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Kristopher Welch
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1 CHEM 101A ARMSTRONG SOLUTIONS TO TOPIC B PROBLEMS 1) We do not need to calculate the original molarity of the solution; all we need is the number of moles of K + in 7.50 g of K 2 CO 3 : 7.50 g K 2 CO 3 1 mol K 2 CO g K 2 CO 3 2 mol K + 1 mol K 2 CO 3 = mol K + The final molarity of K + must be M (i.e moles of K + per of solution). We can use this as a conversion factor to translate the moles of K + into the required volume: mol K + = L = 434 ml mol K This is the total solution volume. Janet originally made 100 ml of solution, so Farid must add 434 ml 100 ml = 334 ml of water. 2) The final solution must contain moles of NO 3 per liter of liquid. The total solution volume is ml = L, so we can calculate the total number of moles of NO 3 in the final solution: L mol NO - 3 = mol NO 3 - Now, let s see how much NO 3 Gerardo put in: g Al(NO 3 ) 3 1 mol Al(NO 3 ) g Al(NO 3 ) 3-3 mol NO 3 - = mol NO 3 1 mol Al(NO 3 ) 3 So Marciela must add mol mol = mol NO 3. How many grams of Mg(NO 3 ) 2 does she need in order to supply this much NO 3? mol NO 3-1 mol Mg(NO 3) 2 2 mol NO g Mg(NO 3) 2 1 mol Mg(NO 3 ) 2 = g Mg(NO 3 ) 2 3) Calculate the number of moles of Cl in each solute, add them up, then divide by the solution volume g KCl 1 mol KCl g KCl 1 mol Cl- 1 mol KCl = mol Cl g MgCl 2 1 mol MgCl g MgCl 2 2 mol Cl - 1 mol MgCl 2 = mol Cl -

2 1.88 g CrCl 3 1 mol CrCl g CrCl 3 3 mol Cl - 1 mol CrCl 3 = mol Cl - Total Cl = mol mol mol = mol Cl So the concentration of chloride ions in Chantelle s solution is: mol Cl L = M 4) a) Each mole of CrCl 3 produces 3 moles of Cl when it dissolves, so the concentration of CrCl 3 is M 3 = M in Wenzhou s solution. b) The mass of CrCl 3 that Wenzhou used can be calculated as follows: 0.25 L mol CrCl g CrCl 3 1 mol CrCl 3 = 3.21 g CrCl 3 5) For this problem, I will only give the answers. See me or visit the tutors at the LAC if you need assistance with this kind of problem. a) The actual solutes are Ca 2+, Cl, K +, and PO 4 3 Ca 3 (PO 4 ) 2 will form 3 Ca 2+ (aq) + 2 PO 4 3 (aq) Ca 3 (PO 4 ) 2 (s) b) The actual solutes are H +, NO 3, Na +, and OH H 2 O will form H + (aq) + OH (aq) H 2 O(l) c) The actual solutes are Al 3+, NO 3, K +, and OH Al(OH) 3 will form Al 3+ (aq) + 3 OH (aq) Al(OH) 3 (s) d) The actual solutes are Zn 2+, Br, Cu 2+, and SO 4 2 No product will form, so there is no reaction e) The actual solutes are H +, Cl, Na +, and HCO 3 CO 2 and H 2 O will form (when H + reacts with HCO 3 to form H 2 CO 3, which decomposes) H + (aq) + HCO 3 (aq) H 2 O(l) + CO 2 (g) f) The actual solutes are Ag +, NO 3, K +, and PO 4 3 Ag 3 PO 4 will form 3 Ag + (aq) + PO 4 3 Ag 3 PO 4 (s)

