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  93 CHAPTER 4 TYPES OF CHEMICAL REACTIONS AND SOLUTION STOICHIOMETRY    Questions  13. a. Polarity is a term applied to covalent compounds. Polar covalent compounds have an unequal sharing of electrons in bonds that results in unequal charge distribution in the overall molecule. Polar molecules have a partial negative end and a partial positive end. These are not full charges as in ionic compounds but are charges much smaller in magnitude. Water is a polar molecule and dissolves other polar solutes readily. The oxygen end of water (the partial negative end of the polar water molecule) aligns with the  partial positive end of the polar solute, whereas the hydrogens of water (the partial  positive end of the polar water molecule) align with the partial negative end of the solute. These opposite charge attractions stabilize polar solutes in water. This process is called hydration. Nonpolar solutes do not have permanent partial negative and partial positive ends; nonpolar solutes are not stabilized in water and do not dissolve.  b. KF is a soluble ionic compound, so it is a strong electrolyte. KF(aq) actually exists as separate hydrated K  +  ions and hydrated F   ions in solution: C 6 H 12 O 6 is a polar covalent molecule that is a nonelectrolyte. C 6 H 12 O 6 is hydrated as described in part a. c. RbCl is a soluble ionic compound, so it exists as separate hydrated Rb +  ions and hydrated Cl   ions in solution. AgCl is an insoluble ionic compound, so the ions stay together in solution and fall to the bottom of the container as a precipitate. d. HNO 3  is a strong acid and exists as separate hydrated H +  ions and hydrated NO 3   ions in solution. CO is a polar covalent molecule and is hydrated as explained in part a. 14. 2.0 L × 3.0 mol/L = 6.0 mol HCl; the 2.0 L of solution contains 6.0 mol of the solute. HCl is a strong acid; it exists in aqueous solution as separate hydrated H +  ions and hydrated Cl   ions. So the solution will contain 6.0 mol of H + (aq) and 6.0 mol of Cl   (aq). For the acetic acid solution, HC 2 H 3 O 2  is a weak acid instead of a strong acid. Only some of the 6.0 moles of HC 2 H 3 O 2  molecules will dissociate into H + (aq) and C 2 H 3 O 2  (aq). The 2.0 L of 3.0  M   HC 2 H 3 O 2  solution will contain mostly hydrated HC 2 H 3 O 2  molecules but will also contain some hydrated H + ions and hydrated C 2 H 3 O 2   ions. 15 . Only statement b is true. A concentrated solution can also contain a nonelectrolyte dissolved in water, e.g., concentrated sugar water. Acids are either strong   or weak    electrolytes. Some ionic compounds are not soluble in water, so they are not labeled as a specific type of electrolyte.  94 CHAPTER 4 SOLUTION STOICHIOMETRY 16. One mole of NaOH dissolved in 1.00 L of solution will produce 1.00  M   NaOH. First, weigh out 40.00 g of NaOH (1.000 mol). Next, add some water to a 1-L volumetric flask (an instrument that is precise to 1.000 L). Dissolve the NaOH in the flask, add some more water, mix, add more water, mix, etc. until water has been added to 1.000-L mark of the volumetric flask. The result is 1.000 L of a 1.000  M   NaOH solution. Because we know the volume to four significant figures as well as the mass, the molarity will be known to four significant figures. This is good practice, if you need a three-significant-figure molarity, your measurements should be taken to four significant figures. When you need to dilute a more concentrated solution with water to prepare a solution, again make all measurements to four significant figures to ensure three significant figures in the molarity. Here, we need to cut the molarity in half from 2.00  M   to 1.00  M  . We would start with 1 mole of NaOH from the concentrated solution. This would be 500.0 mL of 2.00  M    NaOH. Add this to a 1-L volumetric flask with addition of more water and mixing until the 1.000-L mark is reached. The resulting solution would be 1.00  M  . 17. Use the solubility rules in Table 4.1. Some soluble bromides by Rule 2 would be NaBr, KBr, and NH 4 Br (there are others). The insoluble bromides by Rule 3 would be AgBr, PbBr  2 , and Hg 2 Br  2 . Similar reasoning is used for the other parts to this problem. Sulfates: Na 2 SO 4 , K  2 SO 4 , and (NH 4 ) 2 SO 4  (and others) would be soluble, and BaSO 4 , CaSO 4 , and PbSO 4 (or Hg 2 SO 4 ) would be insoluble. Hydroxides: NaOH, KOH, Ca(OH) 2  (and others) would be soluble, and Al(OH) 3 , Fe(OH) 3 , and Cu(OH) 2  (and others) would be insoluble. Phosphates: Na 3 PO 4 , K  3 PO 4 , (NH 4 ) 3 PO 4  (and others) would be soluble, and Ag 3 PO 4 , Ca 3 (PO 4 ) 2 , and FePO 4  (and others) would be insoluble. Lead: PbCl 2 , PbBr  2 , PbI 2 , Pb(OH) 2 , PbSO 4 , and PbS (and others) would be insoluble. Pb(NO 3 ) 2  would be a soluble Pb 2+  salt. 18. Pb(NO 3 ) 2 (aq) + 2 KI(aq)   PbI 2 (s) + 2 KNO 3 (aq) (formula equation) Pb 2+ (aq) + 2 NO 3  (aq) + 2 K  + (aq) + 2 I  (aq)   PbI 2 (s) + 2 K  + (aq) + 2 NO 3  (aq) (complete ionic equation) The 1.0 mol of Pb 2+  ions would react with the 2.0 mol of I   ions to form 1.0 mol of the PbI 2   precipitate. Even though the Pb 2+  and I   ions are removed, the spectator ions K  +  and NO 3   are still present. The solution above the precipitate will conduct electricity because there are  plenty of charge carriers present in solution. 