4.4
Aqueous 4
4.2 Reactions and Solution Stoichiometry
Visualizing Concepts
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. 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. 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. 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. Diagram I shows spectator ions but no precipitate; this corresponds to reaction (b). Diagram II shows spectator ions and a 1:1 precipitate; this corresponds to reaction (c). Diagram III shows only precipitate; this corresponds to reaction (a). The second product in reaction (a) is H 2 O(l), which is also the solvent. Solvent molecules are not shown in any of the diagrams.
4.6
4.8
4.10
Electrolytes
4.12 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 HC 2 H 3 O 2 dissolves, mostly neutral molecules are dispersed throughout the solution. A few of the dissolved molecules ionize to form H + (aq) and C 2 H 3 O 2 (aq). These few ions carry some charge and the solution is weakly conducting. Ions are hydrated when they are surrounded by H 2 O molecules in aqueous solution.
4.14
41
4 Aqueous Reactions
4.16 (a) (b) (c) (d) 4.18 (a) MgI 2 (aq) Mg 2 +(aq) + 2I (aq) Al(NO 3 ) 3 (aq) HClO 4 (aq) KC 2 H 3 O 2 (aq) H + (aq) + ClO 4 (aq)
Solutions to Black Exercises
Al 3 +(aq) + 3NO 3 (aq)
K + (aq) + C 2 H 3 O 2 (aq)
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
4.20 According to Table 4.1: (a) (b) (c) (d) (e) 4.22 Ni(OH) 2 : insoluble PbBr 2 : insoluble; Ba(NO 3 ) 2 : soluble AlPO 4 : insoluble AgC 2 H 3 O 2 : soluble
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) + Cu(C 2 H 3 O 2 ) 2 (aq) CuS(s) + 2NaC 2 H 3 O 2 (aq)
4.24
Spectator ions are those that do not change during reaction. (a) (b) (c) 2Cr 3 +(aq) + 3CO 3 2 (aq) Ba2+ (aq) + SO 4 2 (aq) Fe2 +(aq) + 2OH (aq) Cr 2 (CO 3 ) 3 (s); spectators: NH 4 + , SO 4 2 BaSO 4 (s); spectators: K + , NO 3 Fe(OH) 2 (s); spectators: K + , NO 3
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. (a) Pb(C 2 H 3 O 2 ) 2 (aq) + Na 2 S(aq) net ionic: Pb (aq) + S (aq) Pb(C 2 H 3 O 2 ) 2 (aq) + CaCl 2 (aq) net ionic: Pb (aq) + 2Cl (aq) Na 2 S(aq) + CaCl 2 (aq) net ionic: no reaction (b) Spectator ions: Na + , Ca 2 +, C 2 H 3 O 2
2+ 2+ 2
4.28
PbS(s) + 2NaC 2 H 3 O 2 (aq) PbS (PbCl 2 )s + Ca(C 2 H 3 PO) 2 (aq) (PbCl 2 )(s)
CaS(aq) + 2NaCl(aq)
42
4 Aqueous Reactions
Acid-Base Reactions
4.30
Solutions to Black Exercises
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 0.5 M NH 3 is less basic than 0.1 M KOH. Ca(OH) 2 has twice as many OH per moles as KOH, so 0.1 M Ca(OH) 2 is more basic than 0.1 M KOH. The most basic solution is 0.1 M Ca(OH) 2 . (a) NH 3 produces OH in aqueous solution by reacting with H 2 O (hydrolysis): NH3(aq) + H2O(1) 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 .
