Chapter04rw-final

Chapter04rw-final - 4 4.1 4.2 Aqueous Reactions and...

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Unformatted text preview: 4 4.1 4.2 Aqueous Reactions and Solution Stoichiometry Visualizing Concepts Analyze. Correlate the formula of the solute with the charged spheres in the diagrams. Plan. Determine the electrolyte properties of the solute and the relative number of cations, anions, or neutral molecules produced when the solute dissolves. Solve. Li 2 SO 4 is a strong electrolyte, a soluble ionic solid that dissociates into separate Li + and SO 4 2 – when it dissolves in water. There are twice as many Li + cations as SO 4 2 – anions. Diagram (c) represents the aqueous solution of a 2:1 electrolyte. Although CH 3 OH and HCl are both molecular compounds, HCl is an acid and strong electrolyte. Strong electrolytes exist in solution almost completely as ions, so an aqueous HCl solution conducts electricity. CH 3 OH is a nonelectrolyte that exists as neutral molecules in aqueous solution. Since there are no charge carriers, aqueous solutions of nonelectrolytes such as CH 3 OH do not conduct electricity. Analyze/Plan. Correlate the neutral molecules, cations, and anions in the diagrams with the definitions of strong, weak, and nonelectrolytes. Solve. (a) (b) (c) 4.4 AX is a nonelectrolyte, because no ions form when the molecules dissolve. AY is a weak electrolyte because a few molecules ionize when they dissolve, but most do not. AZ is a strong electrolyte because all molecules break up into ions when they dissolve. 4.3 The brightness of the bulb in Figure 4.2 is related to the number of ions per unit volume of solution. If 0.1 M HC 2 H 3 O 2 has about the same brightness of 0.001 M HBr, the two solutions have about the same number of ions. Since 0.1 M HC 2 H 3 O 2 has 100 times more solute than 0.001 M HBr, HBr must be dissociated to a much greater extent than HC 2 H 3 O 2 . HBr is one of the few molecular acids that is a strong electrolyte. HC 2 H 3 O 2 is a weak electrolyte; if it were a nonelectrolyte, the bulb in Figure 4.2 wouldn’t glow. Analyze. From the names and/or formulas of 3 substances determine their electrolyte and solubilities properties. Plan. Determine whether the substance is molecular or ionic. If it is molecular, is it a weak acid or base and thus a weak electrolyte, or a nonelectrolyte? If it is ionic, is it soluble? 4.5 77 4 Aqueous Reactions Solutions to Exercises Solve. Glucose is a molecular compound that is neither a weak acid nor a weak base. It is a nonelectrolyte that dissolves to produce a nonconducting solution; it is solid C. NaOH is ionic; the anion is OH – . According to Table 4.1, most hydroxides are insoluble, but NaOH is one of the soluble ones. NaOH is a strong electrolyte that dissolves to form a conducting solution; it is solid A. AgBr is ionic; the anion is Br – . According to Table 4.1, most bromides are soluble, but AgBr is one of the insoluble ones; it is solid B. Check. We know by elimination that AgBr is solid B, which we verified by solubility rules. 4.6 Certain pairs of ions form precipitates because their attraction is so strong that they cannot be surrounded and separated by solvent molecules. That is, the attraction between solute particles is greater than the stabilization offered by interaction of individual ions with solvent molecules. Analyze. Given the formulas of some ions, determine whether these ions ever form precipitates in aqueous solution. Plan. Use Table 4.1 to determine if the given ions can form precipitates. If not, they will always be spectator ions. Solve. (a) (b) (c) (d) (e) Cl – can form precipitates with Ag + , Hg 2 2 +, Pb 2 + . NO 3 – never forms precipitates, so it is always a spectator. NH 4 + never forms precipitates, so it is always a spectator. S 2 – usually forms precipitates. SO 4 2 – usually forms precipitates. 4.7 Check. NH 4 + is a soluble exception for sulfides, phosphates, and carbonates, which usually form precipitates, so all rules indicate that it is a perpetual spectator. 4.8 Use the difference in reactivities with SO 4 2 – to identify Pb 2 +(aq) and Mg 2 +(aq). Test a portion of each solution with H 2 SO 4 (aq). Pb 2 +(aq) is an exception to the soluble sulfates rule, so Pb(NO 3 ) 2 (aq) will form a precipitate, while Mg(NO 3 ) 2 (aq) will not. In a redox reaction, one reactant loses electrons and a different reactant gains electrons; electrons are transferred. Acids ionize in aqueous solution to produce (donate) hydrogen ions (H+, protons). Bases are substances that react with or accept protons (H+). In an acid-base reaction, protons are transferred from an acid to a base. We characterize redox reactions by tracking electron transfer using oxidation numbers. We characterize acid-base reactions by tracking H+ (proton) transfer via molecular formulas of reactants and products. Concentration is a ratio of amount of solute to amount of solution or solvent. Thus there are two ways to double the concentration of a solution: double the amount of solute, keeping volume constant or reduce the volume of solution by half, keeping the amount of solute the same. 4.9 4.10 Electrolytes 4.11 No. Electrolyte solutions conduct electricity because the dissolved ions carry charge through the solution (from one electrode to the other). 78 4 Aqueous Reactions 4.12 Solutions to Exercises When CH 3 OH dissolves, neutral CH 3 OH molecules are dispersed throughout the solution. These electrically neutral particles do not carry charge and the solution is nonconducting. When CH 3 CO OH dissolves, mostly neutral molecules are dispersed throughout the solution. A few of the dissolved molecules ionize to form H + (aq) and CH 3 CO O – (aq). These few ions carry some charge and the solution is weakly conducting. Although H 2 O molecules are electrically neutral, there is an unequal distribution of electrons throughout the molecule. There are more electrons near O and fewer near H, giving the O end of the molecule a partial negative charge and the H end of the molecule a partial positive charge. Ionic compounds are composed of positively and negatively charged ions. The partially positive ends of H 2 O molecules are attracted to the negative ions (anions) in the solid, while the partially negative ends are attracted to the positive ions (cations). Thus, both cations and anions in an ionic solid are surrounded and separated (dissolved) by H 2 O molecules. Ions are hydrated when they are surrounded by H 2 O molecules in aqueous solution. Analyze/Plan. Given the solute formula, determine the separate ions formed upon dissociation. Solve. (a) (b) (c) (d) ZnCl 2 (aq) → Zn 2 +(aq) + 2Cl – (aq) HNO 3 (aq) → H + (aq) + NO 3 – (aq) (NH 4 ) 2 SO 4 (aq) → 2NH 4 + (aq) + SO 4 2 –(aq) Ca(OH) 2 (aq) → Ca 2 +(aq) + 2OH – (aq) MgI 2 (aq) → Mg 2 +(aq) + 2I – (aq) Al(NO 3 ) 3 (aq) → Al 3 +(aq) + 3NO 3 – (aq) HClO 4 (aq) → H + (aq) + ClO 4 – (aq) NaCH 3 CO O(aq) → N a + (aq) + CH 3 COO – (aq) 4.13 4.14 4.15 4.16 (a) (b) (c) (d) 4.17 Analyze/Plan. Apply the definition of a weak electrolyte to HCOOH. Solve. When HCOOH dissolves in water, neutral HCOOH molecules, H + ions and HCO O – ions are all present in the solution. HCOOH (aq) ⇌H + (aq) + HCOO – (aq) 4.18 (a) acetone (nonelectrolyte): CH 3 COCH 3 (aq) molecules only; hypochlorous acid (weak electrolyte): HClO(aq) molecules, H + (aq), ClO – (aq); ammonium chloride (strong electrolyte): NH 4 + (aq), Cl – (aq) NH 4 Cl, 0.2 mol solute particles; HClO, between 0.1 and 0.2 mol particles; CH 3 OCH 3 , 0.1 mol of solute particles (b) Precipitation Reactions and Net Ionic Equations 4.19 Analyze. Given: formula of compound. Find: solubility. Plan. Follow the guidelines in Table 4.1, in light of the anion present in the compound and notable exceptions to the “rules.” Solve. 