CHAPTER 15 ACIDS, BASES, AND SALTS SOLUTIONS TO REVIEW QUESTIONS 1. The Arrhenius definition is restricted to aqueous solutions, while the Brønsted-Lowry definition is not. 2. An electrolyte must be present in the solution for the bulb to glow. 3. Electrolytes include acids, bases, and salts. 4. First, the orientation of the polar water molecules about the and is different. The positive end (hydrogen) of the water molecule is directed towards while the negative end (oxygen) of the water molecule is directed towards the Second, more water molecules will fit around since it is larger than the ion. 5. The pH for a solution with a hydrogen ion concentration of 0.003 M will be between 2 and 3. 6. Tomato juice is more acidic than blood, since its pH is lower. 7. By the Arrhenius theory, an acid is a substance that produces hydrogen ions in aqueous solution. A base is a substance that produces hydroxide ions in aqueous solution. By the Brønsted-Lowry theory, an acid is a proton donor, while a base accepts protons. Since a proton is a hydrogen ion, then the two theories are very similar for acids, but not bases. A chloride ion can accept a proton (producing HCl), so it is a Brønsted-Lowry base, but would not be a base by the Arrhenius theory, since it does not produce hydroxide ions. By the Lewis theory, an acid is an electron pair acceptor, and a base is an electron pair donor. Many individual substances would be similarly classified as bases by Brønsted- Lowry or Lewis theories, since a substance with an electron pair to donate, can accept a proton. But, the Lewis definition is almost exclusively applied to reactions where the acid and base combine into a single molecule. The Brønsted-Lowry definition is usually applied to reactions that involve a transfer of a proton from the acid to the base. The Arrhenius definition is most often applied to individual substances, not to reactions. According to the Arrhenius theory, neutralization involves the reaction between a hydrogen ion and a hydroxide ion to form water. Neutralization, according to the Brønsted-Lowry theory, involves the transfer of a proton to a negative ion. The formation of a covalent bond constitutes a Lewis neutralization. Na + Cl - , Na + . Cl - , Cl - Na + - 207 - HEINS15-207-228v4.qxd 12/30/06 2:40 PM Page 207
Transcript
CHAPTER 15
ACIDS, BASES, AND SALTS
SOLUTIONS TO REVIEW QUESTIONS
1. The Arrhenius definition is restricted to aqueous solutions, while the Brønsted-Lowrydefinition is not.
2. An electrolyte must be present in the solution for the bulb to glow.
3. Electrolytes include acids, bases, and salts.
4. First, the orientation of the polar water molecules about the and is different. Thepositive end (hydrogen) of the water molecule is directed towards while the negativeend (oxygen) of the water molecule is directed towards the Second, more watermolecules will fit around since it is larger than the ion.
5. The pH for a solution with a hydrogen ion concentration of 0.003 M will be between 2 and 3.
6. Tomato juice is more acidic than blood, since its pH is lower.
7. By the Arrhenius theory, an acid is a substance that produces hydrogen ions in aqueoussolution. A base is a substance that produces hydroxide ions in aqueous solution.
By the Brønsted-Lowry theory, an acid is a proton donor, while a base accepts protons.Since a proton is a hydrogen ion, then the two theories are very similar for acids, but notbases. A chloride ion can accept a proton (producing HCl), so it is a Brønsted-Lowrybase, but would not be a base by the Arrhenius theory, since it does not producehydroxide ions.
By the Lewis theory, an acid is an electron pair acceptor, and a base is an electron pairdonor. Many individual substances would be similarly classified as bases by Brønsted-Lowry or Lewis theories, since a substance with an electron pair to donate, can accept aproton. But, the Lewis definition is almost exclusively applied to reactions where the acidand base combine into a single molecule. The Brønsted-Lowry definition is usuallyapplied to reactions that involve a transfer of a proton from the acid to the base. TheArrhenius definition is most often applied to individual substances, not to reactions.According to the Arrhenius theory, neutralization involves the reaction between ahydrogen ion and a hydroxide ion to form water.
Neutralization, according to the Brønsted-Lowry theory, involves the transfer of a protonto a negative ion. The formation of a covalent bond constitutes a Lewis neutralization.
