CHAPTER 4: ANSWERS TO ASSIGNED PROBLEMS
Hauser- General Chemistry I
revised 10/14/08
4.5 You are presented with three white solids, A, B, and C, which are glucose (a sugar
substance), NaOH, and AgBr. Solid A dissolves in water to form a conducting solution.
B is not soluble in water. C dissolves in water to form a nonconducting solution. Identify
A, B, and C.
AgBr is an ionic solid that is NOT soluble in water. Since A and C dissolve in water,
B is AgBr.
C dissolves in water but does not conduct a current, so it cannot be ionic. C is
glucose.
Substance A dissolves in water and is comprised of soluble ions. Substance A is
NaOH.
4.7 Which of the following ions will always be a spectator ion in a precipitation
reaction? Explain briefly.
(a) ClNO. Can form ppts with Ag+, Hg22+ and Pb2+.
(b) NO3
YES. All nitrates are soluble.
(c) NH4+
YES. All ammoniums are soluble.
2(d) S
NO. Sulfides usually form ppts.
(e) SO42NO. Sulfates usually form ppts.
4.19 Using solubility guidelines, predict whether each of the following compounds is
soluble or insoluble in water:
(a) NiCl2
SOLUBLE most chlorides are soluble
(b) Ag2S
INSOLUBLE most sulfides not soluble
(c) Cs3PO4
SOLUBLE while most phosphates are not soluble, Grp IA Cs+ is
(d) SrCO3
INSOLUBLE most carbonates not soluble
(e) PbSO4
INSOLUBLE most sulfates soluble, but not with Pb+2
4.21 Will precipitation occur when the following solutions are mixed? If so, write a
balanced chemical equation for the reaction.
(a) Na2CO3 and AgNO3
PPT.
Na2CO3 (aq) + 2 AgNO3 (aq)  2 NaNO3 (aq) + Ag2CO3 (s) or ↓
(b) NaNO3 and NiSO4
NO PPT.
(c) FeSO4 and Pb(NO3)2
PPT.
FeSO4 (aq) + Pb(NO3)2 (aq)  Fe(NO3)2 (aq) + PbSO4 (s) or ↓
1
4.24 Write balanced net ionic equations for the reactions that occur in each of the
following cases. Identify the spectator ion or ions in each reaction.
(a) Cr2(SO4)3 (aq) + (NH4)2CO3 (aq) 
dissociate compounds into ions
Cr3+ (aq)+ SO42- (aq) + NH4+ (aq) + CO32- (aq)
swap partners and write neutral formulas
Cr3+ (aq)+ SO42- (aq) + NH4+ (aq) + CO32- (aq)  (NH4)2SO4 + Cr2(CO3)3
balance
2 Cr3+ (aq)+ 3 SO42- (aq) + 6 NH4+ (aq) + 3CO32- (aq)  3 (NH4)2SO4 +
Cr2(CO3)3
check solubility rules
2 Cr3+ (aq)+ 3 SO42- (aq) + 6 NH4+ (aq) + 3 CO32- (aq)  3 (NH4)2SO4 (aq) +
Cr2(CO3)3 (s)
write net ionic to show stable product formation
2 Cr3+ (aq) + 3 CO32- (aq)  Cr2(CO3)3 (s)
spectator ions were SO42- (aq) and NH4+ (aq)
(b) Ba(NO3)2 (aq) + K2SO4 (aq) 
Ba2+ (aq) + SO4 2- (aq)  BaSO4 (s)
spectator ions were NO3 – and K+
(c) Fe(NO3)2 (aq) + KOH (aq) 
Fe 2+ (aq) + 2 OH- (aq)  Fe(OH)2 (s)
spectator ions were NO3- and K+
4.39 Complete and balance the following molecular equations, and then write the net
ionic equation for each:
(a) HBr (aq) + Ca(OH)2 (aq) 
2 HBr (aq) + Ca(OH)2 (aq)  CaBr2 (aq) + 2 H2O (l)
NET H+ (aq) + OH- (aq)  H2O (l)
