Chapter 6. Solutions to Supplementary Check for Understanding Problems

Chapter 6 Solutions to Supplementary Check for Understanding Problems Formation of Aqueous Solutions 1. A teaspoon of salt will readily dissolve in a pot of water. What happens to the salt concentration as the solution boils? Explain. Solution The salt solution increases. As the water boils the solution volume decreases while the amount of the salt does not change. This means there is more salt per liter of solution which represents a higher salt concentration. 2. Describe how an ionic compound dissolves in water. How are the strong electrostatic forces in the ionic solid overcome? How are the dissolved cations and anions prevented from recombining to form the solid? Solution When an ionic compound dissolves in water the cations and anions become separated and the individual ions are surrounded by water molecules. The initial electrostatic attractions between cations and anions are overcome by the electrostatic attractions of the ions to many polar water molecules. The dissolved ions are surrounded by layers of water molecules which limit the ability of the cations and anions to interact and reform the solid. 3. Use the model for an ionic compound dissolving in water to explain why smaller solute particles dissolve faster than large ones. Solution When an ionic compound dissolves in water only the cations and anions on the surface of the particles of the solute interact with polar water molecules and eventually become separated and surrounded by solvent. For a given amount of solute, as large particles are broken up into smaller ones new surfaces are exposed and there is an increase in the overall surface area which can interact with the water solvent so these smaller particles dissolve faster than larger ones. 4. Why is a concentrated solution of an ionic compound a good conductor of electricity? Solution Electrical conductivity involves the movement of electric charges in the form of ions or electrons A concentrated solution of an ionic compound has many dissolved ions. This large number of mobile charges enables such a solution to conduct electricity readily. S.6.4 S.6.5 5. CHAPTER 6 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Why is a sugar solution a very poor conductor of electricity? Solution When sugar dissolves the sugar molecules become separated and surrounded by water molecules. However, there are no mobile charges created so the solution does not conduct electricity readily. 6. What does the label “concentrated HNO3" on a bottle mean? Solution A solution labeled concentrated HNO3 means that there is a large number of (often the maximum number of) moles of solute (HNO3) per liter of solution. Solution Mass Percent 1. An aqueous NaCl solution that is 3.2% by mass NaCl contains __________ g NaCl for every __________ g water. Answers: 3.2 g NaCl 96.8 g H2O Solution solute mass solute mass = 100 g solution solvent mass + solute mass Therefore, in 100 g solution there will be 3.2 g NaCl and 96.8 g H2O. Recall that solute mass % = 2. Calculate the mass in grams of KCl in 18.6 g of a solution that is 0.15% KCl by mass. Answer: 0.28 g Solution What we know: g solution; g KCl/100 g solution Desired answer: g KCl The solution map for this problem is: g solution ö g KCl S.6.6 CHAPTER 6 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Remember that the mass percent of KCl refers to the g of KCl per 100 g of solution. This is exactly the conversion factor needed for this calculation. Applying this yields: 18.6 g soln x 3. 0.15g KCl = 0.028g KCl 100 g soln What mass of a solution that is 4.8% by mass NaHCO3 is needed to obtain 75 g NaHCO3? Answer: 1.6 x 103 g Solution What we know: g NaHCO3; g NaHCO3/100 g solution Desired answer: g solution The solution map for this problem is: g NaHCO3 ö g solution The conversion factor needed for this calculation is the mass % in the form Applying this yields: 75 g NaHCO3 x 100 g solution = 1.6 x 103 g solution 4.8 g NaHCO3 100 g solution . 