Na+Cl-,Na+.
Cl-,Cl-Na+
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8. Neutralization reactions:
Arrhenius:Brønsted-Lowry:Lewis:
9.
These ions are considered to be bases according to the Brønsted-Lowry theory, becausethey can accept a proton at any of their unshared pairs of electrons. They are consideredto be bases according to the Lewis acid-base theory, because they can donate anelectron pair.
10. The electrolytic compounds are acids, bases, and salts.
11. Names of the compounds in Table 15.3
sulfuric acid acetic acid
nitric acid carbonic acid
HCl hydrochloric acid nitrous acid
HBr hydrobromic acid sulfurous acid
perchloric acid hydrosulfuric acid
NaOH sodium hydroxide oxalic acid
KOH potassium hydroxide boric acid
calcium hydroxide HClO hypochlorous acid
barium hydroxide ammonia
HF hydrofluoric acid
12. Hydrogen chloride dissolved in water conducts an electric current. HCl reacts with polarwater molecules to produce and ions, which conduct electric current. Hexane is anonpolar solvent, so it cannot pull the HCl molecules apart. Since there are no ions in thehexane solution, it does not conduct an electric current. HCl does not ionize in hexane.
13. In their crystalline structure, salts exist as positive and negative ions in definite geometricarrangement to each other, held together by the attraction of the opposite charges. Whendissolved in water, the salt dissociates as the ions are pulled away from each other by thepolar water molecules.
Cl-H3O+
NH3Ba(OH)2
Ca(OH)2
H3BO3
H2C2O4
H2SHClO4
H2SO3
HNO2
H2CO3HNO3
HC2H3O2H2SO4
–Br(a)
–O H(b)
–C N(c)
ClAlCl
–
–Cl
ClAlCl
ClClCl+
AlCl3 + NaCl ¡ AlCl4
- + Na+(H+ + CN- ¡ HCN)HCl + KCN ¡ HCN + KCl
(H+ + OH- ¡ H2O)HCl + NaOH ¡ NaCl + H2O
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14. Testing the electrical conductivity of the solutions shows that is a nonelectrolyte,while NaOH is an electrolyte. This indicates that the OH group in must becovalently bonded to the group.
15. Molten NaCl conducts electricity because the ions are free to move. In the solid state,however, the ions are immobile and do not conduct electricity.
16. Dissociation is the separation of already existing ions in an ionic compound. Ionization isthe formation of ions from molecules. The dissolving of NaCl is a dissociation, since theions already exist in the crystalline compound. The dissolving of HCl in water is anionization process, because ions are formed from HCl molecules and
17. Strong electrolytes are those which are essentially 100% ionized or dissociated in water.Weak electrolytes are those which are only slightly ionized in water.
18. Ions are hydrated in solution because there is an electrical attraction between the chargedions and the polar water molecules.
19. The main distinction between water solutions of strong and weak electrolytes is thedegree of ionization of the electrolyte. A solution of an electrolyte contains many moreions than does a solution of a nonelectrolyte. Strong electrolytes are essentially 100%ionized. Weak electrolytes are only slightly ionized in water.
20. (a) In a neutral solution, the concentration of and are equal.(b) In an acid solution, the concentration of is greater than the concentration
of (c) In a basic solution, the concentration of is greater than the concentration
of
21. The net ionic equation for an acid-base reaction in aqueous solutions is:
22. The HCl molecule is polar and, consequently, is much more soluble in the polar solvent,water, than in the nonpolar solvent, hexane. There is also a chemical reaction betweenHCl and molecules.
23. Pure water is neutral because when it ionizes it produces equal molar concentrations ofacid and base ions.
24. The fundamental difference between a colloidal dispersion and a true solution lies in thesize of the particles. In a true solution particles are usually ions or hydrated moleculesand are less than 1 nm in size. In colloidals the particles are aggregates of ions ormolecules, ranging in size from 1-1000 nm.
[OH-][H+]
HCl + H2O ¡ H3O+ + Cl-H2O
H+ + OH- ¡ H2O.
H+.OH-
OH-.H+
OH-H+
H2O.