4.32
(b) (c)
4.34
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. 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 . (a) (d) HClO 4 : strong CH 3 OCH 3 : non (b) HNO 3 : strong (e) CoSO 4 : strong KC 2 H 3 O 2 (aq) + H 2 O(l) C 2 H 3 O 2 (aq) H 2 O(l) Cr(NO 3 ) 3 (aq) + 3H 2 O(l) (c) NH 4 Cl: strong (f) C 1 2H 2 2O 1 1: non
4.36
3H 2 O(l) + Cr 3 +(aq)
4.38
4.40
(a)
HC 2 H 3 O 2 (aq) + KOH(aq) HC 2 H 3 O 2 (aq) + OH (aq)
(b)
Cr(OH) 3 (s) + 3HNO 3 (aq) Cr(OH) 3 (s) + 3H + (aq)
(c)
Ca(OH) 2 (aq) + 2HClO(aq) HClO(aq) + OH (aq)
Ca(ClO) 2 (aq) + 2H 2 O(l)
ClO (aq) + H 2 O(l)
4.42
(a) (b)
FeO(s) + 2H + (aq) NiO(s) + 2H + (aq)
H 2 O(l) + Fe 2 +(aq) H 2 O(l) + Ni 2 +(aq) 2OH (aq), net ionic;
4.44
K 2 O(aq) + H 2 O(l)
2KOH(aq), molecular; O 2 (aq) + H 2 O(l)
base: (H + ion acceptor) O 2 (aq); acid: (H + ion donor) H 2 O(aq); spectator: K +
43
4 Aqueous Reactions
Oxidation-Reduction Reactions
4.46
Solutions to Black Exercises
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. 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 )
4.48
A u(s) + 3K + (aq )
(c) +3 (d) 2 (e) +3 (f) +6
4.50 4.52
(a) (a) (b) (c) (d)
+4
(b) +2
acid-base reaction oxidation-reduction reaction; Fe is reduced, C is oxidized precipitation reaction oxidation-reduction reaction; Zn is oxidized, N is reduced 2HCl(aq) + Ni(s) H 2 SO 4 (aq) + Fe(s) NiCl 2 (aq) + H 2 (g); Ni(s) + 2H + (aq) FeSO 4 (aq) + H 2 (g); Fe(s) + 2H + (aq) Ni 2 +(aq) + H 2 (g) Fe 2 +(aq) + H 2 (g)
4.54
(a) (b)
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) Zn(C 2 H 3 O 2 ) 2 (aq) + H 2 (g); Zn 2 +(aq) + 2C 2 H 3 O 2 (aq) + H 2 (g) Mg 2 +(aq) + H 2 (g)
2HC 2 H 3 O 2 (aq) + Zn(s) Zn(s) + 2HC 2 H 3 O 2 (aq)
4.56
(a) (b) (c) (d) (e)
Mn(s) + NiCl 2 (aq)
MnCl 2 (aq) + Ni(s) NR
Cu(s) + Cr(C 2 H 3 O 2 )(aq) 2Cr(s) + 3NiSO 4 (aq) Pt(s) + HBr(aq) NR
Cr 2 (SO 4 ) 3 (aq) + 3Ni(s)
H 2 (g) + CuCl 2 (aq) Br 2 + 2NaI Cl 2 + 2NaBr
Cu(s) + 2HCl(aq)
4.58
(a)
2NaBr + I 2 indicates that Br 2 is more easily reduced than I 2 . 2NaCl + Br 2 shows that Cl 2 is more easily reduced than Br 2 .
The order for ease of reduction is Cl 2 > Br 2 > I 2 . Conversely, the order for ease of oxidation is I > Br > Cl .
44
4 Aqueous Reactions
(b)
Solutions to Black Exercises
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
4.60 (a) 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 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 1 L solution 0.50 mol solute 1 L solution
2.50 mol solute 0.50 mol solute
(b)
x grams solute 4.5 g solute
x g solute 4.5 g solute
; 5.0(4.5 g solute)
23 g solute
The result has 2 sig figs; 22.5 rounds to 23 g solute. 4.62 (a) (b) (c) 4.64
M mol solute 0.145 mol Na 2 SO 4 ; L solution 0.750 L 0.193 M Na 2 SO 4
mol
L
M
L;
0.0850 mol KMnO 4 1L
2.20 10
0.125 L 1.06
2
10
2
mol KMnO 4
mol 0.255 mol HCl ; M 11.6 mol HCl/L
L or 22.0 mL
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 L 4.6 L 0.5428 0.54 mol Na
0.138 mol L
4.6 L
0.6348
0.63 mol Na
(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 mol 5.38 5 g NaCl
45
4 Aqueous Reactions
4.66
M mol ; mol L
Solutions to Black Exercises
g (MM is the symbol for molar mass in this manual.) MM
(a) (b) (c) 4.68 (a)
0.360 mol K 2 Cr2 O7 1L 4.28 g (NH4 ) 2 SO4 2.25 g CuSO4
50.0 mL
1L 1000 mL
294.2 g K 2 Cr2 O7 1 mol K 2 Cr2 O7 1000 mL 1L
5.30 g K 2 Cr2 O7 0.108 M (NH4 ) 2 SO4 58.7 mL solution
1 mol (NH4 ) 2 SO4 132.2 g (NH4 ) 2 SO4
1 300. mL
1 mol CuSO4 159.6 g CuSO4
1L 0.240 mol CuSO4
1000 mL 1L
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.