4.8g NaHCO3 S.6.7 CHAPTER 6 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Molarity 1. Which of the following are needed to calculate the molarity of an aqueous solution? Select all that apply. Explain your answer. A. the mass of the solute B. the molar mass of the solute C. the volume of water added D. the total volume of the solution E. the solution density Answer: A, B and D Solution mass solute molsolute molar mass solute = Solute molarity is calculated by: . Therefore, the soln volume (L) soln volume (L) information indicated in A, B and D is needed. 2. What is the sodium ion molarity in a 0.115 M Na3PO4 solution? Answer: 0.345 M Solution What we know: mol Na3PO4/L solution Desired answer: mol Na+/L solution The solution map for this calculation is: mol Na 3 PO 4 L soln ö mol Na  L soln The formula of the solute indicates that in 1 mole of sodium phosphate there are 3 moles of sodium ions. This relationship can be applied as follows. S.6.8 CHAPTER 6 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS 0.115 mol Na 3 PO 4 3mol Na + 0.345 mol Na + x = = 0.345 M Na + L soln 1 mol Na 3 PO 4 L soln 3. What is the molarity of a solution prepared from 5.62 g NH4NO3 dissolved in water and diluted to 125 mL? Answer: 0.562 M Solution What we know: g NH4NO3; mL solution Desired answer: mol NH4NO3/L solution The solution map for this calculation is: g NH 4 NO3 mL soln ö mol NH 4 NO3 mL soln ö mol NH 4 NO3 L soln The conversion factor needed in the first step is the molar mass of NH4NO3 in the form 1mol NH 4 NO3 . 80.05g NH 4 NO3 The conversion factor for the second step is 1mL soln . 10-3 L soln Putting these together yields: 5.62 g NH 4 NO3 1mol NH 4 NO3 0.562 mol NH 4 NO3 1 mL soln x -3 x = 125 mL soln 80.05 g NH 4 NO3 10 L soln L soln 4. What volume (in mL) of a 0.204 M NaOH solution contains 8.53 g NaOH? Answer: 1.05 x 103 mL Solution What we know: g NaOH; mol NaOH/1000 mL solution Desired answer: mL solution S.6.9 CHAPTER 6 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The solution map for this calculation is: g NaOH ö mol NaOH ö mL solution The conversion factor needed in the first step is the molar mass of NaOH in the form 1mol NaOH . 40.00 g NaOH The conversion factor needed in the second step is the solution concentration in the form 1000 mL soln . 0.204 mol NaOH Putting these together yields: 8.53 g NaOH x 5. 1 mol NaOH 1000 mL soln = 1.05 x103 mL soln x 40.00 g NaOH 0.204 mol NaOH How many moles of NaF are present in 22.9 mL of a 5.16 mM NaF solution? Answer: 1.18 x 10-4 mol Solution What we know: mL solution; mmol NaF/1000 mL solution Desired answer: mol NaF The solution map for this calculation is: mL solution ö mmol NaF ö mol NaF The conversion factor needed in the first step is the solution concentration in the form 5.16 mmol NaF . 1000 mL soln 10-3 mol NaF The conversion factor needed in the second step is . 1mmol NaF S.6.10 CHAPTER 6 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Putting these together yields: 22.9 mL soln x 6. 5.16 mmol NaF 10-3 mol NaF = 1.18 x10-4 mol NaF x 1000 mL soln 1 mmol NaF What mass of Na2C2O4 is needed to prepare 75.0 mL of a 0.226 M Na2C2O4 solution? Answer: 2.27 g Solution What we know: mL solution; mol Na2C2O4/1000 mL solution Desired answer: g Na2C2O4 needed The solution map for this problem is: mL solution ö mol Na2C2O4 ö g Na2C2O4 The conversion factor needed in the first step is the solution concentration in the form 0.226 mol Na 2 C2 O 4 . 1000 mL soln The conversion factor needed in the second step is the molar mass of Na2C2O4 in the form 134.00 g Na 2 C2 O 4 . 1mol Na 2 C2 O 4 Putting these together yields: 75.0 mL soln x 7. 