Chapter 1 - Solutions
CHAPTER 4 Problems: 2, 8, 11, 14, 20, 22, 25, 28, 29, 30, 32, 42, 44, 48, 52, 54, 64, 68, 82, 84, 94, 124, 132, 150
2) What is the difference between a nonelectrolyte and an electrolyte? Between a weak electrolyte and a strong electrolyte?
An electrolyte is a substance that produces ions when dissolved in water. A non-electrolyte is a substance that does not produce ions when dissolved in water (such as sugar). Note that water itself is considered a nonelectrolyte, although it does produce a very small (~ 10-7 mol/L) concentration of ions at room temperature due to self-ionization.
A strong electrolyte completely ionizes when dissolved in water to form two or more ions per formula unit of compound. A weak electrolyte ionizes to only a small extent, producing much less than two ions per formula unit of dissolved compound. Strong acids and ionic compounds are strong electrolytes, while weak acids and weak bases are weak electrolytes. Note that “insoluble” ionic compounds are strong electrolytes, because for whatever amount of the compound dissolves in water two or more ions are formed.
Figure 4.2 on page 129 gives a useful flow chart for classifying substances as strong electrolytes, weak electrolytes, or nonelectrolytes.
8) Identify each of the following substances as a strong electrolyte, a weak electrolyte, or a nonelectrolyte.
a) H2O nonelectrolyte
b) KCl strong electrolyte (ionic compound)
c) HNO3 strong electrolyte (strong acid)
d) HC2H3O2 weak electrolyte (weak acid)
e) C12H22O11 nonelectrolyte
11) Predict and explain which of the following systems are electrically conducting:
a) solid NaCl
Nonconducting. While there are cations and anions present, they hold fixed positions in the crystal structure and so cannot move about and transfer charge.
b) molten (liquid) NaCl
Conducting. The ions are in the molten (liquid) state, and so can move freely and transfer charge.
c) an aqueous solution of NaCl
Conducting. Sodium chloride ionizes in water, and so the cations and anions can move freely through the water and transfer charge.
14) Describe hydration. What properties of water enable its molecules to interact with ions in solution?
Hydration is the process where particles (often ions) are surrounded by water molecules. Because water molecules have a partial positive charge on the hydrogen atoms and a partial negative charge on the oxygen atom, the molecules can strongly interact with anions and cations. Water molecules can pull ions out of a crystal structure, and by surrounding the ions, minimize interactions between cations and anions.
20) Characterize the following compounds as soluble or insoluble in water:
a) CaCO3 Insoluble
b) ZnSO4 Soluble (most sulfate compounds are soluble)
c) Hg(NO3)2 Soluble (nitrate compounds are soluble)
d) HgSO4 Soluble (most sulfate compounds are soluble. Note that
Hg2SO4 (mercury (I) sulfate) is insoluble).
e) NH4ClO4 Soluble (both ammonium and perchlorate compounds are soluble)
22) Write total ionic and net ionic equations for the following reactions:
a) Na2S(aq) + ZnCl2(aq) ( 2 NaCl(aq) + ZnS(s)
TOTAL 2 Na+(aq) + S2-(aq) + Zn2+(aq) + 2 Cl-(aq)
( 2 Na+(aq) + 2 Cl-(aq) + ZnS(s)
NET Zn2+(aq) + S2-(aq) ( ZnS(s)
b) 2 K3PO4(aq) + 3 Sr(NO3)2 ( 6 KNO3(aq) + Sr3(PO4)2(s)
TOTAL 6 K+(aq) + 2 PO43-(aq) + 3 Sr2+(aq) + 6 NO3-(aq)
( 6 K+(aq) + 6 NO3-(aq) + Sr3(PO4)2(s)
NET 3 Sr2+(aq) + 2 PO43-(aq) ( Sr3(PO4)2(s)
c) Mg(NO3)2(aq) + 2 NaOH(aq) ( 2 NaNO3(aq) + Mg(OH)2(s)
TOTAL Mg2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 OH-(aq)
( 2 Na+(aq) + 2 NO3-(aq) + Mg(OH)2(s)
NET Mg2+(aq) + 2 OH-(aq) ( Mg(OH)2
25) Give the Arrhenius and the Bronsted definition of an acid and a base. Why are the Bronsted definitions more useful in describing acid-base properties?