79 4 Aqueous Reactions (a) (b) (c) (d) (e) 4.20 NiCl 2 : soluble Ag 2 S: insoluble Cs 3 PO 4 : soluble (Cs + is an alkali metal cation) SrCO 3 : insoluble Solutions to Exercises PbSO 4 : insoluble, Pb 2 + is an exception to soluble sulfates According to Table 4.1: (a) (b) (c) (d) (e) Ni(OH) 2 : insoluble PbBr 2 : insoluble; Ba(NO 3 ) 2 : soluble AlPO 4 : insoluble AgC H 3 COO : soluble 4.21 Analyze. Given: formulas of reactants. Find: balanced equation including precipitates. Plan. Follow the logic in Sample Exercise 4.3. Solve. In each reaction, the precipitate is in bold type. (a) (b) (c) Na 2 CO 3 (aq) + 2AgNO 3 (aq) → Ag 2 CO 3 (s) + 2NaNO 3 (aq) No precipitate (all nitrates and most sulfates are soluble). FeSO 4 (aq) + Pb(NO 3 ) 2 (aq) → PbSO 4 (s) + Fe(NO 3 ) 2 (aq) 4.22 In each reaction, the precipitate is in bold type. (a) (b) (c) Ni(NO 3 ) 2 (aq) + 2NaOH(aq) → Ni(OH) 2 (s) + 2NaNO 3 (aq) No precipitate, and, therefore, no reaction. There is no chemical change to any of the reactant ions. Na 2 S(aq) + CuCH 3 COO(aq) → CuS(s) + 2NaCH 3 CO O (aq) 4.23 Analyze/Plan. Follow the logic in Sample Exercise 4.4. From the complete ionic equation, identify the ions that don’t change during the reaction; these are the spectator ions. Solve. (a) (b) (c) 2Na + (aq) + CO 3 2 –(aq) + Mg 2 +(aq) + SO 4 2 –(aq) → MgCO 3 (s) + 2Na + (aq) + SO 4 2 –(aq) Spectators: Na + , SO 4 2 – Pb 2 +(aq) + 2NO 3 – (aq) + 2Na + (aq) + S 2 –(aq) → PbS(s) + 2Na + (aq) + 2NO 3 – (aq) Spectators: Na + , NO 3 – 6NH 4 + (aq) + 2PO 4 3 –(aq) + 3Ca 2 +(aq) + 6Cl – (aq) → Ca 3 (PO 4 ) 2 (s) + 6NH 4 + (aq) + 6Cl – (aq) Spectators: NH 4 + , Cl – 4.24 Spectator ions are those that do not change during reaction. (a) (b) (c) 2Cr 3 +(aq) + 3CO 3 2 –(aq) → Cr 2 (CO 3 ) 3 (s); spectators: NH 4 + , SO 4 2 – Ba2+ (aq) + SO 4 2 –(aq) → BaSO 4 (s); spectators: K + , NO 3 – Fe2 +(aq) + 2OH – (aq) → Fe(OH) 2 (s); spectators: K + , NO 3 – 80 4 Aqueous Reactions 4.25 Solutions to Exercises Analyze. Given: reactions of unknown with HBr, H 2 SO 4 , NaOH. Find: The unknown contains a single salt. Is K + or Pb 2 + or Ba 2 + present? Plan. Analyze solubility guidelines for Br – , SO 4 2 – and OH – and select the cation that produces a precipitate with each of the anions. Solve. K + forms no precipitates with any of the anions. BaSO 4 is insoluble, but BaCl 2 and Ba(OH) 2 are soluble. Since the unknown forms precipitates with all three anions, it must contain Pb 2 +. Check. PbBr 2 , PbSO 4 , and Pb(OH) 2 are all insoluble according to Table 4.1, so our process of elimination is confirmed by the insolubility of the Pb 2 + compounds. 4.26 Br – and NO 3 – can be ruled out because the Ba 2 + salts are soluble. (Actually all NO 3 – salts are soluble.) CO 3 2 – forms insoluble salts with the three cations given; it must be the anion in question. Analyze. Given: three possible salts in an unknown solution react with Ba(NO 3 ) 2 and then NaCl. Find: Can the results identify the unknown salt? Do the three possible unknowns give distinctly different results with Ba(NO 3 ) 2 and NaCl? Plan. Using Table 4.1, determine whether each of the possible unknowns will form a precipitate with Ba(NO 3 ) 2 and NaCl. Solve. Compound AgNO 3 (aq) CaCl 2 (aq) Al 2 (SO 4 ) 3 Ba(NO 3 ) 2 result no ppt no ppt BaSO 4 ppt NaCl result AgCl ppt no ppt no ppt 4.27 This sequence of tests would definitively identify the contents of the bottle, because the results for each compound are unique. 4.28 (a) Pb(CH 3 COO) 2 (aq) + Na 2 S(aq) → PbS(s) + 2NaCH 3 COO(aq) net ionic: Pb 2 +(aq) + S 2 –(aq) → PbS Pb(CH 3 COO) 2 (aq) + CaCl 2 (aq) → (PbCl 2 )s + Ca(CH 3 CO O) 2 (aq) net ionic: Pb 2 +(aq) + 2Cl – (aq) → PbCl 2 (s) Na 2 S(aq) + CaCl 2 (aq) → CaS(aq) + 2NaCl(aq) net ionic: no reaction (b) Spectator ions: Na + , Ca 2 +, CH 3 CO O – Acid-Base Reactions 4.29 Analyze. Given: solute and concentration of three solutions. Find: solution with greatest concentration of solvated protons. Plan: See Sample Exercise 4.6. Determine whether solutes are strong or weak acids or bases, or nonelectrolytes. For solutions of equal concentration, strong acids will have greatest concentration of solvated protons. Take varying concentration into consideration when evaluating the same class of solutions. 81 4 Aqueous Reactions Solutions to Exercises Solve. LiOH is a strong base, HI is a strong acid, CH 3 OH is a molecular compound and nonelectrolyte. The strong acid HI will have the greatest concentration of solvated protons. Check. The solution concentrations weren’t needed to answer the question. 4.30 NH 3 (aq) is a weak base, while KOH and Ca(OH) 2 are strong bases. NH 3 (aq) is only slightly ionized, so even 0.6 M NH 3 is less basic than 0.150 M KOH. Ca(OH) 2 has twice as many OH – per moles as KOH, so 0.100 M Ca(OH) 2 is more basic than 0.150 M KOH. The most basic solution is 0.100 M Ca(OH) 2 . (a) (b) (c) A monoprotic acid has one ionizable (acidic) H and a diprotic acid has two. A strong acid is completely ionized in aqueous solution, whereas only a fraction of weak acid molecules are ionized. An acid is an H + donor, a substance that increases the concentration of H + in aqueous solution. A base is an H + acceptor and thus increases the concentration of OH – in aqueous solution. NH 3 produces OH – in aqueous solution by reacting with H 2 O (hydrolysis): NH3(aq) + H2O(l) ⇌ NH4 + (aq) + OH– (aq). The OH – causes the solution to be basic. The term “weak” refers to the tendency of HF to dissociate into H + and F – in aqueous solution, not its reactivity toward other compounds. H 2 SO 4 is a diprotic acid; it has two ionizable hydrogens. The first hydrogen completely ionizes to form H + and HSO 4 – , but HSO 4 – only partially ionizes into H + and SO 4 2 – (HSO 4 – is a weak electrolyte). Thus, an aqueous solution of H 2 SO 4 contains a mixture of H + , HSO 4 – and SO 4 2 –, with the concentration of HSO 4 – greater than the concentration of SO 4 2 –. 4.31 4.32 (a) (b) (c) 4.33 Strong acids are completely ionized, while weak acids are partially or slightly ionized. As strong acids, HCl, HBr and HI are completely ionized in aqueous solution. They exist as H+(aq) and Cl–(aq), H+(aq) and Br–(aq) and H+(aq) and I–(aq), respectively. There are no neutral HX molecules in these solutions. As a weak acid, HF exists as a mixture of H+(aq), F–(aq) and HF(aq). There are at least as many neutral HF molecules as anions or cations. For equal solution concentrations, HCl, HBr and HI will produce a bright light like Figure 4.2(c), while HF will produce a dim light like Figure 4.2(b). All soluble ionic hydroxides from Table 4.1 are listed as strong bases in Table 4.2. Insoluble hydroxides like Cd(OH) 2 are not listed as strong bases. “Insoluble” means that less than 1% of the base molecules exist as separated ions and are dissolved. Thus, insoluble hydroxide salts produce too few OH – (aq) to be considered strong bases. Analyze. Given: chemical formulas. Find: classify as acid, base, salt; strong, weak, or nonelectrolyte. Plan. See Table 4.3. Ionic or molecular? Ionic, soluble: OH – , strong base and strong electrolyte; otherwise, salt, strong electrolyte. Molecular: NH 3 , weak base and weak electrolyte; H-first, acid; strong acid (Table 4.2), strong electrolyte; otherwise weak acid and weak electrolyte. Solve. 