ees3040HW2solns

Ees3040HW2solns - EES 3040 Homework 2 Solutions EES 3040 Homework 2 Solutions Environmental Chemistry Reading Assignment D&M1 Chapter 2 pp 32-39

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EES 3040 Homework 2 Solutions 1 EES 3040 Homework 2 Solutions Environmental Chemistry Reading Assignment: D&M1: Chapter 2, pp. 32-39, 44-46, 49-54, 71-73 D&M2: Chapter 2, pp. 37-41, 43-44, 48-55, 73-75 Problems: 1. a) (1.5 pts.) Chapter 2, Problem 2-23 (d) Given: 0,02000 M of SO 4 2- (note M = mole/L) MW of SO 4 2- = 32 .07 + 4(16.00) = 96.07 g/mole (0.02000 mole/L)(96.07 g/mole)(1,000 mg/g) = 1,921 mg /L b) (1.5 pts.) Chapter 2, Problem 2-24 (d) Given: 0.0080 M of CO 2 MW of CO 2 = 12.01 + 2(16.00) = 44.01 g/mole (0.0080 M)(44.01 g/mole)(10 6 μg/g) = 3.52 x 10 5 μg/L 2. (5 pts.) Chapter 2, Problem 2-7 Given: 10.00 g Mg(OH) 2 , 1 L water, pH 1 = 7, pH 2 = 10.5, assuming temperature is 25º C and ionic strength is negligible From Table 2-1 the reaction is Mg(OH) 2 Mg 2+ + 2OH - (1 pt) Convert pK s to K s : [Mg 2+ ][OH - ] 2 = K s = 10 -11.25 Rearrange equation: [Mg 2+ ] = K s / [OH - ] 2 Since pH = 7.0, [H + ] = 10 -pH [H + ] = 10 -7 M (1 pt) Since [H + ][OH - ] = 10 -14 , [OH - ] = 10 -7 mol/L (1 pt) Substituting into equation: [Mg 2+ ] = K s / [OH - ] 2 = 10 -11.25 / (10 -7 ) 2 = 562.34 mol/L (2 pts)
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EES 3040 Homework 2 Solutions 2 3. (6 pts.) Chapter 2, Problem Problem 2-26. Hint : See text Table 2-1.
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This note was uploaded on 10/29/2011 for the course EES 3040 taught by Professor Chan-hilton during the Fall '10 term at FSU.

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Ees3040HW2solns - EES 3040 Homework 2 Solutions EES 3040 Homework 2 Solutions Environmental Chemistry Reading Assignment D&M1 Chapter 2 pp 32-39

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