19. The Brønsted-Lowry definitions are best for our purposes. An acid is a proton donor, and a  base is a proton acceptor. A proton is an H +  ion. Neutral hydrogen has 1 electron and 1  proton, so an H +  ion is just a proton. An acid-base reaction is the transfer of an H +  ion (a  proton) from an acid to a base. 20. The acid is a diprotic acid (H 2 A), meaning that it has two H +  ions in the formula to donate to a base. The reaction is H 2 A(aq) + 2 NaOH(aq)   2 H 2 O(l) + Na 2 A(aq), where A 2   is what is left over from the acid formula when the two protons (H +  ions) are reacted.  CHAPTER 4 SOLUTION STOICHIOMETRY 95 For the HCl reaction, the base has the ability to accept two protons. The most common examples are Ca(OH) 2 , Sr(OH) 2 , and Ba(OH) 2 . A possible reaction would be 2 HCl(aq) + Ca(OH) 2 (aq)   2 H 2 O(l) + CaCl 2 (aq). 21. a. The species reduced is the element that gains electrons. The reducing agent causes reduc- tion to occur by itself being oxidized. The reducing agent generally refers to the entire formula of the compound/ion that contains the element oxidized.  b. The species oxidized is the element that loses electrons. The oxidizing agent causes oxidation to occur by itself being reduced. The oxidizing agent generally refers to the entire formula of the compound/ion that contains the element reduced. c. For simple binary ionic compounds, the actual charges on the ions are the same as the oxidation states. For covalent compounds, nonzero oxidation states are imaginary charges the elements would have if they were held together by ionic bonds (assuming the bond is  between two different nonmetals). Nonzero oxidation states for elements in covalent compounds are not actual charges. Oxidation states for covalent compounds are a  bookkeeping method to keep track of electrons in a reaction. 22. Reference the Problem Solving Strategy box in Section 4.10 of the text for the steps involved in balancing redox reactions by oxidation states. The key to the oxidation states method is to  balance the electrons gained by the species reduced with the number of electrons lost from the species oxidized. This is done by assigning oxidation states and, from the change in oxidation states, determining the coefficients necessary to balance electrons gained with electrons lost. After the loss and gain of electrons is balanced, the remainder of the equation is balanced by inspection. Exercises   Aqueous Solutions: Strong and Weak Electrolytes  23. a. NaBr(s)   Na + (aq) + Br  - (aq) b. MgCl 2 (s)   Mg 2+ (aq) + 2 Cl  (aq) Your drawing should show equal Your drawing should show twice the number of Na +  and Br  -  ions. number of Cl   ions as Mg 2+  ions.  Na + Br  - Br  -  Na +  Na + Br  - Cl - Mg 2+ Mg 2+ Cl - Cl - Cl - Cl - Cl - Mg 2+  96 CHAPTER 4 SOLUTION STOICHIOMETRY c. Al(NO 3 ) 3 (s)   Al 3+ (aq) + 3 NO 3  (aq) d. (NH 4 ) 2 SO 4 (s)   2 NH 4+ (aq) + SO 42  (aq) For e-i, your drawings should show equal numbers of the cations and anions present because each salt is a 1 : 1 salt. The ions present are listed in the following dissolution reactions. e. NaOH(s)   Na + (aq) + OH  (aq) f. FeSO 4 (s)   Fe 2+ (aq) + SO 42  (aq) g. KMnO 4 (s)   K  + (aq) + MnO 4   (aq) h. HClO 4 (aq)   H + (aq) + ClO 4  (aq) i. NH 4 C 2 H 3 O 2 (s)   NH 4+ (aq) + C 2 H 3 O 2  (aq) 24. a. Ba(NO 3 ) 2 (aq)   Ba 2+ (aq) + 2 NO 3  (aq); picture iv represents the Ba 2+  and NO 3    ions  present in Ba(NO 3 ) 2 (aq).  b. NaCl(aq)   Na + (aq) + Cl  (aq); picture ii represents NaCl(aq). c. K  2 CO 3 (aq)   2 K  + (aq) + CO 32  (aq); picture iii represents K  2 CO 3 (aq). d. MgSO 4 (aq)   Mg 2+ (aq) + SO 42  (aq); picture i represents MgSO 4 (aq). HNO 3 (aq) → H + (aq) + NO 3  (aq). Picture ii best represents the strong acid HNO 3 . Strong acids are strong electrolytes. HC 2 H 3 O 2  only partially dissociates in water; acetic acid is a weak electrolyte. None of the pictures represent weak electrolyte solutions; they all are representations of strong electrolytes. 25. CaCl 2 (s) → Ca 2+ (aq) + 2 Cl  (aq) 26. MgSO 4 (s) → Mg 2+ (aq) + SO 42  (aq); NH 4  NO 3 (s) → NH 4+ (aq) + NO 3  (aq) Solution Concentration: Molarity 27. a. 5.623 g NaHCO 3  × 33  NaHCOg01.84  NaHCOmol1  = 6.693 × 2 10  mol NaHCO 3    M   = LmL1000mL0.250 mol10693.6  2     = 0.2677  M   NaHCO 3   SO 42- SO 42-  NH 4+  NH 4+  NH 4+ SO 42-  NH 4+  NH 4+  NH 4+ Al 3+  NO 3-  NO 3-  NO 3-  NO 3- Al 3+  NO 3-  NO 3-  NO 3- Al 3+  NO 3-  NO 3-

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