3 g) The actual solutes are Ba 2+, OH, Na +, and SO 4 2 BaSO 4 will form Ba 2+ (aq) + SO 4 2 (aq) BaSO 4 (s) h) The actual solutes are HC 2 H 3 O 2, Ba 2+, and OH. A very small amount of H + and C 2 H 3 O 2 will also be present. H 2 O and C 2 H 3 O 2 will form HC 2 H 3 O 2 (aq) + OH (aq) H 2 O(l) + C 2 H 3 O 2 (aq) Note that we do not write H + + OH H 2 O. For any net ionic reaction involving a weak electrolyte, the majority rules : there is far more HC 2 H 3 O 2 than there is free H +, so the dominant reaction will be between HC 2 H 3 O 2 and OH. i) At the start, the actual solutes are K +, CO 3 2, H +, and I HCO 3 will form The first reaction is H + (aq) + CO 3 2 (aq) HCO 3 In the second reaction, the actual solutes are K +, HCO 3, H +, and I H 2 O and CO 2 will form (as in part e) The second reaction is H + (aq) + HCO 3 (aq) H 2 O(l) + CO 2 (g) j) The actual solutes are HC 6 H 5 O, Na +, and OH. A very small amount of H + and C 6 H 5 O will also be present. How did we know how HC 6 H 5 O would behave? Since the chemical formula begins with H, we know that this compound is an acid; and since this is not one of the big six strong acids, it must be a weak acid, dissociating only to a small extent. H 2 O and C 6 H 5 O will form HC 6 H 5 O(aq) + OH (aq) H 2 O(l) + C 6 H 5 O (aq) k) At the start, the actual solutes are H 2 C 4 H 4 O 4, Na +, and OH. Very small amounts of H + and HC 4 H 4 O 4 will also be present, along with an extremely small amount of C 4 H 4 O 4 2. H 2 O and HC 4 H 4 O 4 will form Reaction 1: H 2 C 4 H 4 O 4 (aq) + OH (aq) H 2 O(l) + HC 4 H 4 O 4 (aq) In the second reaction, the actual solutes are HC 4 H 4 O 4, Na +, and OH H 2 O and C 4 H 4 O 4 2 will form Reaction 2: HC 4 H 4 O 4 (aq) + OH (aq) H 2 O(l) + C 4 H 4 O 4 2 (aq) Acids that contain more than one hydrogen atom always lose one hydrogen ion at a time. Essentially all of the acid molecules will lose their first hydrogen ion before any of them will lose their second hydrogen ion. 6) The actual solutes here are Ba 2+, Cl, K +, and CrO 2 4. The possible precipitates are KCl and BaCrO 4, but we know that KCl is water-soluble, so the actual precipitate must be BaCrO 4. The reaction is: Ba 2+ (aq) + CrO 2 4 (aq) BaCrO 4 (s)

4 7) See the handout for information about this type of reaction. The net ionic equations are: a) Mg(OH) 2 (s) + 2 H + (aq) Mg 2+ (aq) + 2 H 2 O(l) b) CuO(s) + 2 H + (aq) Cu 2+ (aq) + H 2 O(l) c) Al(OH) 3 (s) + 3 H + (aq) Al 3+ (aq) + 3 H 2 O(l) d) Cr 2 O 3 (s) + 6 H + (aq) 2 Cr 3+ (aq) + 3 H 2 O(l) 8) a) The reaction that occurs is Pb 2+ (aq) + 2 I (aq) PbI 2 (s) b) You must begin by working out the limiting reactant (Pb 2+ versus I ). This can be done by calculating the amount of product that would be formed if each ion were used completely: mol KI 1 mol I L 1 mol KI 1 mmol PbI 2 2 mmol I - = mol PbI 2 formed (if all of the I - is consumed) L mol Pb(NO 3 ) 2 1 mol Pb 2+ 1 mol Pb(NO 3 ) 2 1 mmol PbI 2 1 mmol Pb 2+ = mol PbI 2 formed (if all of the Pb 2+ is consumed) Since the I produces a smaller amount of product, I is the limiting reactant and the reaction forms moles of PbI 2. The molar mass of PbI 2 is 461 g/mol, so moles equals g of PbI 2. c) The excess reactant here is Pb 2+. First, we calculate the amount of Pb 2+ that is used up: L mol KI 1 mol I 1 mol KI 1 mmol Pb2+ 2 mmol I - = mol Pb 2+ used up We started with mol of Pb 2+ (you should be able to calculate this yourself). Therefore, we are left with mol mol = mol of Pb 2+. The total volume of liquid is 5.00 ml ml = 9.00 ml, so the concentration of Pb 2+ in the final mixture is: mol Pb 2+ = M L d) NO 3 is a spectator ion, so the number of moles of NO 3 does not change during the reaction: L mol Pb(NO ) 2 mol NO = mol NO 1 mol Pb(NO 3 ) 3 2 To calculate the molarity, divide the number of moles by the final volume, which is 5.00 ml ml = 9.00 ml = L mol/ L = M