CH3
CH3OHCH3OH
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25. Dialysis is the process of removing dissolved solutes from a colloidal dispersion by useof a dialyzing membrane. The dissolved solutes pass through the membrane leaving thecolloidal dispersion behind. Dialysis is used in artificial kidneys to remove soluble wasteproducts from the blood.
26. A neutral solution is one in which the concentration of acid is equal to the concentrationof base An acidic solution is one in which the concentration of acid isgreater than the concentration of base A basic solution is one in whichthe concentration of base is greater than the concentration of acid
27. Acid rain is caused by the release of nitrogen and sulfur oxides into the air. Whenthese oxides are carried through the atmosphere they react with water and formsulfuric acid and nitric acid Precipitation (rain or snow) carries theacids to the ground.
28. A titration is used to determine the concentration of a specific substance (often an acid or abase) in a sample. A titration determines the volume of a reagent of known concentrationthat is required to completely react with a volume of a sample of unknown concentration.An indicator is used to help visualize the endpoint of a titration. The endpoint is the pointat which enough of the reagent of known concentration has been added to the sample ofunknown concentration to completely react with the unknown solution. An indicator colorchange is visible when the endpoint has been reached.
0.030 mol HCl reacts with 0.030 mol NaOH and produces 0.030 mol NaCl. Thefinal volume is 0.060 L. Since there isone mole each of sodium and chloride ions per mole of NaCl, the molarconcentration of and will be and
(c)
0.040 mol KOH reacts with 0.040 mol HCl. 0.040 mol HCl remains unreacted and0.040 mol KCl is produced. The final volume is 200.0 mL and contains 0.040 molHCl and 0.040 mol KCl. Moles of ions are: and
[H+] = tap water with a pH = 3.98 * 10-7 [H+] = 1.0 * 10-10
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16. (a) 100.0 mL of 2.0 M KCl and 100.0 mL of are mixed, giving a finalvolume of 200.0 mL and concentrations of 1.0 M KCl and Theconcentration of will be 1.0 M and the concentration of will be 0.50 M.The chloride ion concentration will be 2.0 M (1.0 M from the KCl and 2(0.50 M)from the ).
(b)
Final
The and the react completely producing insoluble andNo ions are present in solution.
(c)
The AgCl is insoluble and produces no ions. The 1.0 mol will produce1.0 mol ions and 1.0 mol ions. The final volume of the solution is 1.5 L.The concentration of the ions are:
17. The reaction of HCl and NaOH occurs on a mole ratio.
At the endpoint in these titration reactions, equal moles of HCl and NaOH will havereacted. At the endpoint, mol NaOH.Therefore, at the endpoint,
MAVA = MBVB MA =MBVB
VA
HCl = molMoles = (molarity)(volume).
HCl + NaOH ¡ NaCl + H2O
1 : 1
1.0 mol NO3
-
1.5 L= 0.67 M NO3
-1.0 mol Na+
1.5 L= 0.67 M Na+
NO3-Na+
NaNO3
NaCl(aq)
1.0 mol
+
AgNO3(aq)
1.0 mol¡
AgCl(s)
1.0 mol
+
NaNO3(aq)
1.0 mol
(1.00 L AgNO3)a1.00 mol
Lb = 1.0 mol AgNO3
(0.500 L NaCl)a2.0 mol
Lb = 1.0 mol NaCl
H2O.BaSO4Ba(OH)2H2SO4
H2SO4
0.0070 mol
+
Ba(OH)2
0.0070 mol¡
BaSO4(s)
0.0070 mol
+
2 H2O
0.014 mol
volume = 35.0 mL + 35.0 mL = 70.0 mL
(35.0 mL)a 1 L
1000 mLb ¢0.20 mol H2SO4
L≤ = 0.0070 mol H2SO4
(35.0 mL)a 1 L
1000 mLb ¢0.20 mol Ba(OH)2
L≤ = 0.0070 mol Ba(OH)2
CaCl2
Ca2+K+0.50 M CaCl2 .