4.34 4.35 82 4 Aqueous Reactions (a) (b) (c) (d) 4.36 NaClO 4 : salt, entirely ions (strong electrolyte) Ba(OH) 2 : base, entirely ions (strong electrolyte) Solutions to Exercises HF: acid, mixture of ions and molecules (weak electrolyte) CH 3 CN: none of the above, entirely molecules (nonelectrolyte) Since the solution does conduct some electricity, but less than an equimolar NaCl solution (a strong electrolyte), the unknown solute must be a weak electrolyte. The weak electrolytes in the list of choices are NH 3 and H 3 PO 3 ; since the solution is acidic, the unknown must be H 3 PO 3 . Analyze. Given: chemical formulas. Find: electrolyte properties. Plan. In order to classify as electrolytes, formulas must be identified as acids, bases, or salts as in Solution 4.35. Solve. (a) (b) (c) (d) (e) H 2 SO 3 : H first, so acid; not in Table 4.2, so weak acid; therefore, weak electrolyte C 2 H 5 OH: not acid, not ionic (no metal cation), contains OH group, but not as anion so not a base; therefore, nonelectrolyte NH 3 : common weak base; therefore, weak electrolyte KClO 3 : ionic compound, so strong electrolyte Cu(NO 3 ) 2 : ionic compound, so strong electrolyte HClO 4 : strong CH 3 OCH 3 : non (b) HNO 3 : strong (e) CoSO 4 : strong Solve. (c) NH 4 Cl: strong (f) C 1 2H 2 2O 1 1: non 4.37 4.38 (a) (d) 4.39 Plan. Follow Sample Exercise 4.7. (a) 2HBr(aq) + Ca(OH) 2 (aq) → CaBr 2 (aq) + 2H 2 O(l) H+(aq) + OH – (aq) → H 2 O(l) (b) Cu(OH) 2 (s) + 2HClO 4 (aq) → Cu(ClO 4 ) 2 (aq) + 2H 2 O(l) Cu(OH) 2 (s) + 2H + (aq) → 2H 2 O(l) + Cu 2 +(aq) (c) Al(OH) 3 (s) + 3HNO 3 (aq) → Al(NO 3 ) 3 (aq) + 3H 2 O(l) Al(OH) 3 (s) + 3H + (aq) → 3H 2 O(l) + Al 3 +(aq) 4.40 (a) HC 2 H 3 O 2 (aq) + KOH(aq) → KC 2 H 3 O 2 (aq) + H 2 O(l) HC 2 H 3 O 2 (aq) + OH – (aq) → C 2 H 3 O 2 – (aq) H 2 O(l) (b) Cr(OH) 3 (s) + 3HNO 3 (aq) → Cr(NO 3 ) 3 (aq) + 3H 2 O(l) Cr(OH) 3 (s) + 3H + (aq) → 3H 2 O(l) + Cr 3 +(aq) (c) Ca(OH) 2 (aq) + 2HClO(aq) → Ca(ClO) 2 (aq) + 2H 2 O(l) HClO(aq) + OH – (aq) → ClO – (aq) + H 2 O(l) 4.41 Analyze. Given: names of reactants. Find: gaseous products. Plan. Write correct chemical formulas for the reactants, complete and balance the metathesis reaction, and identify either H 2 S or CO 2 products as gases. Solve. 83 4 Aqueous Reactions (a) CdS(s) + H 2 SO 4 (aq) → CdSO 4 (aq) + H 2 S(g) CdS(s) + 2H + (aq) → H 2 S(g) + Cd 2 +(aq) (b) Solutions to Exercises MgCO 3 (s) + 2HClO 4 (aq) → Mg(ClO 4 ) 2 (aq) + H 2 O(l) + CO 2 (g) MgCO 3 (s) + 2H + (aq) → H 2 O(l) + CO 2 (g) + Mg 2 +(aq) 4.42 (a) (b) FeO(s) + 2H + (aq) → H 2 O(l) + Fe 2 +(aq) NiO(s) + 2H + (aq) → H 2 O(l) + Ni 2 +(aq) 4.43 Analyze. Given the formulas or names of reactants, write balanced molecular and net ionic equations for the reactions. Plan. Write correct chemical formulas for all reactants. Predict products of the neutralization reactions by exchanging ion partners. Balance the complete molecular equation, identify spectator ions by recognizing strong electrolytes, write the corresponding net ionic equation (omitting spectators). Solve. (a) CaCO 3 (s) + 2HNO 3 (aq) → Ca(NO 3 ) 2 (aq) + H 2 O(l) + CO 2 (g) 2H + (aq) + CaCO 3 (s) → H 2 O(l) + CO 2 (g) + Ca 2 +(aq) (b) FeS(s) + 2HBr(aq) → FeBr 2 (aq) + H 2 S(g) 2H + (aq) + FeS(s) → H 2 S(g) + Fe 2 +(aq) 4.44 K 2 O(aq) + H 2 O(l) → 2KOH(aq), molecular; O 2 –(aq) + H 2 O(l) → 2OH – (aq), net ionic base: (H + ion acceptor) O 2 –(aq); acid: (H + ion donor) H 2 O(aq); spectator: K + Oxidation-Reduction Reactions 4.45 (a) (b) 4.46 In terms of electron transfer, oxidation is the loss of electrons by a substance, and reduction is the gain of electrons (LEO says GER). Relative to oxidation numbers, when a substance is oxidized, its oxidation number increases. When a substance is reduced, its oxidation number decreases. Oxidation and reduction can only occur together, not separately. When a metal reacts with oxygen, the metal atoms lose electrons and the oxygen atoms gain electrons. Free electrons do not exist under normal conditions. If electrons are lost by one substance they must be gained by another, and vice versa. Analyze. Given the labeled periodic chart, determine which region is most readily oxidized and which is most readily reduced. Plan. Review the definition of oxidation and apply it to the properties of elements in the indicated regions of the chart. Solve. Oxidation is loss of electrons. Elements readily oxidized form positive ions; these are metals. Elements not readily oxidized tend to gain electrons and form negative ions; these are nonmetals. Elements in regions A, B, and C are metals, and their ease of oxidation is shown in Table 4.5. Metals in region A, Na, Mg, K, and Ca are most easily oxidized. Elements in region D are nonmetals and are least easily oxidized. 4.47 84 4 Aqueous Reactions 4.48 Solutions to Exercises Elements (metals) from Table 4.5 in region A include Na, Mg, K, and Ca; those from region C are Hg and Au. Let’s consider K and Au. Since metals from region A are more readily oxidized than those from region C, K will be oxidized to K + and Au 3 + will be reduced to Au in the redox reaction. (Choose Au 3 + because it is the Au ion shown in Table 4.5.) In a balanced redox reaction, the number of electrons lost must equal the number of electrons gained. Since K loses 1 electron in forming K + , while Au 3 + gains 3 electrons when forming Au, 3 K atoms must be oxidized for every 1 Au 3 + that is reduced. This relationship dictates the coefficients in the balanced redox reaction. 3K(s) + Au3 + (aq) ⇌ A u(s) + 3 K + (aq) 4.49 Analyze. Given the chemical formula of a substance, determine the oxidation number of a particular element in the substance. Plan. Follow the logic in Sample Exercise 4.8. Solve. (a) +4 +4 (b) +4 (b) +2 (c) +7 (c) +3 (d) +1 (d) –2 (e) 0 (e) +3 (f) –1 (O 2 2 – is peroxide ion) (f) +6 4.50 4.51 (a) Analyze. Given: chemical reaction. Find: element oxidized or reduced. Plan. Assign oxidation numbers to all species. The element whose oxidation number becomes more positive is oxidized; the one whose oxidation number decreases is reduced. Solve. (a) (b) (c) (d) N2(g) [N, 0] → 2NH3(g) [N, −3], N is reduced; 3H2(g) [H, 0] → 2NH3(g) [H, +1], H is oxidized. Fe 2 + → Fe, Fe is reduced; Al → Al 3 +, Al is oxidized Cl 2 → 2Cl, Cl is reduced; 2I – → I 2 , I is oxidized S 2 – → SO 4 2 –(S, +6), S is oxidized; H 2 O 2 (O, –1) → H 2 O (O, –2); O is reduced acid-base reaction oxidation-reduction reaction; Fe is reduced, C is oxidized precipitation reaction oxidation-reduction reaction; Zn is oxidized, N is reduced 4.52 (a) (b) (c) (d) 4.53 Analyze. Given: reactants. Find: balanced molecular and net ionic equations. Plan. Metals oxidized by H + form cations. Predict products by exchanging cations and balance. The anions are the spectator ions and do not appear in the net ionic equations. Solve. (a) Mn(s) + H 2 SO 4 (aq) → MnSO 4 (aq) + H 2 (g); Mn(s) + 2H + (aq) → Mn 2 +(aq) + H 2 (g) Products with the metal in a higher oxidation state are possible, depending on reaction conditions and acid concentration. (b) (c) 2Cr(s) + 6HBr(aq) → 2CrBr 3 (aq) + 3H 2 (g); 2Cr(s) + 6H + (aq) → 2Cr 3 +(aq) + 3H 2 (g) Sn(s) + 2HCl(aq) → SnCl 2 (aq) + H 2 (g); Sn(s) + 2H + (aq) → Sn 2 +(aq) + H 2 (g) 85 4 Aqueous Reactions (d) 4.54 (a) (b) Solutions to Exercises 2Al(s) + 6HCOOH(aq) → 2Al(HCOO)3(aq) + 3H 