Name
Chapter 5 Problem Set
Show your work on all calculations and maintain the correct number of significant figures.
1. Saline solution contains 0.90 g NaCl in 100.0 mL of solution. What is the molarity of NaCl in
saline solution?
0.90 g NaCl × (1 mol NaCl/58.5 g NaCl) = 0.015 mol NaCl
100.0 mL × (1 L/1,000 mL) = 0.1000 L
Molarity = mol/L = 0.015 mol/0.1000 L = 0.15 M NaCl
2. How many grams of NaCl are required to make 750 mL of a 1.5 M NaCl solution?
1.5 M = mol/0.75 L
mol = 1.1 mol NaCl
1.1 mol NaCl × (58.5 g/1 mol NaCl) = 64 g NaCl
3. What is the molarity of the solution made by diluting 45.0 mL of 18 M H2SO4 with enough water
to make 500.0 mL of solution?
M1V1 = M2V2
18 M (0.0450 L) = M2 (0.5000 L)
M2 = 1.6 M
4. How many liters of a 12 M HCl solution would you need to make 12 L of a 0.50 M HCl solution?
M1V1 = M2V2
12 M (V2) = 0.50 M (12 L)
V2 = 0.50 L
5. The concentration of caffeine in 5-Hour Energy Extra Strength is 0.0208 M. How many milliliters
would you have to drink to match the 400.0 mg daily limit of caffeine recommended by the Food
and Drug Administration (FDA)? The molar mass of caffeine is 194.2 g/mol.
400.0 mg × (1 g/1,000 mg) × (1 mol caffeine/194.2 g) = 0.002060 mol caffeine
M = mol/L
0.0208 M = 0.002060 mol/volume
0.0208 M (volume) = 0.002060 mol
volume = 0.002060 mol ÷ 0.0208 M = 0.0990 L = 99.0 mL 5-Hour Energy Extra Strength
6. What is the final concentration of a solution made by diluting 25 mL of an 18 M solution to 75
mL?
Use M1V1 = M2V2. You don’t have to convert mL to L. Both volumes are in mL and the
units will cancel out.
18 M x 25 mL = M2 x 75 mL
M2 = 18 M x 25 mL/75 mL = 6.0 M
7. Consider the following reaction, 2 NaOH(aq) + CuSO4(aq) → Cu(OH)2(s) + Na2SO4(aq).
a. How many milliliters of 1.5 M NaOH(aq) are needed to react completely with 125 mL of 2.5
M CuSO4(aq)?
Solve in three steps.
Use the concentration and volume of CuSO4 to calculate moles of CuSO4.
2.5M = mol/0.125 L
mol = 0.125 L x 2.5 M = 0.31 mol CuSO4
Calculate the moles of NaOH.
0.31 mol CuSO4 x (2 mol NaOH/1 mol CuSO4) = 0.62 mol NaOH
Calculate the volume of NaOH.
1.5 M = 0.62 mol/V
V = 0.62 mol/1.5 M = 0.41 L = 410 mL
b. How many grams of copper(II) hydroxide would form in the reaction?
Use either reactant to calculate the moles of that reactant, convert to moles of Cu(OH)2 and then
convert moles to grams.
0.31 mol CuSO4 x (1 mol Cu(OH)2/1 mol CuSO4) x (97.6 g Cu(OH)2/1 mol Cu(OH)2) = 30. g
or
0.62 mol NaOH x (1 mol Cu(OH)2/ 2 mol NaOH) x (97.6 g Cu(OH)2/1 mol Cu(OH)2) = 30. g
8. Write the balanced molecular, complete ionic and net ionic equations for the following reactions.
Be sure to include (s), (l), (g) or (aq) for each substance.
a. Li2SO4(aq) + Pb(CH3CO2)2(aq) →
To make products, match cations with anions and write ionic charges.
Li2SO4(aq) + Pb(CH3CO2)2(aq) → Li+CH3CO2- + Pb2+SO42Lithium is always +1. Acetate is a polyatomic ion, always -1. Pb is +2 in this case (the charge can
be determined from the compound Pb(CH3CO2)2). Sulfate is a polyatomic ion, always -2.
Exchange/subscript charges to make chemical formulas.
Li2SO4(aq) + Pb(CH3CO2)2(aq) → LiCH3CO2 + PbSO4
Write physical states (see solubility chart). LiCH3CO2 is soluble in water; PbSO4 is not.
Li2SO4(aq) + Pb(CH3CO2)2(aq) → LiCH3CO2(aq) + PbSO4(s)
Balance.
Molecular equation: Li2SO4(aq) + Pb(CH3CO2)2(aq) → 2 LiCH3CO2(aq) + PbSO4(s)
Write the complete ionic equation. Only (aq) split into ions.
2 Li+(aq) + SO42-(aq) + Pb2+(aq) + 2 CH3CO2-(aq) → 2 Li+(aq) + 2 CH3CO2-(aq) + PbSO4(s)
Write the net ionic equation. Identical terms on the left and right of the reaction arrow cancel out.
SO42-(aq) + Pb2+(aq) → PbSO4(s)
b. KHCO3(aq) + H2SO4(aq) →
To make products, match cations with anions and write ionic charges.
