Copyright © 2011 Nelson Education Ltd. Chapter 9: Solutions and Their Reactions 9.5-1
Section 9.5: Stoichiometry of Solutions Tutorial 1 Practice, page 447 1. (a) Given:
V
SrCl2
= 150 mL
cSrCl
2
= 0.25 mol / L
cNa
2CO
3
= 0.500 mol/L
Required: volume of 0.500 mol/L sodium carbonate, V
Na2CO
3
Solution: Step 1. Convert all volumes of the solutions to litres.
VSrCl
2
= 150 mL !1 L
1000 mL
VSrCl
2
= 0.150 L
Step 2. Write a dissociation equation listing the calculated amounts and the required value(s). SrCl2(aq) + Na2CO3(aq) → SrCO3(s) + NaCl(aq)
VSrCl
2
= 0.150 L
cSrCl
2
= 0.25 mol / L
cNa
2CO
3
= 0.500 mol/L
VNa
2CO
3
Step 3. Determine the amount of strontium chloride by rearranging the equation.
c =n
V
nSrCl
2
= cSrCl
2
VSrCl
2
=0.25 mol
1L ! 0.15 L
nSrCl
2
= 0.037 50 mol [2 extra digits carried]
Step 4. Determine the amount of sodium carbonate.
nNa
2CO
3
= 0.037 50 molSrCl
2
!
1 molNa
2CO
3
1molSrCl
2
nNa
2CO
3
= 0.037 50 mol
Step 5. Determine the volume of sodium carbonate by rearranging the equation.
VNa
2CO
3
=
nNa
2CO
3
cNa
2CO
3
=0.037 50 mol
0.500mol
L
VNa
2CO
3
= 0.075 L
Statement: The volume of 0.500 mol/L sodium carbonate required is 0.075 L or 75 mL.
Copyright © 2011 Nelson Education Ltd. Chapter 9: Solutions and Their Reactions 9.5-2
(b) Given: n
SrCl2
= 0.037 50 mol (from (a))
Required: mass of precipitate strontium carbonate, m
SrCO3
Solution: Step 1. Write a dissociation equation listing the calculated amounts and the required value(s). SrCl2(aq) + Na2CO3(aq) → SrCO3(s) + NaCl(aq)
n
SrCl2
= 0.037 50 mol
Step 2. Determine the amount of precipitate of strontium carbonate.
nSrCO
3
= 0.037 50 molSrCl
2
!
1 molSrCO
3
1molSrCl
2
nSrCO
3
= 0.037 50 molSrCO
3
Step 3. Calculate the mass of precipitate of strontium carbonate.
mSrCO
3
= 0.037 50 mol !147.63 g
1 mol
mSrCO
3
= 5.5 g
Statement: The mass of strontium carbonate precipitate is 5.5 g. 2. (a) Given:
V
KOH= 70.0 mL
cKOH
= 0.80 mol / L
cFe(NO
3)3
= 0.50 mol/L
VFe(NO
3)3
= 40.0 L
Required: mass of iron(III) hydroxide precipitate, m
Fe(OH)3
Solution: Step 1. Convert all volumes of the solutions to litres.
VKOH
= 70.0 mL !1 L
1000 mL
VKOH
= 0.070 L
VFe(NO
3)3
= 40.0 mL !1 L
1000 mL
VFe(NO
3)3
= 0.040 L
Step 2. Write the dissociation equation listing the calculated amounts and the required value(s). Fe(NO3)3(aq) + 3 KOH (aq) → Fe(OH)3(s) + 3 KCl(aq)
c
Fe(NO3
)3
= 0.50 mol/L c
KOH= 0.80 mol / L
m
Fe(OH)3
V
Fe(NO3
)3
= 0.040 L V
KOH= 0.070 L
Copyright © 2011 Nelson Education Ltd. Chapter 9: Solutions and Their Reactions 9.5-3
Step 3. Determine the amounts.
c =n
V
nFe(NO
3)3
= cFe(NO
3)3
VFe(NO
3)3
=0.50 mol
1L ! 0.040 L
nFe(NO
3)3
= 0.020 mol
nKOH
= cKOH
VKOH
=0.80 mol
1L ! 0.070 L
nKOH
= 0.056 mol
Step 4. Determine the limiting reagent.
nFe(NO
3)3
= 0.056 molKOH
!
