solut04 - CHAPTER FOUR TYPES OF CHEMICAL REACTIONS AND...

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37 CHAPTER FOUR TYPES OF CHEMICAL REACTIONS AND SOLUTION STOICHIOMETRY Aqueous Solutions: Strong and Weak Electrolytes 10. The electrolyte designation refers to what happens to a substance when it dissolves in water, i.e., does it produce a lot of ions or a few ions or no ions when the substance dissolves. A weak elec- trolyte is a substance that only partially dissociates in water to produce only a few ions. Solubility refers to how much substance can dissolve in a solvent. "Slightly soluble" refers to substances that dissolve only to a small extent, whether it is an electrolyte or a nonelectrolyte. A weak electrolyte may be very soluble in water, or it may be slightly soluble. Acetic acid is an example of a weak electrolyte that is very soluble in water. 11. Measure the electrical conductivity of a solution and compare it to the conductivity of a solution of equal concentration containing a strong electrolyte. 12. MgSO 4 (s) Mg 2+ (aq) + SO 4 2- (aq); NH 4 NO 3 (s) NH 4 + (aq) + NO 3 - (aq) Solution Concentration: Molarity 13. For all parts to this problem, we will need: 1.00 L solution × L solute mol 0.50 = 0.50 mol solute a. 0.50 mol H 2 SO 4 × SO H mol 18 L 1 4 2 = 2.8 × 10 -2 L conc. H 2 SO 4 or 28 mL Dilute 28 mL of concentrated H 2 SO 4 to a total volume of 1.00 L with water. b. We need 0.50 mol HCl.
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38 0.50 mol HCl × HCl mol 12 L 1 = 4.2 × 10 -2 L = 42 mL Dilute 42 mL of concentrated HCl to a final volume of 1.00 L. c. We need 0.50 mol NiCl 2 . 0.50 mol NiCl 2 × O H 6 _ NiCl mol O H 6 _ NiCl g 237.7 _ NiCl mol O H 6 _ NiCl mol 1 2 2 2 2 2 2 2 = 120 g NiCl 2 6H 2 O Dissolve 120 g NiCl 2 6H 2 O in water, and add water until the total volume of the solution is 1.00 L. d. We need 0.50 mol HNO 3 . 0.50 mol HNO 3 × HNO mol 16 L 1 3 = 0.031 L = 31 mL Dissolve 31 mL of concentrated HNO 3 in water. Dilute to a total volume of 1.00 L. e. We need 0.50 mol Na 2 CO 3 . 0.50 mol Na 2 CO 3 × CO Na mol CO Na g 106.0 3 2 3 2 = 53 g Na 2 CO 3 Dissolve 53 g Na 2 CO 3 in water, dilute to 1.00 L. 14. a. M ) NO Ca( 2 3 = L 0.100 ) NO Ca( mol 0.100 2 3 = 1.00 M Ca(NO 3 ) 2 (s) Ca 2+ (aq) + 2 NO 3 - (aq); M Ca + 2 = 1.00 M ; M NO - 3 = 2(1.00) = 2.00 M b. M SO Na 4 2 = L 1.25 SO Na mol 2.5 4 2 = 2.0 M Na 2 SO 4 (s) 2 Na + (aq) + SO 4 2- (aq); M Na + = 2(2.0) = 4.0 M ; M SO - 2 4 = 2.0 M
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CHAPTER 4 SOLUTION STOICHIOMETRY 39 c. 5.00 g NH 4 Cl × Cl NH g 53.49 Cl NH mol 1 4 4 = 0.0935 mol NH 4 Cl M Cl NH 4 = 0.187 = L 0.5000 Cl NH mol 0.0935 4 M NH 4 Cl(s) NH 4 + (aq) + Cl - (aq); M NH + 4 = M Cl - = 0.187 M d. 1.00 g K 3 PO 4 × g 212.27 PO K mol 1 4 3 = 4.71 × 10 -3 mol K 3 PO 4 M PO K 4 3 = L 0.2500 mol 10 _ 4.71 -3 = 0.0188 M K 3 PO 4 (s) 3 K + (aq) + PO 4 3- (aq); M K + = 3(0.0188) = 0.0564 M ; M PO - 3 4 = 0.0188 M 15. Molar mass of NaHCO 3 = 22.99 + 1.008 + 12.01 + 3(16.00) = 84.01 g/mol Volume = 0.350 g NaHCO 3 × NaHCO mol 0.100 L 1 _ NaHCO g 84.01 NaHCO mol 1 3 3 3 = 0.0417 L = 41.7 mL 41.7 mL of 0.100 M NaHCO 3 contains 0.350 g NaHCO 3 . 16. 25.0 g (NH 4 ) 2 SO 4 × g 132.15 mol 1 = 1.89 × 10 -1 mol (NH 4 ) 2 SO 4 Molarity = L mL 1000 _ mL 100.0 mol 10 _ 1.89 -1 = 1.89 M (NH 4 ) 2 SO 4 Moles of (NH 4 ) 2 SO 4 in final solution = 10.00 × 10 -3 L × L mol 1.89 = 1.89 × 10 -2 mol (NH 4 ) 2 SO 4
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40 CHAPTER 4 SOLUTION STOICHIOMETRY Molarity of final solution = L mL 1000 _ mL 50.00) + (10.00 mol 10 _ 1.89 -2 = 0.315 M (NH 4 ) 2 SO 4 (NH 4 ) 2 SO 4 (s) 2 NH 4 + (aq) + SO 4 2- (aq); M NH + 4 = 2(0.315) = 0.630 M ; M SO - 2 4 = 0.315 M 17. 75.0 mL × g 46.1 mol 1 _ mL g 0.79 = 1.3 mol C 2 H 5 OH; Molarity = L 0.250 mol 1.3 = 5.2 M C 2 H 5 OH 18. Stock solution = mL steroid g 10 _ 2.00 =
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This homework help was uploaded on 04/06/2008 for the course CHEM 142 taught by Professor Zoller,williamh during the Fall '07 term at University of Washington.

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solut04 - CHAPTER FOUR TYPES OF CHEMICAL REACTIONS AND...

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