2 (g); 2Al(s) + 6HCOOH(aq) → 2Al 3 +(aq) + 6HCOO– (aq) + 3H 2 (g) 2HCl(aq) + Ni(s) → NiCl 2 (aq) + H 2 (g); Ni(s) + 2H + (aq) → Ni 2 +(aq) + H 2 (g) H 2 SO 4 (aq) + Fe(s) → FeSO 4 (aq) + H 2 (g); Fe(s) + 2H + (aq) → Fe 2 +(aq) + H 2 (g) Products with the metal in a higher oxidation state are possible, depending on reaction conditions and acid concentration. (c) (d) 2HBr(aq) + Mg(s) → MgBr 2 (aq) + H 2 (g); Mg(s) + 2H + (aq) → Mg 2 +(aq) + H 2 (g) 2CH3COOH(aq) + Zn(s) → ZnCH3COO(aq) + H 2 (g); Zn(s) + 2CH3COOH(aq) → Zn 2 +(aq) + 2CH3COO– (aq) + H 2 (g) 4.55 Analyze. Given: a metal and an aqueous solution. Find: balanced equation. Plan. Use Table 4.5. If the metal is above the aqueous solution, reaction will occur; if the aqueous solution is higher, NR. If reaction occurs, predict products by exchanging cations (a metal ion or H + ), then balance the equation. Solve. (a) (b) (c) (d) (e) Fe(s) + Cu(NO 3 ) 2 (aq) → Fe(NO 3 ) 2 (aq) + Cu(s) Zn(s) + MgSO 4 (aq) → NR Sn(s) + 2HBr(aq) → SnBr 2 (aq) + H 2 (g) H 2 (g) + NiCl 2 (aq) → NR 2Al(s) + 3CoSO 4 (aq) → Al 2 (SO 4 ) 3 (aq) + 3Co(s) Mn(s) + NiCl 2 (aq) → MnCl 2 (aq) + Ni(s) Cu(s) + Cr(C 2 H 3 O 2 )(aq) → NR 2Cr(s) + 3NiSO 4 (aq) → Cr 2 (SO 4 ) 3 (aq) + 3Ni(s) Pt(s) + HBr(aq) → NR H 2 (g) + CuCl 2 (aq) → Cu(s) + 2HCl(aq) i. Zn(s) + Cd 2 +(aq) → Cd(s) + Zn 2 +(aq) 4.56 (a) (b) (c) (d) (e) 4.57 (a) ii. Cd(s) + Ni 2 +(aq) → Ni(s) + Cd 2 +(aq) (b) According to Table 4.5, the most active metals are most easily oxidized, and Zn is more active than Ni. Observation (i) indicates that Cd is less active than Zn; observation (ii) indicates that Cd is more active than Ni. Cd is between Zn and Ni on the activity series. Place an iron strip in CdCl 2 (aq). If Cd(s) is deposited, Cd is less active than Fe; if there is no reaction, Cd is more active than Fe. Do the same test with Co if Cd is less active than Fe or with Cr if Cd is more active than Fe. Br 2 + 2NaI → 2NaBr + I 2 indicates that Br 2 is more easily reduced than I 2 . Cl 2 + 2NaBr → 2NaCl + Br 2 shows that Cl 2 is more easily reduced than Br 2 . (c) 4.58 (a) 86 4 Aqueous Reactions (b) Solutions to Exercises The order for ease of reduction is Cl 2 > Br 2 > I 2 . Conversely, the order for ease of oxidation is I – > Br – > Cl – . Since the halogens are nonmetals, they tend to form anions when they react chemically. Nonmetallic character decreases going down a family and so does the tendency to gain electrons during a chemical reaction. Thus, the ease of reduction of the halogen, X 2 , decreases going down the family and the ease of oxidation of the halide, X – , increases going down the family. Cl 2 + 2KI → 2KCl + I 2 ; Br 2 + LiCl → no reaction (c) Solution Composition; Molarity 4.59 (a) Concentration is an intensive property; it is the ratio of the amount of solute present in a certain quantity of solvent or solution. This ratio remains constant regardless of how much solution is present. The term 0.50 mol HCl defines an amount (~18 g) of the pure substance HCl. The term 0.50 M HCl is a ratio; it indicates that there are 0.50 mol of HCl solute in 1.0 liter of solution. This same ratio of moles solute to solution volume is present regardless of the volume of solution under consideration. The concentration of the remaining solution is unchanged, assuming the original solution was thoroughly mixed. Molar concentration is a ratio of moles solute to liters solution. Although there are fewer moles solute remaining in the flask, there is also less solution volume, so the ratio of moles solute/solution volume remains the same. The concentration of the remaining solution is increased. The moles solute remaining in the flask are unchanged, but the solution volume decreases after evaporation. The ratio of moles solute/solution volume increases. The second solution is five times as concentrated as the first. An equal volume of the more concentrated solution will contain five times as much solute (five times the number of moles and also five times the mass) as the 0.50 M solution. Thus, the mass of solute in the 2.50 M solution is 5 × 4.5 g = 22.5 g. Mathematically: 2.50 mol solute x grams solute 1 L solution = 4.5 g solute 0.50 mol solute 1 L solution x g solute 2.50 mol solute = ; 5.0(4.5 g solute) = 23 g solute 0.50 mol solute 4.5 g solute (b) 4.60 (a) (b) (c) The result has 2 sig figs; 22.5 rounds to 23 g solute. 4.61 Analyze/Plan. Follow the logic in Sample Exercises 4.11 and 4.13. Solve. 87 4 Aqueous Reactions (a) M= Solutions to Exercises mol solute 0.0250 mol NH 4 Cl 1000 mL ; × = 0.0500 M NH 4 Cl L solution 500 mL 1L Check. (0.025 × 2) ≈ 0.050 M (b) mol = M × L; 2.50 mol HNO 3 × 0.0500 L = 0.125 mol HNO 3 1L Check. (2.5 × .05) ≈ 0.125 mol (c) L= mol 0.275 mol KOH ; = 0.183 L or 183 mL of 1.50 M KOH M 1.50 mol KOH/L Check. (0.275/1.5) is greater than 0.15 and less than 0.20, ≈ 0.18 L. 4.62 (a) (b) (c) 4.63 M= 1 mol Na 2 SO 4 mol solute 0.750 g Na 2 SO 4 ; × = 6.21 × 10 − 3 M Na 2 SO 4 L solution 0.850 L 142.04 g Na 2 SO 4 0.0475 mol KMnO 4 × 0.250 L = 1.19 × 10 −2 mol KMnO 4 1L mol = M × L; L= mol 0.250 mol HCl ; = 2.16 × 10 − 2 L or 21.6 mL M 11.6 mol HCl/L Analyze. Given molarity, M, and volume, L, find mass of Na + (aq) in the blood. Plan. Calculate moles Na + (aq) using the definition of molarity: M = mol ; mol = M × L. L Calculate mass Na+(aq) using the definition moles: mol = g/MM/g = mol × MM. (MM is the symbol for molar mass in this manual.) Solve. 23.0 g Na + 0.135 mol × 5.0 L × = 15.525 = 16 g Na + (aq) L mol Na + Check. Since there are more than 0.1 mol/L and we have 5.0 L, there should be more than half a mol (11.5 g) of Na + . The calculation agrees with this estimate. 4.64 Calculate the mol of Na + at the two concentrations; the difference is the mol NaCl required to increase the Na + concentration to the desired level. 0.118 mol × 4.6 L = 0.5428 = 0.54 mol Na + L 0.138 mol × 4.6 L = 0.6348 = 0.63 mol Na + L (0.6348 – 0.5428) = 0.092 = 0.09 mol NaCl (2 decimal places and 1 sig fig) 0.092 mol NaCl × 58.5 g NaCl = 5.38 = 5 g NaCl mol 4.65 Analyze. Given: g alcohol/100 mL blood; molecular formula of alcohol. Find: molarity (mol/L) of alcohol. Plan. Use the molar mass (MM) of alcohol to change (g/100 mL) to (mol/100 mL) then mL to L. Solve. MM of alcohol = 2(12.01) + 6(1.1008) + 1(16.00) = 46.07 g alcohol/mol 88 4 Aqueous Reactions BAC = 4.66 0.08 g alcohol 100 mL blood × 1 mol alcohol 46.07 g alcohol × 1000 mL 1L Solutions to Exercises = 0.0174 = 0.02 M alcohol Analyze. Given: BAC (definition from Exercise 4.65), vol of blood. Find: mass alcohol in bloodstream. Plan. Change BAC (g/100 mL) to (g/L), then times vol of blood in L. Solve. BAC = 0.10 g/100 mL 0.10 g alcohol 100 mL blood × 1000 mL 1L × 5.0 L blood = 5.0 g alcohol 4.67 Plan. Proceed as in Sample Exercises 4.11 and 4.13. M= g mol ; mol = L M (MM is the symbol for molar mass in this manual.) Solve. (a) 119.0 g KBr 0.175 M KBr × 0.250 L × = 5.21 g KBr 1L 1 mol KBr Check. (0.25 × 120) ≈ 30; 30 × 0.18 ≈ 5.4 g KBr (b) 14.75 g Ca(NO 3 ) 2 × 1 mol Ca(NO 3 ) 2 1 × = 0.06537 M Ca(NO 3 ) 2 1.375 L 164.09 g Ca(NO 3 ) 2 Check. (15/1.5) ≈ 10; 10/160 = 5/80 ≈ 0.06 M Ca(NO 3 ) 2 (c) 2.50 g Na 3 PO 4 × 1 mol Na 3 PO 4 1L 1000 mL × × 163.9 g Na 3 PO 4 1.50 mol Na 3 PO 4 1L = 10.2 mL solution Check. [25/(160 × 1.5)] ≈ 2.5/240 ≈ 1/100 ≈ 0.01 L = 10 mL 4.68 M= g mol (MM is the symbol for molar mass in this manual.) ; mol = MM L 294.2 g K 2 Cr2 O 7 0.488 mol K 2 Cr2 O 7 1L × 50.0 mL × × = 7.18 g K 2 Cr2 O 7 1L 1000 mL 1 mol K 2 Cr2 O 7 4.00 g (NH 4 ) 2 SO 4 × 1 mol (NH 4 ) 2 SO 4 1 1000 mL × × 132.2 g (NH 4 ) 2 SO 4 400. mL 1L (a) (b) = 0.0756 M (NH4)2SO4 1 mol CuSO 4 1L 1000 mL × × = 439 mL solution 159.6 g CuSO 4 0.0250 mol CuSO 4 1L (c) 1.75 g CuSO 4 × 4.69 Analyze. Given: formula and concentration of each solute. Find: concentration of K + in each solution. Plan. Note mol K + /mol solute and compare concentrations or total moles. Solve. 