0.226 mol Na 2 C2 O 4 134.00 g Na 2 C2 O 4 x = 2.27 g Na 2 C 2 O 4 1000 mL soln 1 mol Na 2 C2 O 4 What is the nitrate ion molarity in a solution prepared from 26.8 mg aluminum nitrate dissolved in water and diluted to 0.750 L? Answer: 5.03 x 10-4 M S.6.11 CHAPTER 6 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Solution What we know: mg Al(NO3)3; L solution Desired answer: mol NO3-/L solution The solution map for this calculation is: mg Al(NO3 )3 L soln ö g Al(NO3 )3 L soln ö mol Al(NO3 )3 L soln ö mol NO3 L soln 10-3 g Al(NO3 )3 The conversion factor needed in the first step is . 1mg Al(NO3 )3 The conversion factor needed in the second step is the molar mass of Al(NO3)3 in the form 1mol Al(NO3 )3 . 213.01g Al(NO3 )3 3mol NO3The conversion factor for the last step is . 1mol Al(NO3 )3 Putting these together yields: 26.8 mg Al(NO3 )3 10-3 g Al(NO3 )3 1 mol Al(NO3 )3 3mol NO35.03 x10-4 mol NO3x x x = 1 mg Al(NO3 )3 213.01 g Al(NO3 )3 1 mol Al(NO3 )3 0.750 L soln L soln 8. What is the density of a 15.8 M HNO3 aqueous solution if it is 70.4% by mass HNO3? Answer: 1.41 g/mL Solution What we know: mol HNO3/1000 mL solution; g HNO3/100 g solution Desired answer: g solution/mL solution The solution map for this problem is: mol HNO3 mL soln ö g HNO3 mL soln ö g soln mL soln S.6.12 CHAPTER 6 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The conversion factor needed in the first step is the molar mass of HNO3 in the form 63.02 g HNO3 . 1mol HNO3 The conversion factor needed in the second step is the solute mass percent in the form 100 g soln . 70.4 g HNO3 Putting these together yields: 15.8 mol HNO3 63.02 g HNO3 100 g soln 1.41g soln x = x 1 mol HNO3 70.4 g HNO3 mL soln 1000 mL soln Dilutions 1. Explain why the equation M1V1 = M2V2 works for solving dilution problems. Solution When a solution is diluted the number of moles of solute remains constant while the solution volume increases, thus leading to a decrease in solute concentration. The number of moles of solute equals its molarity (M) times the solution volume (V) so the starting number of solute moles (M1V1) will equal the moles of solute after dilution (M2V2). 2. If a salt solution is diluted by adding a volume of water equal to 50% of the solution volume, by what factor has the salt concentration been diluted? Assume that the volumes are additive. Answer: 2/3 Solution What we know: relationship between the initial volume and final volume Desired answer: dilution factor (V1/V2) S.6.13 CHAPTER 6 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Although no numerical value is given for V1 or V2 we know the relationship between the initial volume: V2 = V1 + ½V1. Substituting for V2 in the dilution factor ratio yields: V1 V1 V1 2 = = = 1 3 V2 3 V1 + V1 V1 2 2 3. How many milliliters of 14.8 M H3PO4 are needed to prepare 2.50 L of 3.0 M H3PO4? Answer: 507 mL Solution What we know: initial concentration of H3PO4 (M1); final concentration of H3PO4 solution (M2); final volume of H3PO4 solution (V2) Desired answer: initial volume of the more concentrated H3PO4 solution (V1) Note that this is a dilution problem. Solve the dilution equation for V1. It is useful to first convert the final volume to milliliters. 2.50 L x 1mL = 2.50 x103 mL -3 10 L  3.0 mol H 3 PO 4  3    2.50 x10 mL dilsoln  1000 mL dilsoln  MV V1  2 2   = 507 mL conc soln M1  14.8 mol H 3 PO 4     1000 mL conc soln  Solution Stoichiometry 1. Barium sulfate is an exception to the rule that sulfates tend to be soluble in water. How many milliters of 0.25 M Na2SO4 are needed to precipitate all of the barium as BaSO4 from 10.0 mL of 0.15 M Ba(NO3)2? Ba(NO3)2(aq) + Na2SO4(g) ÷ BaSO4(s) + 2NaNO3(aq) Answer: 6.0 mL S.6.14 CHAPTER 6 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Solution What we know: mol Na2SO4/1000 mL solution; mL Ba(NO3)2 solution; mol Ba(NO3)2/1000 mL solution; balanced equation relating Na2SO4 and Ba(NO3)2 Desired answer: mL Na2SO4 solution The solution map for this calculation is: mL Ba(NO3)2 soln ö mol Ba(NO3)2 ö mol Na2SO4 ö mL Na2SO4 soln The conversion factor needed in the first step is the Ba(NO3)2 solution concentration in the form 0.15 mol Ba(NO3 )3 . 