Chptr2PrblmSolns - a. [Ca 2+ ] = C tot = [H 2 CO 3 ] + [HCO...

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2.1 – 2.14 The solutions for these problems are the solutions for problems 2.1 - 2.14 in the 2 nd edition Solutions Manual . 2.15 EDTA (C 10 N 2 O 8 H 16 ) has a molecular weight of 292 g/mole Calcium molar concentration: (20 mg/L)(40.1 mol/g) -1 (10 3 mg/g) -1 = 0.000499 M Mass of EDTA: (0.000499 mol/L)(292 g/mol)(44 gal)(3.785 L/gal) = 24.3 g EDTA
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2.16 The balanced equation is: 2Al 0 + 6HCl 2AlCl 3 + 3H 2 (g) So, 2 0 2 2 0 0 2 0 H g Al g 9 H g 2 H mole Al mole Al g 27 H moles 3 Al moles 2 =
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2.17 - 2.21 The solutions for these problems are the solutions for problems 2.15 - 2.19 in the 2 nd edition Solutions Manual . 2.22 From Table 2.3, K sp , CaSO4 is 2 x 10 -5 and K sp = [Ca 2+ ] [SO 4 2- ] = [SO 4 2- ] 2 At saturation: [SO 4 2- ] = K sp ½ = (2 x 10 -5 ) ½ = 0.00447 mol/L Check if saturation is exceeded: mol/L 0.00368 g 136 mole L g 5 . 0 = Since saturation is not exceeded the sulfate concentration is 0.00368 mol/L
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2.23 Assume a closed system.
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Unformatted text preview: a. [Ca 2+ ] = C tot = [H 2 CO 3 ] + [HCO 3-] + [CO 3 2-] eqn. 1 Charge balance: [H + ] + 2[Ca 2+ ] = [HCO 3-] + 2[CO 3 2-] + [OH-] eqn. 2 At pH 8.5: C tot [HCO 3-] [CO 3 2-] (see Figure 2.5) and [H + ] [OH-] So eqn. 2 simplifies to: 2[Ca 2+ ] = C tot + [OH-] eqn. 3 Using eqn. 1 in eqn. 3 yields: [Ca 2+ ] = [OH-] = 10-5.5 = 3.16 x 10-6 Note this is less than the Ca 2+ initially added, so CaCO 3 (s) has precipitated b. [ ] [ ][ ] ( )( ) ( ) M 10 M 10 2.24 10 4.47 10 10 HCO H CO 7.65--8 7--5.5-8.5 1 3 (aq) 2 = = = = + K c. [ ] ( ) ( ) g/L 0.20 g/L 0.19968 mol g 100 M 10-M 10 2 CaCO 5.5-3-(s) 3 = = = 2.24 - 2.36 The solutions for these problems are the solutions for problems 2.20 - 2.32 in the 2 nd edition Solutions Manual ....
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This note was uploaded on 04/03/2008 for the course C E 370 taught by Professor Burgos during the Spring '08 term at Pennsylvania State University, University Park.

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Chptr2PrblmSolns - a. [Ca 2+ ] = C tot = [H 2 CO 3 ] + [HCO...

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