A Bronsted acid acts as a proton donor (H+ donor) in a chemical reaction, while a Bronsted base acts as a proton acceptor. For example, in the equation
HCl(aq) + NH3(aq) ( NH4+(aq) + Cl-(aq)
HCl is donating a proton, and so is a Bronsted acid, and NH3 is accepting a proton, and so is a Bronsted base.
The Bronsted definition of acids and bases is usually (although not always) more useful than the Arrhenius definition because it is less restrictive. It allows classification of reactions as acid-base reactions (in the Bronsted sense) even for processes not taking place in water.
28) What factors qualify a compound as a salt? Specify which of the following compounds are salts:
A salt is an ionic compound that can be formed by an acid-base reaction. In practical terms, all common ionic compounds except hydroxides and oxides are salts.
In the answers below, I have given an acid-base reaction that will form the compounds classified as salts.
a) CH4 not a salt (a molecular compound)
b) NaF a salt ( NaOH + HF ( NaF + H2O)
c) NaOH not a salt (although an ionic compound)
d) CaO not a salt (although an ionic compound)
e) BaSO4 a salt (Ba(OH)2 + H2SO4 ( BaSO4 + 2 H2O)
f) HNO3 not a salt (a strong acid)
g) NH3 not a salt (a weak base)
h) KBr a salt (KOH + HBr ( KBr + H2O)
29) Identify each of the following as a weak or strong acid or base:
a) NH3 weak base
b) H3PO4 weak (polyprotic) acid
c) LiOH strong soluble base
d) HCOOH (formic acid) weak acid
e) H2SO4 strong acid
f) HF weak acid
g) Ba(OH)2 strong soluble base
30) Identify each of the following species as a Bronsted acid, a Bronsted base, or both.
a) HI Bronsted acid
b) C2H3O2- Bronsted base. To see this, consider the reaction of this ion with
water.
C2H3O2-(aq) + H2O(l) ( HC2H3O2(aq) + OH-(aq)
c) H2PO4- Both (and so amphoteric). Consider the following reactions
H2PO4-(aq) + H2O(l) ( HPO42-(aq) + H3O+(aq)
H2PO4-(aq) + HCl(aq) ( H3PO4(aq) + Cl-(aq)
In the first reaction the hydrogen phosphate ion donates a proton and so acts as a Bronsted acid. In the second reaction the ion accepts a proton and so acts as a base.
d) HSO4- Both (and so amphoteric).
32) Balance the following equations, and write the molecular equations, total ionic equations, and net ionic equations (if appropriate):
a)
MOLECULAR HBr(aq) + NH3(aq) ( NH4Br(aq)
TOTAL IONIC H+(aq) + Br-(aq) + NH3(aq) ( NH4+(aq) + Br-(aq)
NET IONIC H+(aq) + NH3(aq) ( NH4+(aq)
b)
MOLECULAR 3 Ba(OH)2(aq) + 2 H3PO4(aq) ( Ba3(PO4)2(s) + 6 H2O(l)
TOTAL IONIC 3 Ba2+(aq) + 6 OH-(aq) + 2 H3PO4(aq)
( Ba3(PO4)2(s) + 6 H2O(l)
NET IONIC the same as the total ionic (no spectator ions)
c)
MOLECULAR 2 HClO4(aq) + Mg(OH)2(s) ( Mg(ClO4)2(aq) + 2 H2O(l)
TOTAL IONIC 2 H+(aq) + 2 ClO4-(aq) + Mg(OH)2(s)
( Mg2+(aq) + 2 ClO4-(aq) + 2 H2O(l)
NET IONIC 2 H+(aq) + Mg(OH)2 ( Mg2+(aq) + 2 H2O(l)
42) For the complete redox reactions given here, write the half-reactions and identify the oxidizing and reducing agents:
a) 4 Fe + 3 O2 ( 2 Fe2O3
oxidation 4 Fe ( 4 Fe3+ + 12 e-
reduction 3 O2 + 12 e- ( 6 O2-
oxidizing agent is O2; reducing agent is Fe
b) Cl2 + 2 NaBr ( 2 NaCl + Br2
oxidation 2 Br- ( Br2 + 2 e-
reduction Cl2 + 2 e- ( 2 Cl-
oxidizing agent is Cl2; reducing agent is Br-
c) Si + 2 F2 ( SiF4
oxidation Si ( Si4+ + 4 e-
reduction 2 F2 + 4 e- ( 4 F-
oxidizing agent is F2; reducing agent is Si
d) H2 + Cl2 ( 2 HCl
oxidation H2 ( 2 H+ + 2 e-
reduction Cl2 + 2 e- 2 Cl-
oxidizing agent is Cl2; reducing agent is H2
44) Phosphorus forms many oxoacids (ternary acids containing H, O, and P). Indicate the oxidation number of phosphorus in each of the following acids:
Note that in all of the compounds below hydrogen has an oxidation number of +1 and oxygen has an oxidation number of -2.