(b) Cu(OH)2 (s) + HClO4 (aq) 
Cu(OH)2 (s) + 2 HClO4 (aq)  Cu(ClO4)2 (aq) + 2 H2O (l)
NET H+ (aq) + OH- (aq)  H2O (l) (USE LOWEST RATIO)
2
(c) Al(OH)3 (s) + HNO3 (aq) 
Al(OH)3 (s) + 3 HNO3 (aq)  Al(NO3)3 (aq) + 3 H2O (l)
NET H+ (aq) + OH- (aq)  H2O (l)
4.43 Write a balanced molecular equation and a net ionic equation for the reaction that
occurs when
(a) solid CaCO3 reacts with an aqueous solution of nitric acid
CaCO3 (s) + 2 HNO3 (aq)  Ca(NO3)2 (aq) + H2O (l) + CO2 (g)
acceptable answer for NET IONIC is…
NET CO3 2- (aq) + 2 H+ (aq)  H2O (l) + CO2 (g) (STABLE PRODUCTS)
[officially 2H+ (aq) + CaCO3 (s)  H2O (l) + CO2 (g) + Ca2+ (aq) since not all
aqueous]
(b) solid iron(II) sulfide reacts with an aqueous solution of hydrobromic acid
FeS (s) + HBr (aq)  FeBr2 (aq) + H2S (g)
acceptable answer for NET IONIC is…
S 2- (aq) + 2 H+ (aq)  H2S (g)
[officially 2H+ (aq) + FeS (s)  H2S (g) + Fe2+ (aq) since not all aqueous]
4.45 Define oxidation and reduction in terms of
(a) electron transfer OXIDATION IS LOSS of eREDUCTION IS GAIN of e-
"
OIL RIG"
(b) oxidation numbers OXIDATION IS INCREASE IN OX #
REDUCTION IS DECREASE IN OX #
4.49 Determine the oxidation number for the indicated element in each of the
following substances:
(a) S in SO2 S = +4
(b) C in COCl2 (note: NOT Co) C = +4 (assign O, then Cl, then C)
(c) Mn in MnO4- Mn = +7 (take charge of ion into account)
(d) Br in HBrO Br = +1 (assign O, then Br)
4.51 Which element is oxidized and which is reduced in the following reactions?
(a) N2 (g) + 3 H2 (g)  2 NH3 (g)
assign OX #'s first
0
0
-3 +1
N2 (g) + 3 H2 (g)  2 NH3 (g)
ignore coefficients for this application
H atom (0 to +1) oxidized
N atom (0 to -3) reduced
3
(b) 3 Fe(NO3)2 (aq) + 2 Al (s)  3 Fe (s) + 2 Al(NO3)3 (aq)
+2 +5 -2
0
0
+3 +5 -2
3 Fe(NO3)2 (aq) + 2 Al (s)  3 Fe (s) + 2 Al(NO3)3 (aq)
Al atom (0 to +3) oxidized
Fe atom (+2 to 0) reduced
ADDITIONAL EXERCISE #2
Examine the following reactions and identify whether or not a REDOX (oxidationreduction) reaction has occurred.
If REDOX has occurred, report the following:
- the atom oxidized
- the atom reduced
- the oxidizing agent
- the reducing agent
+4 -1
0
+2 -1
0
A) SiCl4 + 2 Mg  2 MgCl2 + Si
Mg atom ( 0 to +2) oxidized, Mg is reducing agent
Si atom (+4 to 0) reduced, SiCl4 is oxidizing agent
+4 -1
+1 -2
+1 -1 +4 -2
B) SiCl4 + 2 H2O  4 HCl + SiO2
NO REDOX
ADDITIONAL EXERCISE #3
Use the Activity Series (Table 4.5) to predict the outcome (if any) of each of the
following reactions:
ANY METAL ON LIST CAN BE OXIDIZED BY THE IONS OF AN ELEMENT
BELOW IT.
a) Cr (s) + 3 AgNO3  Cr(NO3)3 (aq) + Ag (s) because Ag ion is below Cr on list
b) Fe (s) + Mn+2  NR because Mn ion is above Fe on list
4.61 (a) Calculate the molarity of a solution that contains 0.0250 mol NH4Cl in exactly
500. mL of solution.