89 4 Aqueous Reactions (a) KCl → K + + Cl – ; 0.20 M KCl = 0.20 M K + Solutions to Exercises K 2 CrO 4 → 2 K + + CrO 4 2 –; 0.15 M K 2 CrO 4 = 0.30 M K + K 3 PO 4 → 3 K + + PO 4 3 –; 0.080 M K 3 PO 4 = 0.24 M K + 0.15 M K 2 CrO 4 has the highest K + concentration. (b) K 2 CrO 4 : 0.30 M K + × 0.0300 L = 0.0090 mol K + K 3 PO 4 : 0.24 M K + × 0.0250 L = 0.0060 mol K + 30.0 mL of 0.15 M K 2 CrO 4 has more K + ions. 4.70 (a) 0.1 M CaCl 2 = 0.2 M Cl – ; 0.15 M KCl = 0.15 M Cl – 0.1 M CaCl 2 has the higher Cl – concentration. (b) (c) 0.1 M KCl has a higher Cl – concentration than 0.080 M LiCl. Total volume does not affect concentration. 0.050 M HCl = 0.050 M Cl – ; 0.020 M CdCl 2 = 0.040 M Cl – 0.050 M HCl has the higher Cl – concentration. 4.71 Analyze. Given: molecular formula and solution molarity. Find: concentration (M) of each ion. Plan. Follow the logic in Sample Exercise 4.12. Solve. (a) NaNO3 → Na+ + NO3 , 0.25 M NaNO3 = 0.25 M Na+, 0.25 M NO3 − − (b) (c) (d) MgSO4 → Mg2+, SO42−, 1.3 × 10−2 M MgSO4 = 1.3 × 10−2 M Mg2+, 1.3 × 10−2 M SO42− C6H12O6 → C6H12O6 (molecular solute); 0.0150 M C6H12O6 = 0.0150 M C6H12O6 Plan. There is no reaction between NaCl and (NH4)2CO3, so this is just a dilution problem, M1V1 = M2V2. Then account for ion stoichiometry. Solve. 45.0 mL + 65.0 mL = 110.0 mL total volume 0.272 M NaCl × 45.0 mL = 0.111 M NaCl; 0.111 M Na+, 0.111 M Cl− 110.0 mL 0.247 M (NH 4 ) 2 CO 3 × 65.0 mL = 0.0146 M (NH4)2CO3 110.0 mL 2 × (0.0146 M) = 0.0292 M NH4+, 0.0146 M CO32− Check. By adding the two solutions (with no common ions or chemical reaction), we have approximately doubled the solution volume, and reduced the concentration of each ion by approximately a factor of two. 90 4 Aqueous Reactions 4.72 (a) Solutions to Exercises Plan. These two solutions have common ions. Find the ion concentration resulting from each solution, then add. Solve. total volume = 42.0 mL + 37.6 mL = 79.6 mL 0.170 M NaOH × 42.0 mL = 0.08970 = 0.0897 M NaOH; 79.6 mL 0.0897 M Na+, 0.0897 M OH− 0.400 M NaOH × 37.6 mL = 0.18894 = 0.189 M NaOH; 79.6 mL 0.189 M Na+, 0.189 M OH− M Na+ = 0.08970 M + 0.18894 M = 0.27864 = 0.2786 M Na+ M OH− = M Na+ = 0.2786 M OH− (b) Plan. No common ions; just dilution. Solve. 44.0 mL + 25.0 mL = 69.0 mL 0.100 M Na 2 SO 4 × 44.0 mL = 0.06377 = 0.0638 M Na2SO4 69.0 mL 2 × (0.06377 M) = 0.1275 = 0.128 M Na+; 0.0638 M SO42− 0.150 M KCl × 25.0 mL = 0.054348 = 0.0543 M KCl 69.0 mL 0.0543 M K+, 0.0543 M Cl− (c) Plan. Calculate concentration of K+ and Cl− due to the added solid. Then sum to get total concentration of Cl−. 3.60 g KCl 75.0 mL so ln 1 mol KCl 74.55 g KCl 1000 mL 1L Solve. × × = 0.6439 = 0.644 M KCl 0.250 M CaCl2; 2(0.250 M) = 0.500 M Cl– total Cl = 0.644 M + 0.500 M = 1.144 M Cl , 0.644 M K+ , 0.250 M Ca2+ 4.73 Analyze/Plan. Follow the logic of Sample Exercise 4.14. Solve. (a) V1 = M 2 V2 /M 1 ; 0.250 M NH 3 × 1000.0 mL = 16.89 = 16.9 mL 14.8 M NH 3 14.8 M NH 3 − − Check. 250/15 ≈ 15 M 91 4 Aqueous Reactions (b) M 2 = M 1 V1 / V2 ; Solutions to Exercises 14.8 M NH 3 × 10.0 mL = 0.296 M NH 3 500 mL Check. 150/500 ≈ 0.30 M 4.74 (a) (b) 4.75 (a) V1 = M 2 V2 / M 1 ; 0.500 M HNO 3 × 0.450 mL 10.0 M HNO 3 = 0.02250 L = 22.5 mL conc. HNO 3 M 2 = M 1 V1 / V2 ; 10.0 M HNO 3 × 25.0 mL 500 mL = 0.500 M HNO 3 Plan/Solve. Follow the logic in Sample Exercise 4.13. The number of moles of 0.250 mol × 0.250 L = 0.06250 = 0.0625 mol sucrose needed is 1L Weigh out 0.0625 mol C 12 H 22 O 11 × 342.3 g C 12 H 22 O 11 1 mol C 12 H 22 O 11 = 21.4 g C 12 H 22 O 11 Add this amount of solid to a 250 mL volumetric flask, dissolve in a small volume of water, and add water to the mark on the neck of the flask. Agitate thoroughly to ensure total mixing. (b) Plan/Solve. Follow the logic in Sample Exercise 4.14. Calculate the moles of solute present in the final 400.0 mL of 0.100 M C 1 2H 2 2O 1 1 solution: 0.100 mol C 12 H 22 O 11 × 0.3500 L = 0.0350 mol C 12 H 22 O 11 1L Calculate the volume of 1.50 M glucose solution that would contain 0.04000 mol C 1 2H 2 2O 1 1: moles C 12 H 22 O 11 = M × L = L = moles/ M ; 0.0350 mol C 12 H 22 O 11 × 0.02333 L × 1000 mL = 23.3 mL 1L 1L = 0.02333 = 0.0233 L 1.50 mol C 12 H 22 O 11 Thoroughly rinse, clean, and fill a 50 mL buret with the 1.50 M C 1 2H 2 2O 1 1. Dispense 23.3 mL of this solution into a 350 mL volumetric container, add water to the mark, and mix thoroughly. (23.3 mL is a difficult volume to measure with a pipette.) 4.76 (a) The amount of AgNO 3 needed is: 0.150 M × 0.1750 L = 0.02625 = 0.263 mol AgNO 3 0.02625 mol AgNO 3 × 169.88 g AgNO 3 1 mol AgNO 3 = 4.4594 = 4.46 g AgNO 3 Add this amount of solid to a 175 mL volumetric container, dissolve in a small amount of water, bring the total volume to exactly 175 mL, and agitate well. (b) Dilute the 3.6 M HNO 3 to prepare 100 mL of 0.50 M HNO 3 . To determine the volume of 3.6 M HNO 3 needed, calculate the moles HNO 3 present in 100 mL of 0.50 M HNO 3 and then the volume of 6.0 M solution that contains this number of moles. 92 4 Aqueous Reactions 0.100 L × 0.50 M = 0.050 mol HNO 3 needed; L= Solutions to Exercises mol 0.050 mol needed ; L 3.6 M HNO 3 = = 0.01389 L = 14 mL M 3.6 M Thoroughly clean, rinse, and fill a buret with the 3.6 M HNO 3 , taking precautions appropriate for working with a relatively concentrated acid. Dispense 14 mL of the 3.6 M acid into a 100 mL volumetric flask, add water to the mark, and mix thoroughly. 4.77 Analyze. Given: density of pure acetic acid, volume pure acetic acid, volume new solution. Find: molarity of new solution. Plan. Calculate the mass of acetic acid, HC 2 H 3 O 2 , present in 20.0 mL of the pure liquid. Solve. 20.00 mL acetic acid × 20.98 g HC2H 3O 2 × 1.049 g acetic acid 1 mL acetic acid = 20.98 g acetic acid 1 mol HC2H 3O 2 = 0.349375 = 0.3494 mol HC2H 3O 2 60.05 g HC2H 3O 2 M = mol/L = 0.349375 mol HC 2 H 3 O 2 = 1.39750 = 1.398 M HC 2 H 3 O 2 0.2500 L solution Check. (20 × 1) ≈ 20 g acid; (20/60) ≈ 0.33 mol acid; (0.33/0.25 = 0.33 × 4) ≈ 1.33 M 4.78 50.000 mL glycerol × 63.280 g C 3 H 8 O 3 × 1.2656 g glycerol 1 mL glycerol = 63.280 g glycerol 1 mol C 3 H 8 O 3 = 0.687124 = 0.68712 mol C 3 H 8 O 3 92.094 g C 3 H 8 O 3 M= 0.687124 mol C 3H8O 3 = 2.7485 M C 3H8O 3 0.25000 L solution Solution Stoichiometry; Titrations 4.79 Analyze. Given: volume and molarity AgNO 3 . Find: mass KCl. Plan. M × L = mol AgNO 3 = mol Ag + ; balanced equation gives ratio mol NaCl/mol Solve. AgNO 3 ; mol NaCl → g NaCl. 0.200 mol AgNO 3 1L × 0.0150 L = 3.00 × 10 − 3 mol AgNO 3 (aq ) AgNO 3 (aq) + KCl(aq) → AgCl(s) + KNO 3 (aq) mol KCl = mol AgNO 3 = 3.00 × 10 – 3 mol NaCl 3.00 × 10 − 3 mol NaCl × 74.55 g KCl 1 mol KCl = 0.227 g KCl Check. (0.2 × 0.015) = 0.003 mol; (0.003 × 75) ≈ 0.225 g KCl 93 4 Aqueous Reactions 4.80 Solve. Solutions to Exercises Plan. M × L = mol Cd(NO 3 ) 2 ; balanced equation → mol ratio → mol NaOH → g NaOH 0.500 mol Cd(NO 3 ) 2 × 0.0350 L = 0.0175 mol Cd(NO 3 ) 2 1L Cd(NO 3 ) 2 (aq) + 2NaOH(aq) → Cd(OH) 2 (s) + 2NaNO 3 (aq) 0.0175 mol Cd(NO 3 ) 2 × 40.00 g NaOH 2 mol NaOH × = 1.40 g NaOH 1 mol Cd(NO 3 ) 2 1 