a) HPO3 +5
b) H3PO2 +1
c) H3PO3 +3
d) H3PO4 +5
e) H4P2O7 +5
f) H5P3O10 +5
48) Give the oxidation number of the underlined atoms in each of the following molecules and ions:
a) Mg3N2 Mg is +2, and so N is -3
b) CsO2 Cs is +1, and so O is -1/2 (weird, but fractional oxidation
numbers occasionally occur)
c) CaC2 Ca is +2, and so C is -1
d) CO32- O is -2, and so C is +4
e) C2O42- O is -2, and so C is +3
f) ZnO22- Zn is +2, and so O is -2
g) NaBH4 Na is +1, H is -1, and so B is +3 (I got this answer using a
more sophisticated method for finding oxidation numbers,
and so would not expect you to be able to do this one)
h) WO42- O is -2, and so W is +6
52) Predict the outcomes of the reactions represented by the following equations by using the activity series, and balance the equations:
a) Cu(s) + HCl(aq) ( no reaction (copper is below hydrogen in the
activity series)
b) Au(s) + NaBr(aq) ( no reaction (gold is below sodium in the
activity series)
c) Mg(s) + CuSO4(aq) ( MgSO4(aq) + Cu(s)
d) Zn(s) + KBr(aq) ( no reaction
54) Classify the following redox reactions as combination, decomposition, or displacement:
a) P4 + 10 Cl2 ( 4 PCl5 combination
b) 2 NO ( N2 + O2 decomposition
c) Cl2 + KI ( 2 KCl + I2 displacement (Cl displaces I)
d) 3 HNO2 ( HNO3 + H2O + 2 NO decomposition
64) Calculate the molarity of each of the following solutions:
a) 6.57 g of methanol (CH3OH) in 1.50 x 102 mL of solution.
M(CH3OH) = 32.04 g/mol
Moles CH3OH = 6.57 g 1 mol = 0.2051 mol
32.04 g
[CH3OH] = 0.2051 mol = 1.37 mol/L
0.150 L
b) 10.4 g of calcium chloride (CaCl2) in 2.20 x 102 mL of solution
M(CaCl2) = 110.98 g/mol
Moles CaCl2 = 10.4 g 1 mol = 0.09371 mol
110.98 g
[CaCl2] = 0.09371 mol = 0.426 mol/L
0.220 L
c) 7.82 g of naphthalene (C10H8) in 85.2 mL of benzene solution
M(C10H8) = 128.2 g/mol
Moles C10H8 = 7.82 g 1 mol = 0.0700 mol
128.2 g
[C10H8] = 0.0700 mol = 0.716 mol/L
0.0852 L
68) Water is added to 25.0 mL of a 0.866 M KNO3 solution until the volume of the solution is exactly 500. mL. What is the concentration of the final solution?
McLc = MdLd
Md = (Lc/Ld) Mc = (25.0 mL/500. mL) (0.866M) = 0.0433 M
82) A sample of 0.6760 g of an unknown compound containing barium ions (Ba2+) is dissolved in water and treated with excess Na2SO4. If the mass of the BaSO4 precipitate formed is 0.4105 g, what is the percent by mass of Ba in the original unknown compound?
The reaction taking place is
Ba2+(aq) + SO42-(aq) ( BaSO4(s)
M(BaSO4) = 233.4 g/mol M(Ba2+) = 137.3 g/mol
Moles precipitate = 0.4105 g 1 mol =1.759 x 10-3 mol
233.4 g
grams Ba2+ ion = 1.759 x 10-3 mol BaSO4 1 mol Ba2+ 137.3 g =0.2415 g Ba2+
1 mol BaSO4 1 mol
% Ba by mass = 0.2415 g 100. % = 35.7 %
0.6760 g
84) Calculate the concentration (in molarity) of an NaOH solution if 25.0 mL of the solution is needed to neutralize 17.4 mL of a 0.312 M HCl solution.