1.0 M CaCl2
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(a)
(b)
(c)
18. The reaction of HCl and NaOH occurs on a mole ratio.
At the endpoint in these titration reactions, equal moles of HCl and NaOH will havereacted. At the endpoint, mol NaOH.Therefore, at the endpoint,
(a)
(b)
(c)
19. (a)
(b)
(c)
20. (a)
(b)
(c)
21. The more acidic solution is listed followed by an explanation.
(a) 1 molar The concentration of in is greater than 1 M sincethere are two ionizable hydrogens per mole of In HCl the concentration of
will be 1 M, since there is only one ionizable hydrogen per mole HCl.
(b) 1 molar HCl. HCl is a strong electrolyte, producing more than whichis a weak electrolyte.
HC2H3O2H+H+
H2SO4.1 M H2SO4H+H2SO4.
Al3+(aq) + PO4
3-(aq) ¡ AlPO4(s)
Zn(s) + 2 H+(aq) ¡ Zn2+(aq) + H2(g)
H2S(g) + Cd2+(aq) ¡ CdS(s) + 2 H+(aq)
Mg(s) + 2 HC2H3O2(aq) ¡ Mg2+(aq) + 2 C2H3O2
-(aq) + H2(g)
CaCO3(s) + 2 H+(aq) ¡ Ca2+(aq) + CO2(g) + H2O(l)
SO4
2-(aq) + Ba2+(aq) ¡ BaSO4(s)
(13.13 mL)(1.425 M)
39.39 mL= 0.4750 M NaOH
(48.04 mL)(0.482 M)
24.02 mL= 0.964 M NaOH
(37.19 mL)(0.126 M)
31.91 mL= 0.147 M NaOH
MB =MAVA
VBMAVA = MBVB
HCl = molMoles = (molarity)(volume).
HCl + NaOH ¡ NaCl + H2O
1 : 1
(18.00 mL)(0.555 M)
27.25 mL= 0.367 M HCl
(33.66 mL)(0.306 M)
19.00 mL= 0.542 M HCl
M HCl =(37.70 mL)(0.728 M)
40.13 mL= 0.684 M HCl
(37.70 mL)(0.728 M) = (40.13 mL)(M HCl)
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22. The more acidic solution is listed followed by an explanation.
(a) 2 molar HCl. 2 M HCl will yield concentration. 1 M HCl will yield concentration.
(b) 1 molar Both are strong acids. The concentration of in isgreater than in because has two ionizable hydrogens per molewhereas has only one ionizable hydrogen per mole.
23.
0.245 M HCl contains 0.245 mol
of 0.245 M HCl
24.
0.245 M HCl contains
of 0.245 M HCl
25.
First calculate the grams of NaOH in the sample.
in the sample
26.
in the samplea 40.00 g
molb = 0.930 g NaOH(0.04990 L HCl)a0.466 mol
Lb a 1 mol NaOH
1 mol HClb
L HCl ¡ mol HCl ¡ mol NaOH ¡ g NaOH
NaOH + HCl ¡ NaCl H2O
a0.1756 g NaOH
0.200 g sampleb(100) = 87.8% NaOH
a 40.00 g
molb = 0.1756 g NaOH(0.01825 L HCl)a0.2406 mol
Lb a 1 mol NaOH
1 mol HClb
L HCl ¡ mol HCl ¡ mol NaOH ¡ g NaOH
NaOH + HCl ¡ NaCl + H2O
= 40.8 mL
(0.0500 L Ca(OH)2)a0.100 mol
Lb ¢ 2 mol HCl
1 mol Ca(OH)2≤ a 1000 mL
0.245 molb
0.245 mol HCl>1000 mL
M Ca(OH)2 ¡ mol Ca(OH)2 ¡ mol HCl ¡ mL HCl
2 HCl + Ca(OH)2 ¡ CaCl2 + 2 H2O
= 1.57 * 103 mL
(10.0 g Al(OH)3)a 1 mol
78.00 gb ¢ 3 mol HCl
1 mol Al(OH)3≤ a 1000 mL
0.245 molb
HCl>1000 mL
g Al(OH)3 ¡ mol Al(OH)3 ¡ mol HCl ¡ mL HCl
3 HCl + Al(OH)3 ¡ AlCl3 + 3 H2O
HNO3
H2SO41 M HNO3
1 M H2SO4H+H2SO4.
1 M H+2 M H+
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in the sample
in the sample
27.