1000 mL Ba(NO3 )3 soln The conversion factor needed in the second step is the Na2SO4/Ba(NO3)2 mole ratio from the balanced equation. The conversion factor in the last step is the Na2SO4 solution concentration in the form 1000 mL Na 2SO 4 soln 0.25 mol Na 2SO 4 Putting these together yields: 10.0 mL Ba(NO3 ) 2 soln x 2. 0.15 mol Ba(NO3 ) 2 1 mol Na 2SO 4 1000 mL Na 2SO 4 soln x x = 6.0 mL Na 2SO 4 soln 0.25 mol Na 2SO 4 1000 mL Ba(NO3 ) 2 soln 1 mol Ba(NO3 ) 2 If 35.0 mL of a 0.162 M CaCl2 solution are added to 20.0 mL of 0.211 M Na2CO3, what is the maximum number of moles of CaCO3 that can form? Answer: 0.00422 mol Solution What we know: mol CaCl2/1000 mL solution; mL CaCl2 solution; mol Na2CO3/1000 mL solution; mL Na2CO3 solution; Desired answer: maximum mol CaCO3 produced S.6.15 CHAPTER 6 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS A balanced equation for this double displacement reaction is needed. CaCl2(aq) + Na2CO3(aq) ÷ CaCO3(s) + 2NaCl(aq) First determine the limiting reactant by calculating how many moles of CaCO3 can possibly be produced from each starting amount of reactant. The solution maps for these calculations are: mL CaCl2 soln ö mol CaCl2 ö mol CaCO3 mL Na2CO3 soln ö mol Na2CO3 ö mol CaCO3 For the first calculation the conversion factors needed are the molarity of the CaCl2 solution and the CaCO3/CaCl2 mole ratio. Putting these together yields: 35.0 mL CaCl2 soln x 1mol CaCO3 0.162 mol CaCl 2 x = 0.00567 mol CaCO3 1000 mL CaCl2 soln 1 mol CaCl2 For the second calculation the conversion factors needed are the molarity of the Na2CO3 solution and the CaCO3/Na2CO3 mole ratio. Putting these together yields: 20.0 mL Na 2 CO3 soln x 0.211 mol Na 2 CO3 1mol CaCO3 x = 0.00422 mol CaCO3 1000 mL Na 2 CO3 soln 1 mol Na 2 CO3 Since the starting amount of Na2CO3 solution produces the smaller amount of CaCO3, Na2CO3 is the limiting reactant and the maximum yield is 0.00422 mol CaCO3. Acid-Base Neutralizations 1. If 33.3 mL of 0.150 M HCl are needed to neutralize 0.0250 L of a NaOH solution, what is the molarity of the NaOH? Answer: 0.200 M Solution What we know: mol HCl/1000 mL solution; mL HCl solution; mL NaOH solution Desired answer: mol NaOH/L solution S.6.16 CHAPTER 6 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS A balanced equation for the neutralization reaction is needed. HCl(aq) + NaOH(aq) ÷ H2O(l) + NaCl(aq) The solution map for this calculation is: mL HCl soln ö mol HCl ö mol NaOH ö mol NaOH L soln The conversion factor needed in the first step is the HCl solution concentration in the form 0.0984 mol KOH . 1000 mL soln The conversion factor needed in the second step is the NaOH/HCl mole ratio from the balanced equation. Putting these together yields: 33.3 mL HClsoln x 0.150 mol HCl 1mol NaOH = 0.00500 mol NaOH 1000 mL HClsoln 1 mol HCl The final step is: 0.00500 mol NaOH 0.200 mol NaOH = 0.0250 L soln L soln 2. How many milliliters of 0.150 M HCl are needed to completely react 0.245 g Na2CO3? Na2CO3(s) + 2HCl(aq) ÷ 2NaCl(aq) + H2O(l) + CO2(g) Answer: 30.8 mL Solution What we know: mol HCl/1000 mL solution; g Na2CO3; balanced equation relating HCl and Na2CO3 Desired answer: mL HCl solution The solution map for this calculation is: g Na2CO3 ö mol Na2CO3 ö mol HCl ö mL HCl soln S.6.17 CHAPTER 6 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The conversion factor needed in the first step is the molar mass of Na2CO3 in the form 1mol Na 2 CO3 . 105.99 g Na 2 CO3 The conversion factor needed in the second step is the HCl/Na2CO3 mole ratio from the balanced equation. The conversion factor needed in the last step is the HCl solution concentration in the form 1000 mL soln . 0.150 mol HCl Putting these together yields: 0.245 g Na 2 CO3 x 1 mol Na 2 CO3 2 mol HCl 1000 mL HClsoln = 30.8 mL HClsoln x x 105.99 g Na 2 CO3 1 mol Na 2 CO3 0.150 mol HCl