3.HW2.chemistry.solution - CEE 330, Spring 2011 Solution...

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1 CEE 330, Spring 2011 Solution for HW#2 Due Feb. 3, 2011 Solution 1 As the description in the problem, one EDTA molecule complexes one metal atom, 2 [] EDTA Ca Ca EDTA + +→ Consuming one Ca 2+ molecule would need the equal molecule of EDTA. The Calcium concentration: 23 11 [ ] 20 / 0.5 10 / 1000 40 / g Ca mg L mol L mg g mol +− × = × Total Ca 2+ in 44-gallon drums: 2 3.785 [ ] 0.5 10 / 44 0.08327 1 EDTA Ca L M M Ca Volume mol L gallons mol gallon + ==⋅ = × × × = Mass equals to molecules multiplied by molecules weight: 0.08327 (12 10 14 2 16 8 1 16) / 24.3 EDTA EDTA m M EDTA mol g mol g =⋅= × × + × + × + × = Solution 2 The carbonate system in the ocean is comprised of the following chemical species: Aqueous carbon dioxide CO 2 (aq) Carbonic acid H 2 CO 3 Bicarbonate ion HCO - 3 Carbonate ion CO 3 2- Calcium cation Ca 2+ Aqueous CO 2 is formed when atmospheric CO 2 dissolves in water; we can find its concentration in fresh water using Henry’s law: 65 2 [ ( )] 0.033363 / 360 10 1.2 10 / Hg CO aq K P mol L atm atm mol L −− == × × = × Aqueous CO 2 then forms carbonic acid (H 2 CO 3 ). This aqueous CO 2 reacts quickly with water to form carbonic acid, a diprotic acid, which can lose two protons (hydrogen ion, H + ). The first proton is released when it ionizes to form bicarbonate (HCO 3 - ): 22 3 3 7 1 2 () 4.47 10 / CO aq H O H HCO HH C O K mol L CO aq ++ ⎡⎤ ⎣⎦ × R The bicarbonate (HCO 3 -
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3.HW2.chemistry.solution - CEE 330, Spring 2011 Solution...

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