mol NaOH 4.81 (a) Analyze. Given: M and vol base, M acid. Find: vol acid Plan/Solve. Write the balanced equation for the reaction in question: HClO 4 (aq) + NaOH(aq) → NaClO 4 (aq) + H 2 O(l) Calculate the moles of the known substance, in this case NaOH. 0.0875 mol NaOH × 0.0500 L = 0.004375 = 0.00438 mol NaOH 1L Apply the mole ratio (mol unknown/mol known) from the chemical equation. moles NaOH = M × L = 0.004375 mol NaOH × 1 mol HClO 4 = 0.004375 mol HClO 4 1 mol NaOH Calculate the desired quantity of unknown, in this case the volume of 0.115 M HClO 4 solution. L = mol/M ; L = 0.004375 mol HClO 4 × 1L = 0.0380 L = 38.0 mL 0.115 mol HClO 4 Check. (0.09 × 0.045) = 0.0045 mol; (0.0045/0.11) ≈ 0.040 L ≈ 40 mL (b) Following the logic outlined in part (a): 2HCl(aq) + Mg(OH) 2 (s) → MgCl 2 (aq) + 2H 2 O(l) 2.87 g Mg(OH) 2 × 1 mol Mg(OH) 2 58.32 g Mg(OH) 2 = 0.049211 = 0.0492 mol Mg(OH) 2 0.0492 mol Mg(OH) 2 × 2 mol HCl = 0.0984 mol HCl 1 mol Mg(OH) 2 1 L HCl = 0.769 L = 769 mL 0.128 mol HCl L = mol/ M = 0.09840 mol HCl × (c) AgNO 3 (aq) + KCl(aq) → AgCl(s) + KNO 3 (aq) 785 mg KCl × 1 × 10 −3 g 1 mg × 1 mol KCl 1 mol AgNO 3 × = 0.01053 = 0.0105 mol AgNO 3 74.55 g KCl 1 mol KCl M = mol/L = (d) 0.01053 mol AgNO 3 0.0258 L = 0.408 M AgNO 3 HCl(aq) + KOH(aq) → KCl(aq) + H 2 O(l) 1 mol KOH 56.11 g KOH 0.108 mol HCl × 0.0453 L × × = 0.275 g KOH 1 mol HCl 1 mol KOH 1L 94 4 Aqueous Reactions 4.82 (a) Solutions to Exercises 2HCl(aq) + Ba(OH) 2 (aq) → BaCl 2 (aq) + 2H 2 O(l) 0.101 mol Ba(OH) 2 2 mol HCl × 0.0500 L Ba(OH) 2 × 1 L Ba(OH) 2 1 mol Ba(OH) 2 1 L HCl × = 0.0842 L or 84.2 mL HCl soln 0.120 mol HCl (b) H 2 SO 4 (aq) + 2NaOH(aq) → Na 2 SO 4 (aq) + 2H 2 O(l) 0.200 g NaOH × 1 L H 2 SO 4 1 mol NaOH 1 mol H 2 SO 4 × × 40.00 g NaOH 2 mol NaOH 0.125 mol H 2 SO 4 = 0.0200 L or 20.0 mL H 2 SO 4 soln (c) BaCl 2 (aq) + Na 2 SO 4 (aq) → BaSO 4 (s) + 2NaCl(aq) 752 mg = 0.752 g Na 2 SO 4 × 1 mol Na 2 SO 4 1 mol BaCl 2 1 × × 142.1 g Na 2 SO 4 1 mol Na 2 SO 4 0.0558 L = 0.0948 M BaCl 2 (d) 2HCl(aq) + Ca(OH) 2 (aq) → CaCl 2 (aq) + 2H 2 O(l) 0.0427 L HCl × 4.83 0.208 mol HCl 1 mol Ca(OH) 2 74.10 g Ca(OH) 2 × × = 0.329 g Ca(OH) 2 1 L HCl 2 mol HCl 1 mol Ca(OH) 2 Solve. Analyze/Plan. See Exercise 4.81(a) for a more detailed approach. 2 mol NaHCO 3 84.01 g NaHCO 3 6.0 mol H 2 SO 4 × 0.027 L × × = 27 g NaHCO 3 1L 1 mol NaHCO 3 1 mol H 2 SO 4 4.84 See Exercise 4.81(a) for a more detailed approach. 1 mol CH 3 COOH 60.05 g CH 3 COOH 0.115 mol NaOH × 0.0425 L × × 1L 1 mol NaOH 1 mol CH 3 COOH = 0.29349 = 0.293 g CH 3 COOH in 3.45 mL 1L 1000 mL 0.29349 g CH 3 COOH 1.00 qt vinegar × × × = 80.5 g CH 3 COOH/qt 1.057 qt 1L 3.45 mL vinegar 4.85 Analyze. Given: M and vol HBr, vol Ca(OH) 2 . Find: M Ca(OH) 2 , g Ca(OH) 2 /100 mL soln Plan. Write balanced equation; mol mol HBr ⎯⎯⎯→ mol Ca(OH) 2 → M Ca(OH) 2 ; → g Ca(OH) 2 / 100 mL ratio Solve. The neutralization reaction here is: 2HBr(aq) + Ca(OH) 2 (aq) → CaBr 2 (aq) + 2H 2 O(l) 0.0488 L HBr soln × 5.00 × 10 −2 mol HBr 1 mol Ca(OH) 2 1 × × 1 L soln 2 mol HBr 0.100 L of Ca(OH) 2 = 1.220 × 10 − 2 = 1.22 × 10 − 2 M Ca(OH) 2 From the molarity of the saturated solution, we can calculate the gram solubility of Ca(OH) 2 in 100 mL of H 2 O. 95 4 Aqueous Reactions 0.100 L soln × Solutions to Exercises 1.220 × 10 −2 mol Ca(OH) 2 74.10 g Ca(OH) 2 × 1 L soln 1 mol Ca(OH) 2 = 0.0904 g Ca(OH) 2 in 100 mL soln Check. (0.05 × 0.05/0.2) = 0.0125 M; (0.1 × 0.0125 × 64) ≈ 0.085 g/100 mL 4.86 The balanced equation for the titration is: Sr(NO 3 ) 2 (aq) + Na 2 CrO 4 (aq) → SrCrO 4 (s) + 2NaNO 3 (aq) Beginning with a 0.100 L sample, we can do the following conversions: volume soln → g Sr(NO 3 ) 2 → mol Sr(NO 3 ) 2 → mol Na 2 CrO 4 → vol Na 2 CrO 4 soln 0.100 L soln × 6.82 g Sr(NO 3 ) 2 1 mol Sr(NO 3 ) 2 1 mol Na 2 CrO 4 × × 0.500 L soln 211.6 g Sr(NO 3 ) 2 1 mol Sr(NO 3 ) 2 × 1 L soln = 0.263 L Na 2 CrO 4 soln 0.0245 mol Na 2 CrO 4 4.87 (a) (b) (c) NiSO 4 (aq) + 2KOH(aq) → Ni(OH) 2 (s) + K 2 SO 4 (aq) The precipitate is Ni(OH) 2 . Plan. Compare mol of each reactant; mol = M × L Solve. 0.200 M KOH × 0.1000 L KOH = 0.0200 mol KOH 0.150 M NiSO 4 × 0.2000 L KOH = 0.0300 mol NiSO 4 1 mol NiSO 4 requires 2 mol KOH, so 0.0300 mol NiSO 4 requires 0.0600 mol KOH. Since only 0.0200 mol KOH is available, KOH is the limiting reactant. (d) Plan. The amount of the limiting reactant (KOH) determines amount of product, in this case Ni(OH) 2 . Solve. 0.0200 mol KOH × 1 mol Ni(OH) 2 92.71 g Ni(OH) 2 × = 0.927 g Ni(OH) 2 2 mol KOH 1 mol Ni(OH) 2 (e) Plan/Solve. Limiting reactant: OH – : no excess OH – remains in solution. Excess reactant: Ni 2 +: M Ni 2 + remaining = mol Ni 2 + remaining/L solution 0.0300 mol Ni 2 + initial – 0.0100 mol Ni 2 + reacted = 0.0200 mol Ni 2 + remaining 0.0200 mol Ni 2 +/0.3000 L = 0.0667 M Ni 2 +(aq) Spectators: SO 4 2 –, K + . These ions do not react, so the only change in their concentration is dilution. The final volume of the solution is 0.3000 L. M 2 = M 1 V 1 /V 2 : 0.200 M K + × 0.1000 L/0.3000 L = 0.0667 M K + (aq) 0.150 M SO 4 2 – × 0.2000 L/0.3000 L = 0.100 M SO 4 2 – (aq) 4.88 (a) (b) HNO 3 (aq) + NaOH(s) → NaNO 3 (aq) + H 2 O(l) Determine the limiting reactant, then the identity and concentration of ions remaining in solution. Assume that the H 2 O(l) produced by the reaction does not increase the total solution volume. 96 4 Aqueous Reactions 12.0 g NaOH × Solutions to Exercises 1 mol NaOH = 0.300 mol NaOH 40.00 g NaOH 0.200 M HNO 3 × 0.0750 L HNO 3 = 0.0150 mol HNO 3 . The mol ratio is 1:1, so HNO 3 is the limiting reactant. No excess H + remains in solution. The remaining ions are OH – (excess reactant), Na + , and NO 3 – (spectators). OH – : 0.300 mol OH – initial – 0.0150 mol OH – react = 0.285 mol OH – remain 0.285 mol OH – /0.0750 L soln = 3.80 M OH – (aq) Na + : 0.300 mol Na + /0.0750 L soln = 4.00 M Na + (aq) NO 3 – : 0.0150 mol NO 3 – /0.0750 L = 0.200 M NO 3 – (aq) (c) 4.89 The resulting solution is basic because of the large excess of OH – (aq). Analyze. Given: mass impure Mg(OH) 2 ; M and vol excess HCl; M and vol NaOH. Find: mass % Mg(OH) 2 in sample. Plan/Solve. Write balanced equations. Mg(OH) 2 (s) + 2HCl(aq) → MgCl 2 (aq) + 2H 2 O(l) HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l) Calculate total moles HCl = M HCl × L HCl 0.2050 mol HCl × 0.1000 L = 0.02050 mol HCl total 1 L soln mol excess HCl = mol NaOH used = M NaOH × L NaOH 0.1020 mol NaOH × 0.01985 L = 0.0020247 = 0.002025 mol NaOH 1 L soln mol HCl reacted with Mg(OH) 2 = total mol HCl – excess mol HCl 0.02050 mol total – 0.0020247 mol excess = 0.0184753 = 0.01848 mol HCl reacted (The result has 5 decimal places and 4 sig. figs.) Use mol ratio to get mol Mg(OH) 2 in sample, then molar mass of Mg(OH) 2 to get g pure Mg(OH) 2 . 