4.70
(a)
H :
0.130 M 16.0 mL 0.600 M 12.0 mL 28.0 mL
0.331 M H
Cl : concentration Cl = concentration H + = 0.331 M Cl (b)
Na : 2(0.200 M 18.0 mL) 33.0 mL 0.218 M ; K : 0.150 M 15.0 mL 33.0 mL 0.0682 M
SO 4 2 :
0.200 M 18.0 mL 33.0 mL
0.109 M ; Cl :
0.150 M 15.0 mL 33.0 mL
0.0682 M
(c)
Na :
2.38 g NaCl 1 mol 0.0500 L 58.44 g
0.815 M; Ca 2 : 0.400 M
Cl : 0.815 M (from NaCl(s)) + 0.800 M (from CaCl 2 (aq)) = 1.615 M 4.72 (a) (b) 4.74 (a)
V1 M 2 V2 / M1 ; 0.400 M HNO3 0.350 mL 10.0 M HNO3 0.0140 L 14.0 mL conc. HNO3
M2
M1V1 / V2 ;
10.0 M HNO3 25.0 mL 500 mL
0.500 M HNO3
The amount of AgNO 3 needed is: 0.150 M
0.0375 mol AgNO 3 169.88 g AgNO 3 1 mol AgNO 3
0.2500 L = 0.0375 mol AgNO 3
6.37 g AgNO 3
6.3705
Add this amount of solid a to 250 mL volumetric flask, dissolve in a small amount of water, bring the total volume to exactly 250 mL, and agitate well. (b) Dilute the 6.0 M HNO 3 to prepare 100 mL of 0.50 M HNO 3 . To determine the volume of 6.0 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.
46
4 Aqueous Reactions
0.100 L
L mol ; L 6.0 M HNO 3 M
Solutions to Black Exercises
0.050 mol needed 6.0 M
0.50 M = 0.050 mol HNO 3 needed;
0.00833 L 8.3 mL
Thoroughly clean, rinse, and fill a buret with the 6.0 M HNO 3 , taking precautions appropriate for working with a relatively concentrated acid. Dispense 8.3 mL of the 6.0 M acid into a 100 mL volumetric flask, add water to the mark, and mix thoroughly. 4.76
50.000 mL glycerol 1.2656 g glycerol 1 mL glycerol 1 mol C 3 H 8 O 3 92.094 g C 3 H 8 O 3 63.280 g glycerol
63.280 g C 3 H 8 O 3
0.687124
0.68712 mol C 3 H 8 O 3
M
0.687124 mol C3H8O3 0.25000 L solution
2.7485 M C3H8O3
Solution Stoichiometry; Titrations
4.78 Plan. M Solve. L = mol Cd(NO 3 ) 2 ; balanced equation
0.02500 L
mol ratio
mol NaOH
g NaOH
0.500 mol Cd(NO 3 ) 2 1L
0.0125 mol Cd(NO 3 ) 2
Cd(NO 3 ) 2 (aq) + 2NaOH(aq)
0.0125 mol Cd(NO 3 ) 2
Cd(OH) 2 (s) + 2NaNO 3 (aq)
40.00 g NaOH 1 mol NaOH 1.00 g NaOH
2 mol NaOH 1 mol Cd(NO 3 ) 2
4.80
(a)
2HCl(aq) + Ba(OH) 2 (aq)
0.101 mol Ba(OH)2 1 L Ba(OH)2
BaCl 2 (aq) + 2H 2 O(l)
2 mol HCl 1 mol Ba(OH)2 0.0842 L or 84.2 mL HCl soln
0.0500 L Ba(OH)2
1 L HCl 0.120 mol HCl
(b)
H 2 SO 4 (aq) + 2NaOH(aq)
0.200 g NaOH
Na 2 SO 4 (aq) + 2H 2 O(l)
1 L H 2 SO 4 0.125 mol H 2 SO 4 0.0200 L or 20.0 mL H 2 SO 4 soln
1 mol NaOH 1 mol H 2 SO 4 40.00 g NaOH 2 mol NaOH
(c)
BaCl 2 (aq) + Na 2 SO 4 (aq)
752 mg 0.752 g Na 2 SO 4
BaSO 4 (s) + 2NaCl(aq)
1 mol Na 2 SO 4 142.1 g Na 2 SO 4 1 mol BaCl2 1 mol Na 2 SO 4 1 0.0558 L 0.0948 M BaCl2
(d)
2HCl(aq) + Ca(OH) 2 (aq)
0.0427 L HCl
CaCl 2 (aq) + 2H 2 O(l)
74.10 g Ca(OH)2 1 mol Ca(OH)2
0.208 mol HCl 1 mol Ca(OH)2 1 L HCl 2 mol HCl
= 0.329 g Ca(OH)2
47
4 Aqueous Reactions
4.82
Solutions to Black Exercises
60.05 g HC 2 H 3 O 2 1 mol HC 2 H 3 O 2 0.29349 0.293 g HC 2 H 3 O 2 in 3.45 mL
80.5 g HC 2 H 3 O 2 /qt
See Exercise 4.79(a) for a more detailed approach.
