CHEM 162 2002 HOURLY EXAM III ANSWERS EQUILIBRIA SOLUBILITY PRODUCT When three

Chem 162 2002 hourly exam iii answers equilibria

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7.CHEM 162-2002 HOURLY EXAM III + ANSWERSEQUILIBRIASOLUBILITY PRODUCTWhen three moles of CaCl2are added to one mole of Na2SO4and two moles of NaF, all contained in one liter, of aqueous solution, which of the following is true? (CaSO4Ksp = 6.1 x 10-5and CaF2Ksp = 5.3 x 10-9) A. Precipitation of CaSO4is not completeB.CaF2and CaSO4will precipitate Chapter 15B practice problems 50
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C . Precipitation of CaSO 4 is complete D. Only CaSO 4 will precipitate E. Only CaF 2 will precipitate CaSO 4 Ca 2+ + SO 4 2- Ksp = 6.1 x 10 -5 X X X x X = 6.1 x 10 -5 X = 0.78 x 10 -2 = [Ca 2+ ] to initiate CaSO 4 precipitation CaF 2 Ca 2+ + 2F - Ksp = 5.3 x 10 -9 X 2X X x (2X) 2 = 4X 3 = 5.3 x 10 -9 X = 1.1 x 10 -3 = [Ca 2+ ] to initiate CaF 2 precipitation Since it takes less Calcium ion to initiate the CaF 2 precipitation, then calcium fluoride will precipitate first. Calcium fluoride will precipitate since it requires only 1.1 x 10 -3 M Calcium ion, and there is 3 molar calcium ion present. If the stoichiometric amount of calcium completely reacts with 1 mole of fluoride to precipitate CaF 2 , then this would leave two moles of calcium to react with the one mole of sulfate. Since it requires only 0.078 x 10 -2 M Ca 2+ to initiate precipitation of sulfate, then calcium sulfate will also precipitate. We can stop here and choose “B” as the answer. However, if we want to investigate whether “C” might also be an answer, then: To determine whether the CaSO 4 precipitation is complete, assume that the CaSO4 precipitation reaction goes to completion, and then work backwards to determine how much sulfate is in solution. Ca 2+ + SO 4 2- CaSO 4 K = 1/Ksp = 1/6.1 x 10 -5 = 1.64 x 10 4 Initial 2 1 0 Change -1 -1 +1 Completion +1 0 +1 Change +X +X -X Equilibrium 1+X +X (1+X) x X = 6.1 x 10 -5 Assume that X is very small; simplify. 1 x X = 6.1 x 10 -5 X = 6.1 x 10 -5 = SO 4 2- concentration remaining in solution after precipitation of CaSO 4 (6.1 x 10 -5 /1) x 100 = 0.0061% SO 4 2- left in solution after precipitation of CaSO 4 . Therefore, the precipitation of CaSO 4 is complete making “C” also a correct answer. Chapter 15B practice problems 51
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Chapter 15B practice problems 52
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25. CHEM 162-2002 HOURLY EXAM III + ANSWERS EQUILIBRIA SOLUBILITY PRODUCT A solution is made by mixing 200 mL 0.10 M CaCl 2 and 200 mL 1.0 M NaF. If the K sp for CaF 2 is 5.3 x 10 -9 , the concentration of Ca 2+ and F - in solution will be 0.02 moles of CaCl 2 & 0.2 moles of NaF in 0.400 L Therefore, 0.05 M CaCl 2 and 0.50 M NaF Considering the excess of F - and the small K sp , assume that the precipitation reaction between the Ca 2+ and the F - goes to completion, i.e., the entire 0.05 M Ca 2+ is used up to form 0.05 M CaF 2 precipitate, and then work backwards to find out how much Ca 2+ and F - are remaining in solution. Therefore, remaining is 0.40 M F - . An equilibrium equation may be written from left to right or right to left; it makes no difference. CaF 2 Ca 2+ + 2F - K sp for CaF 2 is 5.3 x 10 -9 Initial 0 0.05 0.50 Change +0.05 -0.05 -0.10 Completion +0.05 0 +0.40 Change +0.05 - X X 0.40 + 2X X 0.40 + 2X X x (0.40 + 2X) 2 = 5.3 x 10 -9 Simplifying: X x (0.40) 2 = 5.3 x 10 -9 X = 3.31 x 10 -8 M Ca 2+ + 0.40 M F - A . Ca 2+ = 3.3 x 10 -8 , and F - = 0.40 M B. Ca 2+ = 0.10 M, and F - = 6.0 x 10 -6 M C. Ca 2+ = 0.050 M, and F - = 0.050 M D. Ca 2+ = 0.30 M, and F - = 0.040 M E. Ca 2+ = 1.1 x 10 -3 M, and F - = 1.1 x 10 -3 M 1.
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