AP Chemistry Chapter 4 Answers – Zumdahl
4.36
The solubility rules referenced in the following answers are from Table 4.1 of the text. The
phrase “slightly soluble” is interpreted to mean insoluble and the phrase “marginally soluble” is
interpreted to mean soluble.
a.
b.
c.
d.
e.
f.
g.
h.
Soluble (Rule 3)
Soluble (Rule 1)
Insoluble (Rule 4)
Soluble (Rules 2 and 3)
Insoluble (Rule 6)
Insoluble (Rule 5)
Insoluble (Rule 6)
Soluble (Rule 2)
4.38
Use Table 4.1 to predict the solubility of the possible products.
a.
b.
c.
d.
Possible products = Hg2SO4 and Cu(NO3)2; precipitate = Hg2SO4
Possible products = NiCl2 and Ca(NO3)2; Both salts are soluble so no precipitate forms.
Possible products = KI and MgCO3; precipitate = MgCO3
Possible products = NaBr and Al2(CrO4)3; precipitate = Al2(CrO4)3
4.44
a.
CrCl3(aq) + 3NaOH(aq) → Cr(OH)3(s) + 3NaCl(aq)
Cr3+(aq) + 3OH-(aq) → Cr(OH)3(s)
b.
2AgNO3(aq) + (NH4)2CO3(aq) → Ag2CO3(s) + 2NH4NO3(aq)
2Ag+(aq) + CO32-(aq) → Ag2CO3(s)
c.
CuSO4(aq) + Hg2(NO3)2(aq) → Cu(NO3)2(aq) + Hg2SO4(s)
Hg22+(aq) +_SO42-(aq) → Hg2SO4(s)
d.
No reaction occurs because all possible products (SrI2 and KNO3) are soluble.
4.46
Because no precipitates formed upon addition of NaCl or Na2SO4, we can conclude that Hg22+
and Ba2+ are not present in the sample since Hg2Cl2 and BaSO4 are insoluble salts. However,
Mn2+ may be present since Mn2+ does not form a precipitate with either NaCl or Na2SO4. A
precipitate formed with NaOH; the solution must contain Mn2+ because it forms a precipitate
with OH- [Mn(OH)2(s)].
4.47
2AgNO3(aq) + Na2CrO4(aq) → Ag2CrO4(s) + 2NaNO3(aq)
0.0750 L x 0.100 mol AgNO3/1 L x 1 mol Na2CrO4/2 mol AgNO3 x
161.98 g Na2CrO4/1 mol Na2CrO4 = 0.607 g Na2CrO4
4.49
Al(NO3)3(aq) + 3KOH(aq) → Al(OH)3(s) + 3KNO3(aq)
0.0500 L x 0.200 mol Al(NO3)3/1 L = 0.0100 mol Al(NO3)3
0.2000 L x 0.100 mol KOH/1 L = 0.0200 mol KOH
From the balanced equation, 3 mol of KOH are required to react with 1 mol of Al(NO3)3 (3:1
mole ratio). The actual KOH to Al(NO3)3 mole ratio present is 0.0200/0.0100 = 2 (2:1). Since
the actual mole ratio present is less than the required mole ratio, KOH (the numerator) is the
limiting reagent.
0.0200 mol KOH x 1 mol Al(OH)3/3 mol KOH x 78.00 g Al(OH)3/1 mol Al(OH)3 =
0.520 g Al(OH)3
4.51
2 AgNO3(aq) + CaCl2(aq) → 2AgCl(s) + Ca(NO3)2(aq)
mol AgNO3 = 0.1000 L x 0.20 mol AgNO3/1 L = 0.020 mol AgNO3
mol CaCl2 = 0.1000 L x 0.15 mol CaCl2/1 L = 0.015 mol CaCl2
The required mol AgNO3 to mol CaCl2 ratio is 2:1 (from the balanced equation). The actual
mole ratio present is 0.020/0.015 = 1.3 (1.3:1). Therefore, AgNO3 is the limiting reagent.