1 mol HC 2 H 3 O 2 0.115 mol NaOH 0.0425 L 1L 1 mol NaOH
1L 1000 mL 1.057 qt 1L
1.00 qt vinegar
0.29349 g HC 2 H 3 O 2 3.45 mL vinegar
4.84
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 0.500 L soln
1 mol Sr(NO 3 ) 2 211 .6 g Sr(NO 3 ) 2
1 mol Na 2 CrO 4 1 mol Sr(NO 3 ) 2 0.192 L Na 2 CrO 4 soln
1 L soln 0.0335 mol Na 2 CrO 4
4.86 (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.
12.0 g NaOH 1 mol NaOH 40.00 g NaOH 0.300 mol 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) 4.88 The resulting solution is basic because of the large excess of OH (aq). CaCl 2 (aq) + H 2 O(l) + CO 2 (g) NaCl(aq) + H 2 O(l)
Plan. CaCO 3 (s) + 2HCl(aq) HCl(aq) + NaOH(aq)
total mol HCl excess mol HCl = mol HCl reacted; mol CaCO 3 = (mol HCl)/2; g CaCO 3 = mol CaCO 3 Solve. molar mass; mass % = (g CaCO 3 /g sample) 100
48
4 Aqueous Reactions
1.035 mol HCl 1 L soln 0.03000 L 0.031050
Solutions to Black Exercises
0.03105 mol HCl total
1.010 mol NaOH 0.01156 L 1 L soln
0.011676
0.01168 mol HCl excess
0.031050 total 0.011676 excess = 0.019374 = 0.01937 mol HCl reacted
0.019374 mol HCl mass % CaCO 3 1 mol CaCO 3 2 mol HCl g CaCO 3 100 g rock 100.09 g CaCO 3 1 mol CaCO 3 0.96959 100 1.248 0.96959 0.9696 g CaCO 3
77.69%
Additional Exercises
4.89 As soon as the dispersion formed, ion-pairing would begin, and eventually KBr(s) would form. In water, the solute ions stay separated as in solution because they are stabilized by electrostatic attractive forces with the polar water molecules. In a nonpolar solvent like mineral oil, there are no electrostatic attractive forces between the solute ions and the solvent. The oppositely charged ions are attracted to each other, forming KBr solid. 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) (c) (d) (e) Al(OH) 3 (s) + 3H + (aq) Mg(OH) 2 (s) + 2H + (aq) MgCO 3 (s) + 2H + (aq) Al 3 +(aq) + 3H 2 O(l) Mg 2 +(aq) + 2H 2 O(l) Mg 2 +(aq) + H 2 O(l) + CO 2 (g) Na + (aq) + Al 3 +(aq) + 3H 2 O(l) + CO 2 (g)
4.91
4.93
NaAl(CO 3 )(OH) 2 (s) + 4H + (aq) 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).] 4.94 (a) (b) (c) (d) 2H + (aq) + SO 3 2 (aq) H 2 SO 3 (aq) H 2 SO 3 (aq); sulfurous acid
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) SO 3 2 (aq) + 2H + (aq) (ii) Ag 2 SO 3 (s) + 2HCl(aq) 2NaCl(aq) + H 2 O(l) + SO 2 (g)
H 2 O(l) + SO 2 (g) 2AgCl(s) + H 2 O(l) + SO 2 (g) 2AgCl(s) + H 2 O(l) + SO 2 (g)
Ag 2 SO 3 (s) + 2H + (aq) + 2Cl (aq) (iii) KHSO 3 (s) + HCl(aq) KHSO 3 (s) + H + (aq)
KCl(aq) + H 2 O(l) + SO 2 (g) K + (aq) + H 2 O(l) + SO 2 (g)
49
4 Aqueous Reactions
(iv) ZnSO 3 (aq) + 2HCl(aq) SO 3 2 (aq) + 2H + (aq) 4.96