5 9) a) Again, we start by determining the limiting reactant. We will do this by calculating the number of moles of AgI that is formed if each reactant is used up. For Ag 2 O: 1.45 g Ag 2 O g Ag O 2 1 mol Ag 2 O 2 mol AgI = mol AgI 1 mol Ag 2 O For H + : For I : L 2.00 mol HI 1 mol H+ 1 mol HI 2 mol AgI = mol AgI + 2 mol H 2.00 mol HI 1 mol I 2 mol AgI L = mol AgI 1 mol HI 2 mol I Ag 2 O produces less AgI, so Ag 2 O is the limiting reactant and we actually make moles of AgI. The molar mass of AgI is g/mol, so moles of AgI equals 2.94 g of AgI. b) We start with 0.05 moles of H + (that should be an easy calculation). Now we must figure out how much H + is consumed, based on the fact that Ag 2 O is the limiting reactant: 1 mol Ag 1.45 g Ag 2 O 2 O g Ag 2 O 2 mol H + 1 mol Ag 2 O = mol H+ consumed So we end up with 0.05 mol mol = moles of H +. The molarity of H + is then mol/0.025 L = 1.50 M. 10) When you add Na 3 PO 4 solution to AgNO 3 solution, Ag 3 PO 4 precipitates: 3 Ag + (aq) + PO 4 3 (aq) Ag 3 PO 4 (s) You can calculate the volume of the Ag 3 PO 4 solution in a single (lengthy) conversion factor setup: L mol AgNO 3 1 mol Na 3 PO 4 1 mol PO 4 3- Converting this volume to milliliters gives 18.3 ml. If this seems confusing, here is a step-by-step procedure: 1 mol Ag + 1 mol PO 4 1 mol AgNO 3 3 mol Ag mol Na 3 PO 4 = L of the Na 3 PO 4 solution First, calculate the number of moles of AgNO 3 in 25.0 ml ( L) of the M AgNO 3 solution. You should get mol. Second, recognize that each mole of AgNO 3 contains one mole of Ag +, so the AgNO 3 solution actually contains mol of Ag + ions. Third, use the mole ratio between Ag + and PO 4 3 to calculate the number of moles of PO 4 3 that must be added in order to completely consume the Ag + ions. You should get mol. 3-

6 Fourth, recognize that one mole of Na 3 PO 4 contains one mole of PO 4 3, so you must add mol of Na 3 PO 4. Finally, calculate the volume of M Na 3 PO 4 solution that you must add in order to get mol of Na 3 PO 4. You ll get L = 18.3 ml, as above. 11) a) We can determine whether Ag + is present in excess by calculating the amount of Ag + in 10.0 ml of 1 M AgNO 3 and seeing if it is more than the amount of Ag + in g of AgCl. The number of moles of Ag + in the AgNO 3 solution is: 10.0 ml 1000 ml 1 mol AgNO 3 The number of moles of Ag + in the AgCl is: 1 mol Ag + 1 mol AgNO 3 = mol Ag g AgCl 1 mol AgCl g AgCl 1 mol Ag+ 1 mol AgCl = mol Ag + The amount of Ag + in the AgCl (product) is less than the amount of Ag + in the AgNO 3 (reactant), so Ag + was in excess and Cl was the limiting reactant. b) AgCl contains equal numbers of moles of Ag + and Cl, so the AgCl contains moles of Cl. Therefore, the original solution contained moles of Cl. The volume of the original solution was ml = L, so the molarity of Cl in the original solution was: mol L = M 12) We start by calculating the number of moles of OH ions that were added in the titration: L mol NaOH 1 mol OH - 1 mol NaOH = mol OH- At the endpoint, we have added exactly enough OH to neutralize the acid. According to the balanced equation, this requires three moles of OH for each mole of H 3 C 6 H 5 O 7, so mol OH - 1 mol H 3 C 6 H 5 O 7 3 mol OH - = mol H 3 C 6 H 5 O 7 The original solution therefore contained moles of H 3 C 6 H 5 O 7. The volume of the original solution was ml = L, so its molarity was:

7 mol L = M 13) a) CaCO 3 dissolves because the CO 3 2 ion reacts with H + to form HCO 3, and Ca(HCO 3 ) 2 is water-soluble. The net ionic equation is: CaCO 3 (s) + H + (aq) Ca 2+ (aq) + HCO 3 (aq) In this reaction, Cl is a spectator ion. b) The mixture bubbles because the HCO 3 ion reacts with H + to form gaseous CO 2 and water. The net ionic equation is: HCO 3 (aq) + H + (aq) CO 2 (g) + H 2 O(l) In this reaction, Ca 2+ is a spectator ion (as is Cl ). 14) Bottles 1 and 3 are incorrect. For bottle 1, it is not possible to dissolve 9.5 moles of BaI 2 into a single liter of solution. (9.5 moles of BaI 2 weighs 3715 grams and takes up about a liter of space (or more) by itself.) For bottle 3, PbI 2 is insoluble in water, so it is impossible to dissolve any amount of PbI 2 in water. (Actually, not compound is absolutely insoluble, but the solubility of PbI 2 is far too low to make at 0.15 M solution.) 15) In both cases, the H + ions from the sulfuric acid react with the CO 3 2 ions from the metal carbonate, forming CO 2 (the bubbles) and H 2 O. However, when you use BaCO 3, the Ba 2+ ions also combine with SO 4 2 ions, forming insoluble BaSO 4. In effect, you are exchanging one insoluble salt (BaCO 3 ) for another (BaSO 4 ), so you never see the solid go into solution. In the case of Mg 2+, though, MgSO 4 is water-soluble and does not form, so the product mixture does not contain any solids. The reactions are: MgCO 3 (s) + 2 H + (aq) Mg 2+ (aq) + CO 2 (g) + H 2 O(l) BaCO 3 (s) + H + (aq) + HSO 4 (aq) BaSO 4 (s) + CO 2 (g) + H 2 O(l) 16) You need to work backward. First, calculate the number of moles of Na 2 S 2 O 3 that are added in the second reaction: L mol Na 2 S 2 O 3 = mol Na 2 S 2 O 3 Then calculate the number of moles of I 2 that must have been present in the second reaction: mol Na 2 S 2 O 3 1 mol I 2 2 mol Na 2 S 2 O 3 = mol I 2 This must be the amount of I 2 that was formed in the first reaction. Now we can calculate the number of moles of CuSO 4 that were consumed in reaction mol I 2 2 mol CuSO 4 1 mol I 2 = mol CuSO 4

8 Therefore, we can conclude that the original CuSO 4 solution contained moles of CuSO 4. The volume of this solution was ml = L, so its molarity was: mol L = M 17) The answer is The solution definitely contains SO 4 2. The solution definitely does not contain Cl or PO 4 3. You cannot determine whether the solution contains NO 3. Here is the reasoning. In experiment 1, adding Ba(NO 3 ) 2 is equivalent to adding Ba 2+ and NO 3. Ba 2+ forms insoluble compounds with SO 4 2 and PO 4 3, so this experiment tells us that at least one of these two anions must be present in the original solution. The solution could contain SO 4 2, PO 4 3, or both ions. This experiment tells us nothing about Cl and NO 3, because Ba 2+ does not form precipitates with either of these ions. If either Cl or NO 3 are present, they are still in the solution. In experiment 2, adding AgNO 3 is equivalent to adding Ag + and NO 3. The remaining solution from experiment 1 could only contain Cl or NO 3, because adding Ba(NO 3 ) 2 removed any SO 4 2 and PO 4 3 ions that might have been present. Ag + forms an insoluble compound with Cl, but we don t see a precipitate in this experiment, so this experiment tells us that Cl cannot be present. In experiment 3, adding Mg(NO 3 ) 2 is equivalent to adding Mg 2+ and NO 3. Mg 2+ forms an insoluble compound with PO 4 3, but not with any of the other possible ions in the original solution. Since this experiment doesn t produce a precipitate, PO 4 3 cannot be present in the original solution. Putting experiments 1 and 3 together: experiment 1 told us that the solution must contain at least one of SO 4 2 and PO 4 3, and experiment 3 told us that the solution does not contain PO 4 3. Therefore, the solution must contain SO 4 2. None of the chemicals the chemist added would react with NO 3, so we cannot tell whether NO 3 is present or absent. (In fact, there is no simple precipitation test that will tell you whether a solution contains nitrate ions.)

Chapter 4 Reactions in Aqueous Solutions. Copyright McGraw-Hill

Chapter 4 Reactions in Aqueous Solutions Copyright McGraw-Hill 2009 1 4.1 General Properties of Aqueous Solutions Solution - a homogeneous mixture Solute: the component that is dissolved Solvent: the component