1 molFe(NO
3)3
3molKOH
nFe(NO
3)3
= 0.0187 molFe(NO
3)3
[1 extra digit carried]
Therefore, 0.056 mol potassium hydroxide requires 0.019 mol of iron(III) nitrate. Since 0.020 mol of iron(III) nitrate is present, potassium hydroxide is the limiting reagent. Step 5. Use the amount of limiting reagent to determine the amount of iron(III) hydroxide precipitate.
nFe(OH)
3
= 0.056 molKOH
!
1 molFe(OH)
3
3 molKOH
nFe(OH)
3
= 0.0187 molFe(OH)
3
[1 extra digit carried]
Step 6. Determine the mass of iron(III) hydroxide precipitate.
mFe(OH)
3
= 0.0187 mol !106.88 g
1 mol
mFe(OH)
3
= 2.0 g
Statement: The mass of iron(III) hydroxide precipitate is 2.0 g.
Copyright © 2011 Nelson Education Ltd. Chapter 9: Solutions and Their Reactions 9.5-4
Tutorial 2 Practice, page 448 1. (a) Given:
c
BaCl2
= 0.50 mol/L
Required: amount concentration of the anion, c
Cl–
Solution: Step 1: Write the dissociation reaction. BaCl2(s) → Ba2+(aq) + 2 Cl–(aq) Step 2. Convert the concentration of the compound into concentration of anions.
cCl
–= 0.50
molBaCl
2
L!
2 molCl
–
1molBaCl
2
cCl
–= 1.0 mol/L
Statement: The amount concentration of chloride anion is 1.0 mol/L. (b) Given:
c
KOH= 6.0 mol/L
Required: amount concentration of the anion, c
OH–
Solution: Step 1: Write the dissociation reaction. KOH(s) → K+(aq) + OH–(aq) Step 2. Convert the concentration of the compound into concentration of anions.
cOH
–= 6.0
molKOH
L!
1 molOH
–
1molKOH
cOH
–= 6.0 mol/L
Statement: The amount concentration of hydroxide anion is 6.0 mol/L. (c) Given:
c
Al(ClO3
)3
= 0.10 mol/L
Required: amount concentration of the anion,
cClO
3
–
Solution: Step 1: Write the dissociation reaction. Al(ClO3)3(s) → Al3+(aq) + 3 ClO3
–(aq) Step 2. Convert the concentration of the compound into concentration of cations.
cClO
3
–= 0.10
molAl(ClO
3)3
L!
3 molClO
3
–
1molAl(ClO
3)3
cClO
3
–= 0.30 mol/L
Statement: The amount concentration of chlorate anion is 0.30 mol/L.
Copyright © 2011 Nelson Education Ltd. Chapter 9: Solutions and Their Reactions 9.5-5
2. Given: m
(NH4
)2CO
3
= 14.4 g
V
(NH4
)2CO
3
= 100.0 mL
Required: amount concentration of ammonium ion,
cNH
4
+
Solution: Step 1. Convert the volume of the solution to litres.
V(NH
4)2CO
3
= 100.0 mL !1 L
1000 mL
V(NH
4)2CO
3
= 0.1000 L
Step 2. Determine the amount of solute.
n(NH
4)2CO
3
= 14.4 g !1 mol
96.11 g
n(NH
4)2CO
3
= 0.149 83 mol [2 extra digits carried]
Step 3. Determine the amount concentration.
c(NH
4)2CO
3
=
n(NH
4)2CO
3
V(NH
4)2CO
3
=0.149 83 mol
0.1000 L
c(NH
4)2CO
3
= 1.50 mol/L
Step 4. Write a dissociation equation listing the calculated amounts and the required value(s). (NH4)2CO3(s) → 2 NH4
+(aq) + CO32–(aq)
c
(NH4
)2CO
3
= 1.50 mol/L
Step 5. Convert the concentration of the compound into concentration of cations.
cNH
4
+= 1.50
mol(NH
4)2CO
3
L!