Solutions to Black Exercises
ZnCl 2 (aq) + H 2 O(l) + SO 2 (g) H 2 O(l) + SO 2 (g)
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) (b) (c) (d) (e) (f) Zn(s) + Na + (aq) Zn(s) + Pb 2 +(aq) Zn(s) + Mg 2 +(aq) Zn(s) + Fe 2 +(aq) Zn(s) + Cu 2 +(aq) Zn(s) + Al 3 +(aq) A : La 2 O 3 B : La(OH) 3 C : LaCl 3 D : La 2 (SO 4 ) 3 (b) 4La(s) + 3O 2 (g) 2La(s) + 6HOH(l) no reaction Zn 2 +(aq) + Pb(s) no reaction Zn 2 +(aq) + Fe(s) Zn 2 +(aq) + Cu(s) no reaction 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. 2La 2 O 3 (s) 2La(OH) 3 (s) + 3H 2 (g)
4.97
(a)
(There are no spectator ions in either of these reactions.) molecular: La 2 O 3 (s) + 6HCl(aq) net ionic: La 2 O 3 (s) + 6H + (aq) 2LaCl 3 (aq) + 3H 2 O(l) 2La 3 +(aq) + 3H 2 O(l) LaCl 3 (aq) + 3H 2 O(l) La 3 +(aq) + 3H 2 O(l) La 2 (SO 4 ) 3 (s) + 6HCl(aq) La 2 (SO 4 ) 3 (s)
molecular: La(OH) 3 (s) + 3HCl(aq) net ionic: La(OH) 3 (s) + 3H + (aq)
molecular: 2LaCl 3 (aq) + 3H 2 SO 4 (aq) net ionic: (c) 2La 3 +(aq) + 3SO 4 2 (aq)
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.
0.0400 L soln 0.160 mol NaCl 1 L soln 6.40 10
3
4.99
(a)
mol NaCl
0.0650 L soln
0.150 mol NaCl 1 L soln
9.75
10
3
mol NaCl
Total moles NaCl = 1.615
10 2 = 1.62
10 2
Total volume = 0.0400 L + 0.0650 L = 0.1050 L
Molarity 1.615 10 2 mol 0.1050 L 0.154 M
50
4 Aqueous Reactions
(b) 4.101
Solutions to Black Exercises
Both solutions are the same concentration, 0.750 M, so the resulting solution is 0.750 M.
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.
0.010 mol Ca 2 1 L water
1.0 10 3 L
2 mol change 1 mol Ca 2
20 mol of charge
0.0050 mol Mg 2 1 L water
1.0 10 3 L
2 mol charge 1 mol Mg 2
10 mol of charge
30 moles of + charge must be replaced; 30 mol Na + are needed. 4.102 H 2 C 4 H 4 O 6 + 2OH (aq)
0.02262 L NaOH soln
C 4 H 4 O 6 2 (aq) + 2H 2 O(l)
1 0.04000 L H 2 C 4 H 4 O 6
0.2000 mol NaOH 1 mol H 2 C 4 H 4 O 6 1L 2 mol NaOH
=0.05655 M H 2 C 4 H 4 O 6 soln 4.104 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 0.400 L HBr = 0.200 mol H + 0.0985 L NaOH
mol OH from NaOH = M NaOH
L NaOH = 0.500 M NaOH
= 0.04925 = 0.0493 mol OH mol OH from Zn(OH) 2 (s) = 0.200 mol H + 0.04925 mol OH from NaOH = 0.15075 = 0.151 mol OH from Zn(OH) 2
0.15075 mol OH 1 mol Zn(OH)2 2 mol OH 99.41 g Zn(OH)2 1 mol Zn(OH)2 7.49 g Zn(OH)2
Integrative Exercises
4.106 (a) At the equivalence point of a titration, mol NaOH added = mol H + present
M Na OH L Na OH MM acid g acid (for an acid with 1 acidic hydrogen) MM acid 0.2053 g 0.1008 M 0.0150 L 136 g/mol
g acid M Na OH L Na OH
(b)
Assume 100 g of acid.