The neutralization reaction is
HCl + NaOH ( NaCl + H2O
Moles NaOH = 0.0174 mL HCl soln 0.312 mol HCl 1 mol NaOH = 5.429 x 10-3 mol
L soln 1 mol HCl
[NaOH] = 5.429x 10-3 mol = 0.2172 mol/L
0.0250 L
94) Sodium carbonate (Na2CO3) is available in very pure form and can be used to standardize acid solutions. What is the molarity of an HCl solution if 28.3 mL of the solution is required to react with 0.256 g of Na2CO3?
The reaction taking place is
2 HCl + Na2CO3 ( 2 NaCl + H2CO3
M(Na2CO3) = 106.0 g/mol
Moles HCl = 0.256 g Na2CO3 1 mol Na2CO3 2 mol HCl = 4.830 x 10-3 mol
106.0 g 1 mol Na2CO3
[HCl] = 4.830 x 10-3 mol = 0.1707 mol/L
0.0283 L
124) A 0.8870 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO3, to yield 1.913 g of AgCl. Calculate the percent by mass of each compound in the mixture.
This problem requires a bit of thought. The reaction of interest involves chloride ion
Ag+(aq) + Cl-(aq) ( AgCl(s)
Based on the mass of precipitate formed we can determine the grams of Cl-. Since the only sources of chloride ion are the NaCl and KCl in the mixture, it follows that
total mass Cl- = mass from NaCl + mass from KCl
mass Cl- from NaCl = (mass of NaCl) (fraction Cl in NaCl)
mass Cl- from KCl = (mass KCl) (fraction Cl in KCl)
M(AgCl) = 143.4 g/mol M(NaCl) = 58.44 g/mol M(KCl) = 74.55 g/mol
fraction Cl in NaCl) = 35.45 g Cl/mol NaCl = 0.6066
58.44 g NaCl/mol NaCl
fraction Cl in KCl) = 35.45 g Cl/mol KCl = 0.4755
74.55 g NaCl/mol KCl
total mass Cl = 1.913 g AgCl 1 mol AgCl 1 mol Cl 35.45 g Cl = 0.4729 g Cl
143.4 g 1 mol AgCl 1 mol Cl
Now, let x = mass of NaCl. Then (0.8870 - x) = mass of KCl. Using the equation for the total mass of Cl, we get
0.4729 g = x (0.6066) + (0.8870 - x) (0.4755) = 0.4218 g - 0.1311 x
x = (0.4729 g - 0.4218 g) = 0.3898 g
0.1311
The percent by mass NaCl in the sample is
% NaCl(by mass) = 0.3898 g NaCl 100. % = 43.9 % NaCl by mass
0.8870 g sample
132) The recommended procedure for preparing a very dilute solution is not to weigh out a very small mass or measure a very small volume of a stock solution. Instead, it is done by a series of dilutions. A sample of 0.8214 g of KMnO4 was dissolved in water and made up to a volume of 500.0 mL in a volumetric flask. A 2.000 mL sample of this solution was transferred to a 1000.0 mL volumetric flask and diluted to the mark with water. Next, 10.00 mL of the diluted solution was transferred to a 250.0 mL flask and diluted to the mark with water.
a) Calculate the concentration (in molarity) of the final solution.
We may use McLc = MdLd , which gives Md = (Lc/Ld) Mc , in all of the dilution steps.
For the initial concentration
M(KMnO4) = 158.0 g/mol
Moles KMnO4 = 0.8214 g 1 mol = 5.199 x 10-3 mol
158.0 g
Initial concentration = 5.199 x 10-3 mol = 1.0397 x 10-2 mol/L
0.500 L
1st dilution Md = (2.000 mL/1000.0 mL)(1.0397 x 10-2 M) = 2.079 x 10-5 mol/L
2nd dilution Md = (10.00 mL/250.0 mL) (2.079 x 10-5 M) = 8.32 x 10-7 mol/L
This is the molarity of the final solution
b) Calculate the mass of KMnO4 needed to prepare the final solution if it was prepared directly.
The mass of KMnO4 needed to prepare 250.0 mL of a 8.32 x 10-7 M solution directly is
mass KMnO4 = 0.250 L soln 8.32 x 10-7 mol 158.0 g KMnO4 = 3.29 x 10-5 g
L soln mol
It would be difficult to precisely measure out such a small mass of this compound directly, which is why the dilute solution is prepared by serial dilution.
150) The current maximum level of fluoride ion that the EPA allows in U.S. drinking water is 4.0 mg/L. Convert this concentration into molarity.
[F-] = 4.0 x 10-3 g 1 mol = 2.11 x 10-4 M
1. L 19.00 g
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