0.0184753 mol HCl × mass % Mg(OH) 2 = 1 mol Mg(OH) 2 58.32 g Mg(OH) 2 × = 0.5387 Mg(OH) 2 2 mol HCl 1 mol Mg(OH) 2 g Mg(OH) 2 0.5388 g Mg(OH) 2 × 100 = × 100 = 91.40% Mg(OH) 2 g sample 0.5895 g sample 4.90 Plan. CaCO 3 (s) + 2HCl(aq) → CaCl 2 (aq) + H 2 O(l) + CO 2 (g) HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l) total mol HCl – excess mol HCl = mol HCl reacted; mol CaCO 3 = (mol HCl)/2; g CaCO 3 = mol CaCO 3 × molar mass; mass % = (g CaCO 3 /g sample) × 100 97 4 Aqueous Reactions Solve: Solutions to Exercises 1.035 mol HCl × 0.03000 L = 0.031050 = 0.03105 mol HCl total 1 L soln 1.010 mol NaOH × 0.01156 L = 0.011676 = 0.01168 mol HCl excess 1 L soln 0.031050 total – 0.011676 excess = 0.019374 = 0.01937 mol HCl reacted 0.019374 mol HCl × mass % CaCO 3 = 1 mol CaCO 3 100.09 g CaCO 3 × = 0.96959 = 0.9696 g CaCO 3 2 mol HCl 1 mol CaCO 3 g CaCO 3 0.96959 × 100 = × 100 = 77.69% g rock 1.248 Additional Exercises 4.91 If we know the identity of a solute and need to measure its concentration in solution, titration is a good way to do this if three conditions are met. The solute must be a reactant in a chemical reaction of known stoichiometry, the second reactant must be available as a standard solution of well-known concentration, and the equivalence point of the titration must be detectable, either visually or instrumentally. If these conditions are met, a known quantity of solution can be precisely delivered by pipette or buret to a titration vessel. The unknown concentration is then determined by its stoichiometric relationship to the quantity of standard solution required to reach the equivalence point of the titration. Multiple titrations can be performed to ensure precision of the measurement. In summary, titration works well for measuring the unknown concentration of a solute because a well-known quantity of the solute can be measured volumetrically and titrations are repeatable. The precipitate is CdS(s). Na + (aq) and NO 3 – (aq) are spectator ions and remain in solution. Any excess reactant ions also remain in solution. The net ionic equation is: Cd 2 +(aq) + S 2 –(aq) → CdS(s). 4.93 The two precipitates formed are due to AgCl(s) and SrSO 4 (s). Since no precipitate forms on addition of hydroxide ion to the remaining solution, the other two possibilities, Ni 2 + and Mn 2 +, are absent. (a,b) Expt. 1 No reaction Expt. 2 2Ag + (aq) + CrO 4 2 –(aq) → Ag 2 CrO 4 (s) red precipitate Expt. 3 No reaction Expt. 4 2Ag + (aq) + C 2 O 4 2 –(aq) → Ag 2 C 2 O 4 (s) white precipitate Expt. 5 Ca 2 +(aq) + C 2 O 4 2 –(aq) → CaC 2 O 4 (s) white precipitate Expt. 6 Ag + (aq) + Cl – (aq) → AgCl(s) white precipitate (c) The silver salts of both ions are insoluble, but many silver salts are insoluble (Expt. 6). The calcium salt of CrO 4 2 – is soluble (Expt. 3), while the calcium salt of C 2 O 4 2 –(aq) is insoluble (Expt. 5). Thus, chromate salts appear more soluble than oxalate salts. 4.92 4.94 98 4 Aqueous Reactions 4.95 (a) (b) (c) (d) (e) Al(OH) 3 (s) + 3H + (aq) → Al 3 +(aq) + 3H 2 O(l) Mg(OH) 2 (s) + 2H + (aq) → Mg 2 +(aq) + 2H 2 O(l) Solutions to Exercises MgCO 3 (s) + 2H + (aq) → Mg 2 +(aq) + H 2 O(l) + CO 2 (g) NaAl(CO 3 )(OH) 2 (s) + 4H + (aq) → Na + (aq) + Al 3 +(aq) + 3H 2 O(l) + CO 2 (g) CaCO 3 (s) + 2H + (aq) → Ca 2 +(aq) + H 2 O(l) + CO 2 (g) [In (c), (d) and (e), one could also write the equation for formation of bicarbonate, e.g., MgCO 3 (s) + H + (aq) → Mg 2 + + HCO 3 – (aq).] 4.96 (a) (b) (c) (d) 2H + (aq) + SO 3 2 –(aq) → H 2 SO 3 (aq); sulfurous acid H 2 SO 3 (aq) → H 2 O(l) + SO 2 (g); sulfur dioxide The boiling point of SO 2 (g) is –10°C. It is a gas at room temperature (23°C) and pressure (1 atm). (i) Na 2 SO 3 (aq) + 2HCl(aq) → 2NaCl(aq) + H 2 O(l) + SO 2 (g) SO 3 2 –(aq) + 2H + (aq) → H 2 O(l) + SO 2 (g) (ii) Ag 2 SO 3 (s) + 2HCl(aq) → 2AgCl(s) + H 2 O(l) + SO 2 (g) Ag 2 SO 3 (s) + 2H + (aq) + 2Cl – (aq) → 2AgCl(s) + H 2 O(l) + SO 2 (g) (iii) KHSO 3 (s) + HCl(aq) → KCl(aq) + H 2 O(l) + SO 2 (g) KHSO 3 (s) + H + (aq) → K + (aq) + H 2 O(l) + SO 2 (g) (iv) ZnSO 3 (aq) + 2HCl(aq) → ZnCl 2 (aq) + H 2 O(l) + SO 2 (g) SO 3 2 –(aq) + 2H + (aq) → H 2 O(l) + SO 2 (g) 4.97 4NH 3 (g) + 5 O 2 (g ) → 4NO(g) + 6H 2 O(g) . N = −3 (a) O=0 N = +2 O = −2 redox reaction (b) N is oxidized, O is reduced 2NO(g ) + O 2 (g ) → 2NO 2 ( g) . N = +2 O = 0 N = +4, O = +2 (a) redox reaction (b) N is oxidized, O is reduced 3NO 2 (g ) + H 2 O(l) → HNO 3 (aq) + NO(g) . N = +4 (a) (b) redox reaction N = +5 N = +2 N is oxidized (NO 2 → HNO 3 ), N is reduced (NO 2 → NO). A reaction where the same element is both oxidized and reduced is called disproportionation. 4.98 A metal on Table 4.5 is able to displace the metal cations below it from their compounds. That is, zinc will reduce the cations below it to their metals. (a) Zn(s) + Na + (aq) → no reaction 99 4 Aqueous Reactions (b) (c) (d) (e) (f) 4.99 (a) Zn(s) + Pb 2 +(aq) → Zn 2 +(aq) + Pb(s) Zn(s) + Mg 2 +(aq) → no reaction Zn(s) + Fe 2 +(aq) → Zn 2 +(aq) + Fe(s) Zn(s) + Cu 2 +(aq) → Zn 2 +(aq) + Cu(s) Zn(s) + Al 3 +(aq) → no reaction A : La 2 O 3 B : La(OH) 3 C : LaCl 3 D : La 2 (SO 4 ) 3 (b) Solutions to Exercises Metals often react with the oxygen in air to produce metal oxides. When metals react with water (HOH) to form H 2 , OH – remains. Most chlorides are soluble. Sulfuric acid provides SO 4 2 – ions. 4La(s) + 3O 2 (g) → 2La 2 O 3 (s) 2La(s) + 6HOH(l) → 2La(OH) 3 (s) + 3H 2 (g) (There are no spectator ions in either of these reactions.) molecular: La 2 O 3 (s) + 6HCl(aq) → 2LaCl 3 (aq) + 3H 2 O(l) net ionic: La 2 O 3 (s) + 6H + (aq) → 2La 3 +(aq) + 3H 2 O(l) molecular: La(OH) 3 (s) + 3HCl(aq) → LaCl 3 (aq) + 3H 2 O(l) net ionic: La(OH) 3 (s) + 3H + (aq) → La 3 +(aq) + 3H 2 O(l) molecular: 2LaCl 3 (aq) + 3H 2 SO 4 (aq) → La 2 (SO 4 ) 3 (s) + 6HCl(aq) net ionic: (c) 2La 3 +(aq) + 3SO 4 2 –(aq) → La 2 (SO 4 ) 3 (s) La metal is oxidized by water to produce H 2 (g), so La is definitely above H on the activity series. In fact, since an acid is not required to oxidize La, it is probably one of the more active metals. 4.100 Plan. Calculate moles KBr from the two quantities of solution (mol = M × L), then new molarity (M = mol/L). KBr is nonvolatile, so no solute is lost when the solution is evaporated to reduce the total volume. Solve. 1.00 M KBr × 0.0350 L = 0.0350 mol KBr; 0.600 M KBr × 0.060 L = 0.0360 mol KBr 0.0350 mol KBr + 0.0360 mol KBr = 0.0710 mol KBr total 0.0710 mol KBr = 1.42 M KBr 0.0500 L soln 4.101 (a) 50 pg 1 × 10 −12 g 1 × 10 3 mL 1 mol Na × × × = 2.17 × 10 −9 = 2.2 × 10 −9 M Na+ 1 mL 1 pg L 23.0 g Na 2.17 × 10 −9 mol Na 6.02 × 10 23 Na atom 1L × × 3 3 1 L soln 1 mol Na 1 × 10 cm = 1.3 × 10 12 atom or Na + ions/cm 3 (b) 100 4 Aqueous Reactions 4.102 Solutions to Exercises Na + must replace the total positive (+) charge due to Ca 2 + and Mg 2 +. Think of this as moles of charge rather than moles of particles. 0.020 mol Ca 2 + 2 mol + change × 1.5 × 10 3 L × = 60 mol of + charge 1 L water 1 mol Ca 2+ 0.0040 mol Mg 2+ 2 mol + charge × 1.5 × 10 3 L × = 12 mol of + charge 1 L water 1 mol Mg 2 + 72 moles of + charge must be replaced; 72 mol Na + are needed. 4.103 H 2 C 4 H 4 O 6 + 2OH – (aq) → C 4 H 4 O 6 2 –(aq) + 2H 2 O(l) 0.02465 L NaOH soln × 0.2500 mol NaOH 1 mol H 2 C 4 H 4 O 6 1 × × 1L 2 mol NaOH 0.0500 L H 2 C 4 H 4 O 6 =0.06163 M H 2 C 4 H 4 O 6 soln 4.104 Plan. mol MnO 4 – = M × L → mol ratio → mol H 2 O 2 → M H 2 O 2 . 2MnO 4 – (aq) + 5H 2 O 2 (aq) + 6H + → 2Mn 2 +(aq) + 5O 2 (aq) + 8H 2 O(l) 0.134 mol MnO 4 − L × 0.0148 L MnO 4 − × 5 mol H 2 O 2 2 mol MnO 4 − Solve. × 1 0.0100 L H 2 O 2 = 0.4958 mol H 2 O 2 / L = 0.496 M H 2 O 2 4.105 mol OH – from NaOH(aq) + mol OH – from Zn(OH) 2 (s) = mol H + from HBr mol H + = M HBr × L HBr = 0.500 M HBr × 0.350 L HBr = 0.175 mol H + mol OH – from NaOH = M NaOH × L NaOH = 0.500 M NaOH × 0.0885 L NaOH = 0.04425 = 0.0443 mol OH – mol OH – from Zn(OH) 2 (s) = 0.175 mol H + – 0.04425 mol OH – from NaOH = 0.13075 = 0.131 mol OH – from Zn(OH) 2 0.13075 mol OH − × 1 mol Zn(OH) 2 2 mol OH − × 99.41 g Zn(OH) 2 = 6.50 g Zn(OH) 2 1 mol Zn(OH) 2 Integrative Exercises 4.106 (a) At the equivalence point of a titration, mol NaOH added = mol H + present M NaOH × L NaOH = g acid (for an acid with 1 acidic hydrogen) MM acid MM acid = g acid 0.2053 g = = 136 g/mol M NaOH × L NaOH 0.1008 M × 0.0150 L 101 4 Aqueous Reactions (b) Assume 100 g of acid. 70.6 g C × 5.89 g H × 23.5 g O × 1 mol C = 5.88 mol C; 5.88 / 1.47 ≈ 4 12.01 g C 1 mol H = 5.84 mol H; 5.84 / 1.47 ≈ 4 1.008 g H 1 mol O = 1.47 mol O; 1.47 / 1.47 = 1 16.00 g O Solutions to Exercises The empirical formula is C 4 H 4 O. MM 136 = = 2; the molecular formula is 2 × the empirical formula. FW 68.1 The molecular formula is C 8 H 8 O 2 . 