70.6 g C 1 mol C 12.01g C 1 mol H 1.008 g H 1 mol O 16.00 g O 5.88 mol C; 5.88 / 1.47 4
5.89 g H
5.84 mol H; 5.84 / 1.47
4
23.5 g O
1.47 mol O; 1.47 / 1.47
1
51
4 Aqueous Reactions
The empirical formula is C 4 H 4 O.
MM FW 136 68.1
Solutions to Black Exercises
2; the molecular formula is 2 the empirical formula.
The molecular formula is C 8 H 8 O 2 . 4.108 Plan. Write balanced equation.
mass H2 SO4 soln
mass %
mass H2 SO4
mol H2 SO4
mol Na 2 CO3
mass Na 2 CO3
Solve. H 2 SO4 (aq) Na 2 CO 3 (s)
5.0 10 3 kg conc. H 2 SO4
Na 2 SO4 (aq) H 2 O(l) CO 2 (g)
4.75 10 3 4.8 10 3 kg H 2 SO4
0.950 kg H 2 SO4 1.00 kg conc. H 2 SO4
4.75 10 3 kg H 2 SO4
1 10 3 g 1 kg
1 mol H 2 SO4 98.08 g H 2 SO4
1 mol Na 2 CO 3 1 mol H 2 SO4
105.99 g NaHCO3 1 mol NaHCO3
1 kg 1 10 3 g
5.133 10 3
5.1 10 3 kg Na 2 CO 3
4.109
(a) (b)
Mg(OH) 2 (s) + 2HNO 3 (aq)
5.53 g Mg(OH) 2
Mg(NO 3 ) 2 (aq) + 2H 2 O(l)
0.09482 0.0948 mol Mg(OH) 2
1 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 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 0.765 g NaCl 1 mol NaCl 1 mol Cl 1.00 g mixture 58.44 g NaCl 1 mol NaCl
0.065 g MgCl 2 1 mol MgCl 2 1.00 g mixture 95.21 g MgCl 2 0.10842 2 mol Cl 1 mol MgCl
0.09818
0.0982 mol Cl
7.50 mixture mol Cl
0.01024
0.010 mol Cl 0.217 M Cl
0.09818 0.01024
0.108 mol Cl ; M
0.10842 mol Cl 0.5000 L
4.112
Plan. M
mol Br ; mg Br L seawater
g
de nsity
g Br
mL water
mol Br ;
L sea water
1 kg seawater
52
4 Aqueous Reactions
Solve. 65 mg Br
Solutions to Black Exercises
1 mol Br 79.90 g Br 8.135 10
4
1 g Br 1000 mg Br
8.1 10
4
mol Br
1 kg seawater
1000 g 1 mL water 1L 1 kg 1.025 g water 1000 mL
8.3 10
4
0.9756 L
M Br
8.135 10 4 mol Br 0.9756 L seawater
M Br
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 L soln 1 mol Ag 3 AsO 4 3 mol Ag 1 mol As 1 mol Ag 3 AsO 4 0.06368 74.92 g As 1 mol As
0.0637 g As
mass percent
0.06368 g As 100 1.22 g sample
5.22% As
4.116
(a)
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 = 7.70 10 4 = 7.70 10 4 mol NaOH
10 4 mol HCl remain
mol HCl initial mol HCl remaining = mol NH 3 from air (0.0105 M HCl 10.5 0.100 L) 7.703 10 4 mol HCl = mol NH 3 10 4 = 2.8
3
10 4 mol HCl 7.703
4
10 4 mol HCl = 2.80
4.77 10
3
10 4 mol NH 3
2.8 10
mol NH3
17.03 g NH3 1 mol NH3
4.8 10
g NH3
(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 mol
1.686 10 20 1.7 10 20 NH 3 molecules
Calculate total volume of air processed, then g air using density, then molecules air using molar mass.
10.0 L 1.20 g air 10.0 min 1 min 1 L air 1 mol air 29.0 g air 6.022 10 23 molecules 1 mol
2.492 1024
2.5 1024 air molecules
ppm NH3
(c)
1.686 10 20 NH3 molecules 2.492 10 24 air molecules
1 10 6
68 ppm NH3
68 ppm > 50 ppm. The manufacturer is not in compliance.
53