This is a limiting reactant problem. First find the moles of Zn and HCl from the givendata and then identify the limiting reactant.
Therefore Zn is in excess and HCl is the limiting reactant.
produced in the reaction
28.
This is a limiting reactant problem. First find moles of Zn and HCl from the given dataand then identify the limiting reactant.
Zn is in excess and HCl is the limiting reactant.
V =nRT
P=
(0.0350 mol H2)(0.0821 L atm>mol K)(300. K)
0.921 atm= 0.936 L H2
PV = nRT
T = 27°C = 300. K P = (700. torr)a 1 atm
760 torrb = 0.921 atm
(0.0700 mol HCl)¢ 1 mol H2
2 mol HCl≤ = 0.0350 mol H2
(0.200 L HCl)a0.350 mol
Lb = 0.0700 mol HCl
g Zn ¡ mol Zn (5.00 g Zn)a 1 mol
65.39 gb = 0.0765 mol Zn
Zn + 2 HCl ¡ ZnCl2 + H2
V =nRT
P=
(0.0175 mol H2)(0.0821 L atm>mol K)(300. K)
0.921 atm= 0.468 L H2
PV = nRT
T = 27°C = 300. K P = (700. torr)a 1 atm
760 torrb = 0.921 atm
(0.0350 mol HCl)¢ 1 mol H2
2 mol HCl≤ = 0.0175 mol H2
(0.100 L HCl)a0.350 mol
Lb = 0.0350 mol HCl
(5.00 g Zn)a 1 mol
65.39 gb = 0.0765 mol Zng Zn ¡ mol Zn
Zn + 2 HCl ¡ ZnCl2 + H2
a 0.070 g NaCl
1.00 g sampleb (100) = 7.0% NaCl
1.00 g sample - 0.930 g NaOH = 0.070 g NaCl
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29. Calculation of the pH solutions:
(a)
(b)
(c)
30. (a)
(b)
(c)
31. (a)
(b)
32. (a)
(b)
33. (a)
(b)
(c)
(d)
34. (a)
(b)
(c)
(d)
35. (a) (d) basic;
(b) basic (e) neutral;
(c) basic (f) acidic
pH = 7
pH = 9acidic; pH = 4
HBr is a strong acid HBr(aq) ¡ H+(aq) + Br-(aq)
HClO4 is a strong acid HClO4(aq) ¡ H+(aq) + ClO-4 (aq)
Ba(OH)2 is a strong base Ba(OH)2 ¡ Ba2+(aq) + 2OH-(aq)
H2C2O4 is a weak acid H2C2O4(aq) ∆ 2 H+(aq) + C2O2-4 (aq)
HC2H3O2 is a weak acid HC2H3O2(aq) ∆ H+(aq) + C2H3O-2 (aq)
H+ = 0.00010 M = 1.0 * 10-4 M; pH = - log11.0 * 10-42 = 4.00
H+ = 0.50 M; pH = - log15.0 * 10-12 = 0.30
H+ = 1 * 10-7 M; pH = - log11 * 10-72 = 7.0
H+ = 6.5 * 10-9 M; pH = - log16.5 * 10-92 = 8.19
H+ = 1.0 M; pH = - log 1.0 = 0
H+ = 0.01 M = 1 * 10-2 M; pH = - log11 * 10-22 = 2.0
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H2O
H2O
H2O
H2O
H2O
H2O
H2O
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36. (a)
For each ionic compound, 1 calcium ion and 2 chloride ions result.
(b)
For each KF ionic compound, 1 potassium ion and 1 fluoride ion result.
(c)
For each ionic compound, 1 aluminum ion and 3 bromide ions result.
37.
38.
M HCl ¡ mol HCl ¡ mol Ba(OH)2 ¡ M Ba(OH)2
Ba(OH)2 + 2 HCl ¡ BaCl2 + 2 H2O
¢0.520 mol I-
L≤ ¢1 mol Ca2+
2 mol I≤ = ¢ 0.260 mol Ca2+
L≤ = 0.260 M Ca2+
CaI2 ¡ Ca2+ + 2 I-
Al3+ Br– Br– Br–
AlBr3
AlBr3(s) ¡ Al3+(aq) + 3 Br -(aq)
K+ F –
KF(s) ¡ K+(aq) + F-(aq)
Ca2+ Cl– Cl–
CaCl2
CaCl2(s) ¡ Ca2+(aq) + 2 Cl-(aq)
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39. The acetic acid solution freezes at a lower temperature than the alcohol solution. Theacetic acid ionizes slightly while the alcohol does not. The ionization of the acetic acidincreases its particle concentration in solution above that of the alcohol solution, resultingin a lower freezing point for the acetic acid solution.