M = mol solute / L sol'n = 0.0250 mol / 0.500 L = 0.0500 mol/L or M (Did you
convert mL to L?)
(b) How many moles of HNO3 are present in 50.0 mL of a 2.50 M solution of nitric acid?
convert mL to L while maintaining SF = 50.0 mL ( 1 L / 1000 mL) =0.0500 L
M = mol solute / L sol'n so rearrange to mol solute = M L = 2.50 M ( 0.0500 L);
convert M to its units of mol/L; L cancels ANSWER: 0.125 mol
4
(c) How many milliliters of 1.50 M KOH solution are needed to provide 0.275 mol of
KOH?
M = mol solute / L sol'n so rearrange to L = mol solute / M
L = 0.275 mol / (1.50 mol/L) = 0.183 L; convert to 183 mL
4.67 Calculate (b) the molar concentration of a solution containing 14.75 g of Ca(NO3)2
in 1.375 L
M = mol solute / L sol'n
convert grams to moles
Ca 1 X 40.08 = 40.08
N 2 X 14.01 = 28.02
O 6 X 16.00 = 96.00
sum = 164.10 g / mol
(14.75 g) ( 1 mol Ca(NO3)2 / 164.10 g)
--------------------------------------------------------- = 0.065370 = 0.06537 mol / L (4 SF)
1.375 L
4.69 (a) Which will have the highest concentration of potassium ion: 0.20 M KCl, 0.15 M
K2CrO4, or 0.080 M K3PO4?
Beware of dissociation!
0.20 M KCl
1 X 0.20 mol/L = 0.20 M K+
0.15 M K2CrO4
2 X 0.15 mol/L = 0.30 M K+
0.080 M K3PO4
3 X 0.080 mol/L = 0.24 M K+
0.15 M K2CrO4 has the highest potassium ion concentration
4.73 (a) You have a stock solution of 14.8 M NH3. How many milliliters of this solution
should you dilute to make 1000.0 mL of 0.250 M NH3?
CiVi = CfVf
Ci = 14.8 M
Vf = 1000.0 mL
Cf = 0.250 M
SOLVE THE ALGEBRA! Vi = CfVf / Ci
Vi = (0.250 M) (1000.0 mL) / 14.8 M = 16.891 mL = 16.9 mL (3 SF)
Check work: You have more volume and lesser concentration at end.
5
4.81 (a) What volume of 0.115 M HClO4 solution is needed to neutralize 50.00 mL of
0.0875 M NaOH?
Start your calculation by writing a balanced chemical equation:
1 NaOH + 1 HCLO4  H2O + NaCLO4
Start the math with the item you know the most about (both volume and molarity
here). Don't forget to convert mL to L so you can use molarity concepts. FOLLOW
THE LABELS! RELATE MOLES OF ACID TO MOLES OF BASE.
50.00 mL (1 L / 1000 mL) = 0.05000 L
(0.0875 mol NaOH)
( 1mol HCLO4)
0.05000 L NaOH ___________________ ______________
(1 L NaOH)
(1 mol NaOH)
(1 L HCLO4)
______________
0.115 mol HCLO4
= 0.0380 L or 38.0 mL
(b) What volume of 0.128 M HCl is needed to neutralize 2.87 g of Mg(OH)2?
BEWARE! "DIBASIC" COMPOUND!
Start your calculation by writing a balanced chemical equation:
1 Mg(OH)2 + 2 HCL  2H2O + MgCL2
You will be converting grams to moles of Mg(OH)2:
Mg 1 X 24.31 = 24.31
O 2 X 16.00 = 32.00
H 2 X 1.01 = 2.02
sum 58.33 g /mol
(1 mol Mg(OH)2) (2 mol HCL)
( 1 L HCL)
(1000 mL)
2.87 g Mg(OH)2 ______________ _____________ _____________ ________ =
(58.33 g Mg(OH)2) (1 mol Mg(OH)2) 0.128 mol HCL
(1L)
= 0.76879 L = 0.769 L or 769 mL (3 SF)
6