4.107 Ba 2 +(aq) + SO 4 2 –(aq) → BaSO 4 (s) 0.2815 g BaSO 4 × mass % = 137.3 g Ba = 0.16560 = 0.1656 g Ba 233.4 g BaSO 4 g Ba 0.16560 g Ba × 100 = × 100 = 4.793 % Ba g sample 3.455 g sample 4.108 Plan. Write balanced equation. mass % mass H 2 SO 4 soln ⎯⎯ ⎯ ⎯ → mass H 2 SO 4 → mol H 2 SO 4 → mol Na 2 CO 3 → mass Na 2 CO 3 ⎯ Solve. H 2 SO 4 (aq) + Na 2 CO 3 (s) → Na 2 SO 4 (aq ) + H 2 O(l) + CO 2 (g ) 5.0 × 10 3 kg conc. H 2 SO 4 × 4.75 × 10 3 kg H 2 SO 4 × 0.950 kg H 2 SO 4 = 4.75 × 10 3 = 4.8 × 10 3 kg H 2 SO 4 1.00 kg conc. H 2 SO 4 1 × 10 3 g 1 mol H 2 SO 4 1 mol Na 2 CO 3 × × 1 kg 98.08 g H 2 SO 4 1 mol H 2 SO 4 × 105.99 g NaHCO 3 1 kg × = 5.133 × 10 3 = 5.1 × 10 3 kg Na 2 CO 3 1 mol NaHCO 3 1 × 10 3 g 4.109 (a) (b) Mg(OH) 2 (s) + 2HNO 3 (aq) → Mg(NO 3 ) 2 (aq) + 2H 2 O(l) 5.53 g Mg(OH) 2 × 1 mol Mg(OH) 2 = 0.09482 = 0.0948 mol Mg(OH) 2 58.32 g Mg(OH) 2 0.200 M HNO 3 × 0.0250 L = 0.00500 mol HNO 3 The 0.00500 mol HNO 3 would neutralize 0.00250 mol Mg(OH) 2 and much more Mg(OH) 2 is present, so HNO 3 is the limiting reactant. (c) Since HNO 3 limits, 0 mol HNO 3 is present after reaction. 0.00250 mol Mg(NO 3 ) 2 is produced. 0.09482 mol Mg(OH) 2 initial – 0.00250 mol Mg(OH) 2 react = 0.0923 mol Mg(OH) 2 remain 102 4 Aqueous Reactions 4.110 (a) (b) Calculate mol of each reactant and compare. 1.50 g Pb(NO 3 ) 2 × Solutions to Exercises Na 2 SO 4 (aq) + Pb(NO 3 ) 2 (s) → PbSO 4 (s) + 2NaNO 3 (aq) 1 mol Pb(NO 3 ) 2 = 0.004529 = 4.53 × 10 − 3 mol Pb(NO 3 ) 2 331.2 g Pb(NO 3 ) 2 0.100 M Na 2 SO 4 × 0.125 L = 0.0125 mol Na 2 SO 4 Since the reactants combine in a 1:1 mol ratio, Pb(NO 3 ) 2 is the limiting reactant. (c) Pb(NO 3 ) 2 is the limiting reactant, so no Pb 2 + remains in solution. The remaining ions are: SO 4 2 – (excess reactant), Na + and NO 3 – (spectators). SO 4 2 –: 0.0125 mol SO 4 2 – initial – 0.00453 mol SO 4 2 – reacted = 0.00797 = 0.0080 mol SO 4 2 – remain 0.00797 mol SO 4 2 – / 0.125 L soln = 0.064 M SO 4 2 – Na + : Since the total volume of solution is the volume of Na 2 SO 4 (aq) added, the concentration of Na + is unchanged. 0.100 M Na 2 SO 4 × (2 mol Na + / 1 mol Na 2 SO 4 ) = 0.200 M Na + NO 3 – : 4.53 × 10 – 3 mol Pb(NO 3 ) 2 × 2 mol NO 3 – / 1 mol Pb(NO 3 ) 2 = 9.06 × 10 – 3 mol NO 3 – 9.06 × 10 – 3 mol NO 3 – / 0.125 L = 0.0725 M NO 3 – 4.111 Plan. Cl – is present in NaCl and MgCl 2 ; using mass %, calculate mass NaCl and MgCl 2 in mixture, mol Cl – in each, then molarity of Cl – in 0.500 L solution. Solve. 7.50 mixture × 7.50 mixture × 0.765 g NaCl 1 mol NaCl 1 mol Cl − × × = 0.09818 = 0.0982 mol Cl − 1.00 g mixture 58.44 g NaCl 1 mol NaCl 0.065 g MgCl 2 1 mol MgCl 2 2 mol Cl − × × = 0.01024 = 0.010 mol Cl − 1.00 g mixture 95.21 g MgCl 2 1 mol MgCl 0.10842 mol Cl − = 0.217 M Cl − 0.5000 L mol Cl − = 0.09818 + 0.01024 = 0.10842 = 0.108 mol Cl − ; M = mol Br − ; mg Br − → g Br − → mol Br − ; L seawater ⎯⎯ ⎯ → mL water → L seawater ⎯ 1 g Br − 1000 mg Br − density 4.112 Plan. M = 1 kg seawater → g Solve. 65 mg Br − × 1 kg seawater × M Br − = × 1 mol Br − 79.90 g Br − = 8.135 × 10 − 4 = 8.1 × 10 − 4 mol Br − 1000 g 1 mL water 1L × × = 0.9756 L 1 kg 1.025 g water 1000 mL 8.135 × 10 −4 mol Br − = 8.3 × 10 − 4 M Br − 0.9756 L seawater 103 4 Aqueous Reactions 4.113 Ag + (aq) + Cl – (aq) → AgCl(s) Solutions to Exercises 1 mol Cl − 35.453 g Cl − 0.2997 mol Ag + × 0.04258 L × × = 0.45242 = 0.4524 g Cl − 1L 1 mol Ag + 1 mol Cl − 25.00 mL seawater × mass % Cl − = 1.025 g = 25.625 = 25.63 g seawater mL 0.45242 g Cl − × 100 = 1.766% Cl − 25.625 g seawater 4.114 (a) (b) (c) AsO 4 3 –; +5 Ag 3 PO 4 is silver phosphate; Ag 3 AsO 4 is silver arsenate 0.0250 L soln × 0.102 mol Ag + 1 mol Ag 3 AsO 4 1 mol As 74.92 g As × × × + 1 L soln 1 mol Ag 3 AsO 4 1 mol As 3 mol Ag = 0.06368 = 0.0637 g As mass percent = 0.06368 g As × 100 = 5.22% As 1.22 g sample 4.115 Analyze. Given 10 ppb AsO43– , find mass Na 3 AsO 4 in 1.00 L of drinking water. Plan. Use the definition of ppb to calculate g AsO43– in 1.0 L of water. Convert g AsO43– → g Na 3 AsO 4 using molar masses. Assume the density of H 2 O is 1.00 g/mL. Solve. 1 billion = 1 × 10 9 ; 1 ppb = 1 g solute 1 × 10 9 g solution 10 ppb AsO 4 3 − = × 1 g solute 1 × 10 9 g solution × 1 × 10 3 mL 1 L solution = g AsO 4 3− 1 × 10 6 L H 2 O 1 g solution 1 mL solution 10 g AsO 4 3 − × 1 L H 2 O = 1.0 × 10 − 5 g AsO 4 3 − /L. 1 × 10 6 L H 2 O 1.0 × 10 − 5 g AsO 4 3 − × 1 mol AsO 4 3 − 1 mol Na 3 AsO 4 207.89 g Na 3 AsO 4 × × 3− 1 mol Na 3 AsO 4 138.92 g AsO 4 1 mol AsO 4 3 − = 1.5 × 10 −5 g Na 3 AsO 4 in 1.00 L H 2 O 4.116 (a) Analyze. Given: 55 gal of 50 ppb AsO43– . Find. g Na3AsO4. Plan. Use the definition of ppb to calculate g AsO43– in 55 gal of water. Then change AsO43– to g Na3SO4 using molar masses. Assume the density of H2O is 1.00 g/mL. 1 g solute Solve. 1 billion = 1 × 10 9 ; 1 ppb = 1 × 10 9 g solution 1 g solute 1 × 10 9 g solution × 1 g solution 1 mL solution × 1 × 10 3 mL 1 L solution = g AsO 4 3 − 1 × 10 6 L H 2 O 50 ppb AsO43– = 50 g AsO43– / 1 × 106 L H2O 104 4 Aqueous Reactions 50 g AsO 4 3− 1 × 10 L H 2 O 6 Solutions to Exercises 1 gal × 55 gal × 207.89 g Na 3 AsO 4 1 38.92 g AsO 4 3 − × 3.7854 L = 0.015578 = 0.016 g Na3AsO4 (b) Parts per billion, ppb, is a mass ratio. If 90% of the arsenic, As, in the form of arsenate, AsO43– , is retained in the bucket, then 90% of the mass of As (or AsO43– ) is in the bucket and 10% of the As (or AsO43– ) passes thru. Assuming 1 × 106 L of drinking water, 10% of the 500 g AsO43– in the water is 50 g AsO43– . Thus, only one pass through the bucket is required to reduce the AsO43– concentration in drinking water from 500 ppb to 50 ppb, the acceptable standard. mol HCl initial – mol NH 3 from air = mol HCl remaining = mol NaOH required for titration mol NaOH = 0.0588 M × 0.0131 L = 7.703 × 10 – 4 = 7.70 × 10 – 4 mol NaOH = 7.70 × 10 – 4 mol HCl remain mol HCl initial – mol HCl remaining = mol NH 3 from air (0.0105 M HCl × 0.100 L) – 7.703 × 10 – 4 mol HCl = mol NH 3 10.5 × 10 – 4 mol HCl – 7.703 × 10 – 4 mol HCl = 2.80 × 10 – 4 = 2.8 × 10 – 4 mol NH 3 2.8 × 10 − 4 mol NH 3 × 17.03 g NH 3 = 4.77 × 10 − 3 = 4.8 × 10 − 3 g NH 3 1 mol NH 3 4.117 (a) (b) ppm is defined as molecules of NH 3 /1 × 10 6 molecules in air. Calculate molecules NH 3 from mol NH 3 . 2.80 × 10 − 4 mol NH 3 × 0.022 × 20 23 molecules = 1.686 × 10 20 1 mol = 1.7 × 10 20 NH 3 molecules Calculate total volume of air processed, then g air using density, then molecules air using molar mass. 1.20 g air 1 mol air 6.022 × 10 23 molecules 10.0 L × 10.0 min × × × 1 L air 29.0 g air 1 mol 1 min = 2.492 × 10 24 = 2.5 × 10 24 air molecules ppm NH 3 = 1.686 × 10 20 NH 3 molecules 2.492 × 10 24 air molecules × 1 × 10 6 = 68 ppm NH 3 (c) 68 ppm > 50 ppm. The manufacturer is not in compliance. 105 ...
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This note was uploaded on 04/04/2009 for the course CHEM 121 taught by Professor Wyzlouzil during the Spring '07 term at Ohio State.

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