40. It is more economical to purchase at the same cost per pound as Because has a lower molar mass than the solution willcontain more particles per pound in a given solution and therefore, have a greater effecton the freezing point of the radiator solution.Assume 100. g of each compound.
41. A hydronium ion is a hydrated hydrogen ion.
42. Freezing point depression is directly related to the concentration of particles in thesolution.
HClHighest freezing point Lowest freezing point
1 mol mol 2 mol 3 mol (particles in solution)
43. (a)
(b) so, concentration is higher at 100 C.
(c) The water is neutral at both temperatures, because the ionizes into equalconcentrations of and at any temperature.OH-H+
H2O
°H+1 * 10-6 7 1 * 10-7
25°C pH = - log11 * 10-72 = 7.0 pH of H2O is greater at 25°C
100°C pH = - log11 * 10-62 = 6.0
1+
CaCl277HC2H3O27C12H22O11
H+
(hydrogen ion)
+
H2O
¡
H3O+
(hydronium ion)
CH3CH2OH: 100. g
46.07 g>mol= 2.17 mol
CH3OH: 100. g
34.04 g>mol= 2.84 mol
CH3OHC2H5OH,CH3OHC2H5OH.CH3OH
0.00630 mol Ba(OH)2
0.02040 L= 0.309 M Ba(OH)2
(0.0126 mol HCl)a1 mol Ba(OH)2
2 mol HClb = 0.00630 mol Ba(OH)2
a0.430 mol HCl
Lb a 1 L
1000 mLb (29.26 mL) = 0.0126 mol HCl
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44. As the pH changes by 1 unit, the concentration of in solution changes by a factor of 10.For example, the pH of 0.10 M HCl is 1.00, while the pH of 0.0100 M HCl is 2.00.
45. A 1.00 m solution contains 1 mol solute plus We need to find the total numberof moles and then calculate the mole percent of each component.
46.
47.
48.
49. pH of 1.0 L solution containing 0.1 mL of 1.0 M HCl
(g) 50.00 mL of 0.2000 M NaOH neutralizes 50.00 mL of 0.2000 M HCl. No excessacid or base is in the solution. Therefore, the solution is neutral with a
54. (a)
(b)
(c)
55.
56. Yes, adding water changes the concentration of the acid, which changes the concentrationof the and changes the pH. The pH will rise.
No, the solution theoretically will never reach a pH of 7, but it will approach pH 7 aswater is added.
[H+],
M1 = 12 M HNO3 (original solution)
(M1)(10.0 mL) = (1.2 M)(100.00 mL)
Dilution problem M1V1 = M2V2
MA = 1.2 M (diluted solution)
(MA)(25 mL) = (0.60 M)(50.0 mL)
MAVA = MBVB
HNO3 + KOH ¡ KNO3 + H2O
(0.0050 mol H2SO4)¢ 1 mol Na2SO4
1 mol H2SO4≤ a142.1 g
molb = 0.71 g Na2SO4
(0.0050 mol H2SO4)¢ 2 mol NaOH
1 mol H2SO4≤ a 1000 mL
0.10 molb = 1.0 * 102 mL NaOH
mol H2SO4 ¡ mol NaOH ¡ mL NaOH
2 NaOH(aq) + H2SO4(aq) ¡ Na2SO4(aq) + 2 H2O(l)
0
mL NaOH
1
2
3
4
5
6
7
8
0 10 20 30 40 50
pHpH = 7.0
[H+] =2 * 10-6 mol
0.09999 L pH = - log¢ 2 * 10-6
9.999 * 10-2 ≤ = 4.7
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57. First determine the molarity of the two HCl solutions. Take the antilog of the pH value toobtain the