KHCO3(aq) + H2SO4(aq) → K+SO42- + H+HCO3Potassium is always +1. Sulfate is a polyatomic ion, always -2. H is always +1. HCO3- is a
polyatomic ion, always -1. Subscripts only carry over to products if they are a part of polyatomic
ions, i.e. H+HCO3-, not H2+HCO3Exchange/subscript charges to make chemical formulas.
KHCO3(aq) + H2SO4(aq) → K2SO4 + HHCO3
Rewrite HHCO3 as H2CO3. H2CO3 is not a stable compound. It immediately decomposes to
CO2(g) + H2O(l). Rewrite products.
KHCO3(aq) + H2SO4(aq) → K2SO4 + CO2(g) + H2O(l)
Write physical states (see solubility chart). K2SO4 is soluble in water.
KHCO3(aq) + H2SO4(aq) → K2SO4(aq) + CO2(g) + H2O(l)
Balance.
Molecular equation: 2 KHCO3(aq) + H2SO4(aq) → K2SO4(aq) + 2 CO2(g) + 2 H2O(l)
Write the complete ionic equation. Only (aq) split into ions.
2 K+(aq) + 2 HCO3-(aq) + 2 H+(aq) + SO42-(aq) → 2 K+(aq) + SO42-(aq) + 2 CO2(g) + 2 H2O(l)
Write the net ionic equation. Identical terms on the left and right of the reaction arrow cancel out.
2 HCO3-(aq) + 2 H+(aq) → 2 CO2(g) + 2 H2O(l)
If the coefficients can be simplified, then simplify them.
HCO3-(aq) + H+(aq) → CO2(g) + H2O(l)
9. Complete the table and answer the questions.
CH4(g) + F2(g) → CF4(l) + HF(g)
Oxidation
number
C
-4
(see
below)
Reactants
H
F
+1
0
(always) (free
element)
C
+4
(see below)
Products
H
+1 (always)
F
-1 (always)
CH4. There is no rule for carbon, but hydrogen is +1.
C + 4H = 0 (because this molecule’s net charge is 0)
C+ 4(+1) = 0
C+4=0
C = -4
CF4. There is no rule for carbon, but fluorine is -1.
C + 4F = 0 (because this molecule’s net charge is 0)
C+ 4(-1) = 0
C-4=0
C=4
Which element is oxidized? Carbon (oxidation number becomes more positive, loses electrons)
Which element is reduced? Fluorine (oxidation number becomes more negative, gains electrons).
10. Complete and balance this reaction. Be sure to include (s), (l), (g) or (aq) for each substance.
Element + compound is a single replacement reaction. Zinc replaces copper.
Zn(s) + CuSO4(aq) →
Products are Zn2+SO42- = ZnSO4. Zinc sulfate is soluble in water, ZnSO4(aq).
Copper is a solid at room temperature and pressure, Cu(s).
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
11. Consider these reactions. If the reaction will occur, complete and balance. Be sure to include (s),
(l), (g) or (aq) for each substance.
a. Mg(s) + CuSO4(aq) →
This reaction will occur because Mg is above Cu in the activity series. Magnesium replaces
copper.
Mg(s) + CuSO4(aq) → MgSO4(aq) + Cu(s)
The ionic charge on magnesium is +2; sulfate is -2. Mg2+SO42- = MgCl2. Magnesium
sulfate is soluble in water, MgSO4(aq).
Copper is a solid at room temperature and pressure so it is Cu(s).
b. Ag(s) + CuSO4(aq) →
This reaction will not occur because Ag is not above Cu in the activity series.
c. Mg(s) + HCl(aq) →
This reaction will occur because Mg is above H in the activity series.
Magnesium replaces hydrogen. The ionic charge on magnesium is +2; chlorine is -1.
Mg2+Cl- = MgCl2. Magnesium chloride is soluble in water, MgCl2(aq).
Hydrogen is a diatomic element, H2. Hydrogen is a gas at room temperature and pressure,
H2(g).
Balance the equation.
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
12. Write the complete, balanced chemical equation for the double replacement reaction that occurs
when two aqueous solutions are combined to produce aqueous sodium chloride and liquid water.
Products are NaCl(aq) and H2O(l) so the chemical equation is:
reactants → NaCl(aq) + H2O(l)
The format of double replacement reactions is AB + CD → AD + CB where A and C are cations and
B and D are anions. In this case, A = Na+, B = OH-, C = H+ and D = Cl-. B = OH- and C = H+ because
the water is formed from H+OH-.
AB + CD → NaCl(aq) + H2O(l)
Na+OH-(aq) + H+Cl-(aq) → NaCl(aq) + H2O(l)
Use the ionic charges to make the chemical formulas. Balance.
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
13. Write the complete, balanced chemical equation for the single replacement reaction that occurs
when a solid element and an aqueous solution are combined to produce aqueous magnesium
nitrate and hydrogen gas.
Products are Mg(NO3)2(aq) and H2(g) so the chemical equation is:
reactants → Mg(NO3)2(aq) + H2(g)
The format of single replacement reactions is A + BC → AC + B.
In the reactants, A is a free element, B is a cation and C is an anion.
In the products, A is a cation, C is an anion and B is free element.
Therefore, A = Mg (reactant)/Mg+2 (product), B = H+(reactant)/H2(product) and C = NO3-.
Mg(s) + H+NO3-(aq) → Mg(NO3)2(aq) + H2(g)
Use the ionic charges to make the chemical formulas. Balance.
Mg(s) + 2 HNO3(aq) → Mg(NO3)2(aq) + H2(g)