mass AgCl = 0.020 mol AgNO3 x 1 mol AgCl/1 mol AgNO3 x 143.4 g AgCl/1 mol AgCl =
2.9 g AgCl
The net ionic equation is: Ag+(aq) + Cl-(aq) → AgCl(s). The ions remaining in solution are the
unreacted Cl- ions and the spectator ions, NO3- and Ca2+ (all of the Ag+ is used up in forming
AgCl). The mole of each ion present initially (before reaction) can be easily determined from
the moles of each reactant. 0.020 mol AgNO3 dissolves to form 0.020 mol Ag+ and 0.020 mol
NO3-. 0.015 mol CaCl2 dissolves to form 0.015 mol Ca2+ and 2(0.015) = 0.030 mol Cl-.
mol unreacted Cl- = 0.030 mol Cl- initially = 0.020 mol Cl- reacted = 0.010 mol Cl- unreacted
M(Cl-) = 0.010 mol Cl-/total volume = 0.010 mol Cl-/0.1000 L + 0.1000 L = 0.050 M ClThe molarity of the spectator ions are:
M(NO3-) = 0.20 mol NO3-/0.2000 L = 0.10 M NO3M(Ca2+) = 0.015 mol Ca2+/0.2000 L = 0.075 M Ca2+
4.56
a.
3HNO3(aq) + Al(OH)3(s) → 3H2O(l) + Al(NO3)3(aq)
3H+(aq) + 3NO3-(aq) + Al(OH)3(s) → 3H2O(l) + Al3+(aq) + 3NO3-(aq)
3H+ (aq) + Al(OH)3(s) → 3H2O(l) + Al3+(aq)
b.
HC2H3O2(aq) + KOH(aq) → H2O(l) + KC2H3O2(aq)
HC2H3O2(aq) + K+(aq) + OH-(aq) → H2O(l) + K+(aq) + C2H3O2-(aq)
HC2H3O2(aq) + OH-(aq) → H2O(l) + C2H3O2-(aq)
c.
Ca(OH)2(aq) + 2HCl(aq) → 2H2O(l) + CaCl2(aq)
Ca2+(aq) + 2OH-(aq) + 2H+(aq) + 2Cl-(aq) → 2H2O(l) + Ca2+(aq) + 2Cl-(aq)
2H+(aq) + 2OH-(aq) → 2H2O(l) or H+(aq) + OH-(aq) → H2O(l)
4.59
If we begin with 50.00 mL of 0.200 M NaOH, then:
50.00 x 10-3 L x 0.200 mol/1 L = 1.00 x 10-2 mol NaOH is to be neutralized.
a.
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
1.00x 10-2 mol NaOH x 1 mol HCl/1 mol NaOH x 1 L/0.100 mol = 0.100 L or 100. mL
b.
HNO3(aq) + NaOH (aq) → H2O(l) + NaNO3(aq)
1.00 x 10-2 mol NaOH x 1 mol HNO3/1 mol NaOH x 1 L/0.150 mol HNO3 =
6.67 x 10-2 L or 66.7 mL
c.
HC2H3O2(aq) + NaOH (aq) → H2O(l) + NaC2H3O2(aq)
1.00 x 10-2 mol NaOH x 1 mol HC2H3O2/1 mol NaOH x 1 L/0.200 mol HC2H3O2
= 5.00 x 10-2 L = 50.0 mL
4.63
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
24.16 x 10-3 L NaOH x 0.106 mol NaOH/1 L NaOH x 1 mol HCl/1 mol NaOH =
2.56 x 10-3 mol HCl
Molarity of HCl = 2.56 x 10-3 mol/25.00 x 10-3 L = 0.102 M HCl
4.65
KHP is a monoprotic acid: NaOH(aq) + KHP(aq) → H2O(l) + NaKP(aq)
Mass KHP = 0.02046 L NaOH x 0.1000 mol NaOH/1 L NaOH x 1 mol KHP/1 mol
NaOH x 204.22 g KHP/1 mol KHP = 0.4178 g KHP