 taught by Professor Gottfried during the Fall '11 term at University of Michigan.

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230_Sp10_L34_post - CHEM 230 Sp10 Lecture 34 Review for...

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