Wet Chemistry: The Science of Substances in Solution

Chapter 6 Check for Understanding 6.1 1. Indicate whether each of the following is a good electrical conductor. a) mixture of CaCl2(s) and H2O(l) b) Na(s) c) mixture of CH3OH(l) and H2O(l) d) Br2(l) e) mixture of CaCO3(s) and H2O(l) f) HCl(aq) Answers: a) good electrical conductor d) nonconductor b) good electrical conductor e) nonconductor c) nonconductor f) good electrical conductor Solutions a) This mixture contains a soluble ionic compound that will produce many mobile ions when it dissolves in water. b) This is a metallic solid which has mobile electrons. c) This is a mixture of two molecular compounds, neither of which is an acid, so there are very few mobile charges present. d) This is a pure molecular substance so there are no mobile charges present. e) This mixture contains an insoluble ionic compound and a molecular compound so there are very few mobile charges present. f) There are many mobile charges present because when this acid dissolves in water many mobile H+ and Cl- ions are produced. Check for Understanding 6.2 1. Cow’s milk typically contains 4.6 % lactose sugar (C12H22O11) by mass. What is the mass of lactose present in 8 oz of milk having a mass of 245 g? Answer: 11 g Solution What we know: g milk; g lactose/100 g milk Desired answer: g lactose The solution map for this problem is: g milk ö g lactose Remember that the mass percent of lactose in milk refers to the g of lactose per 100 g of milk. This is exactly the conversion factor needed for this calculation. Applying this yields: 245 g milk x 4.6 g lactose = 11g lactose 100 g milk Check for Understanding 6.3 1. How many moles of HCl are in 23.18 mL of 0.1006 M HCl(aq)? Answer: 2.332 x 10-3 mol Solution What we know: mL solution; mol HCl/1000 mL solution Desired answer: mol HCl The solution map for this problem is: mL solution ö mol HCl The conversion factor needed is the solution concentration in the form 0.1006 mol HCl . 1000 mL soln Applying this yields: 23.18 mL soln x 2. 0.1006 mol HCl = 2.332 x10-3 mol HCl 1000 mL soln How many grams of KMnO4 are needed to prepare 250. mL of a 0.0988 M KMnO4 solution? Answer: 3.90 g Solution What we know: mL solution; mol KMnO4/1000 mL solution Desired answer: g KMnO4 needed The solution map for this problem is: mL solution ö mol KMnO4 ö g KMnO4 The conversion factor needed in the first step is the solution concentration in the form 0.0988 mol KMnO 4 . 1000 mL soln The conversion factor needed in the second step is the molar mass of KMnO4 in the form 158.04 g KMnO 4 . 1mol KMnO 4 Putting these together yields: 250. mL soln x 3. 0.0988 mol KMnO 4 158.04 g KMnO 4 x = 3.90 g KMnO 4 1000 mL soln 1 mol KMnO 4 What volume (in mL) of 0.2006 M NaOH contains 0.0150 mol NaOH? Answer: 74.8 mL Solution What we know: mol NaOH; mol NaOH/1000 mL solution Desired answer: mL NaOH solution The solution map for this problem is: mol NaOH ö mL solution The conversion factor needed is the solution concentration in the form 1000 mL soln . 0.2006 mol NaOH Applying this yields: 0.0150 mol NaOH x 1000 mL soln = 74.8 mL soln 0.2006 mol NaOH Check for Understanding 6.4 1. A 14.8 M NH3 aqueous solution has a density of 0.90 g/mL. What is the mass % NH3 in this solution? Answer: 28% Solution What we know: mol NH3/1000 mL solution; g solution/mL solution Desired answer: g NH3/100 g solution The solution map for this problem is: mol NH 3 mL solution ö g NH 3 mL solution ö g NH 3 g solution ö g NH 3 100 g solution The conversion factor needed in the first step is the molar mass of NH3 in the form 17.03g NH3 . 1mol NH 3 The conversion factor needed in the second step is the solution density in the form 1mL soln . 0.90 g soln Putting these together yields: 14.8 mol NH 3 17.03g NH 3 1 mL soln 0.28g NH 3 x = x 1000 mL soln 1 mol NH 3 0.90 g soln g soln The final step is: mass % NH3 = 0.28g NH 3 x 100 = 28% g soln Check for Understanding 6.5 1. What is the solution concentration if 3.00 mL of 1.83 x 10-3 M FeSO4 is diluted to 250. mL? Answer: 2.20 x 10-5 M Solution What we know: initial concentration of FeSO4 (M1); initial volume of FeSO4 solution (V1); final volume of FeSO4 solution (V2) Desired answer: concentration of diluted FeSO4 solution (M2) Note that this is a dilution problem. Solve the dilution equation for M2. It is useful to first convert the final volume to liters. 250. mL x 10-3 L = 0.250 L 1 mL  1.83 x10-3 mol FeSO 4    (3.00 mL conc soln ) 1000 mL conc soln  M 1V1 2.20 x10-5 mol FeSO 4  M2 = = = = 2.20 x10-5 M V2 0.250 L dilsoln L dilsoln Check for Understanding 6.6 1. How many mL of 0.244 M HCl(aq) are needed to completely neutralize 20.0 mL of 0.0984 M KOH? Answer: 8.07 mL Solution What we know: mL KOH solution; mol KOH/1000 mL solution; mol HCl/1000 mL solution Desired answer: mL HCl solution A balanced equation for the neutralization reaction is needed. HCl(aq) + KOH(aq) ÷ H2O(l) + KCl(aq) The solution map for this calculation is: mL KOH soln ö mol KOH ö mol HCl ö mL HCl soln The conversion factor needed in the first step is the KOH solution concentration in the 0.0984 mol KOH form . 1000 mL soln The conversion factor needed in the second step is the HCl/KOH mole ratio from the balanced equation. The conversion factor needed in the last step is the HCl solution concentration in the 1000 mL soln form . 0.244 mol HCl Putting these together yields: 20.0 mL KOH soln x 0.0984 mol KOH 1 mol HCl 1000 mL HClsoln = 8.07 mL HClsoln x x 0.244 mol HCl 1000 mL KOH soln 1 mol KOH Chapter 6 1. No. I2 is a nonpolar molecule and does not have a strong affinity for polar water molecules. 2. The faint light suggests the presence of some mobile ions. Tap water is not pure water. It contains small amounts of dissolved solutes, including some salts which will produce mobile ions such as Ca2+, Mg2+ and Cl- in solution. 3. Vegetable oil is composed of large molecules that are mainly nonpolar. Vinegar is primarily an aqueous solution of acetic acid and hence a mixture of polar molecules and ions. The two substances do not have a strong affinity for each other and hence do not mix readily. 4. A 6. E 7. 1.30 mol 8. 69.9 g 9. 7.2% 10. 43.3 g 11. 0.229 M 12. 224 mL 13. 75 mL 14. 55.6 L 15. 1.82 M 16. 18.0 M 17. 13 mL 18. a) KOH b) 0.0142 mol 19. 3.26 g 20. 262 mL