2 molNH
4
+
1mol(NH
4)2CO
3
cNH
4
+= 3.00 mol/L
Statement: The amount concentration of the ammonium ion is 3.00 mol/L.
Copyright © 2011 Nelson Education Ltd. Chapter 9: Solutions and Their Reactions 9.5-6
3. Given: V
Na3PO
4
= 1.75 L
c
Na+= 0.25 mol / L
Required: mass of sodium phosphate, m
Na3PO
4
Solution: Step 1. Write a dissociation equation listing the calculated amounts and the required value(s). Na3PO4(s) → 3 Na+(aq) + PO4
3–(aq)
V
Na3PO
4
= 1.75 L c
Na+= 0.25 mol / L
m
Na3PO
4
Step 2. Convert the concentration of the sodium ion into concentration of sodium phosphate.
cNa
3PO
4
= 0.25
molNa
+
L!
1 molNa
3PO
4
3molNa
+
cNa
3PO
4
= 0.083 mol/L
Step 3. Determine the amount of solute.
c =n
V
nNa
3PO
4
= cNa
3PO
4
VNa
3PO
4
=0.083 mol
1L !1.75 L
nNa
3PO
4
= 0.146 mol [1 extra digit carried]
Step 4. Determine the mass of sodium phosphate.
mNa
3PO
4
= 0.146 mol !163.95 g
1 mol
mNa
3PO
4
= 24 g
Statement: The mass of sodium phosphate is 24 g. Section 9.5 Questions, page 449 1. (a) Given:
V
AgNO3
= 32.0 mL
cAgNO
3
= 0.100 mol / L
VNaCl
= 25 mL
Required: amount concentration of sodium ions, c
Na+
Solution: Step 1. Convert all volumes of the solutions to litres.
VAgNO
3
= 32.0 mL !1 L
1000 mL
VAgNO
3
= 0.032 L
Copyright © 2011 Nelson Education Ltd. Chapter 9: Solutions and Their Reactions 9.5-7
VNaCl
= 25 mL !1 L
1000 mL
VNaCl
= 0.025 L
Step 2. Write a dissociation equation listing the calculated amounts and the required value(s). AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
VAgNO
3
= 0.032 L
cAgNO
3
= 0.100 mol / L
VNaCl
= 0.025 L
cNaCl
Step 3. Determine the amount of solute.
c =n
V
nAgNO
3
= cAgNO
3
VAgNO
3
=0.100 mol
1L ! 0.032 L
nAgNO
3
= 0.003 20 mol [1 extra digit carried]
Step 4. Determine the amount of sodium chloride.
nNaCl
= 0.003 20 molAgNO
3
!1 mol
NaCl
1 molAgNO
3
nNaCl
= 0.003 20 molNaCl
Step 5. Determine the amount concentration of sodium chloride.
cNaCl
=n
NaCl
VNaCl
=0.003 20 mol
0.025 L
cNaCl
= 0.13 mol/L
Step 6. Write the dissociation reaction for sodium ions. NaCl(aq) → Na+(aq) + Cl–(aq) Step 7. Convert the concentration of the compound into concentration of sodium cations.
cNa
+= 0.13
molNaCl
L!
1 molNa
+
1molNaCl
cNa
+= 0.13 mol/L
Statement: The amount concentration of sodium ions is 0.13 mol/L.
Copyright © 2011 Nelson Education Ltd. Chapter 9: Solutions and Their Reactions 9.5-8
(b) Given: V
NaCl= 0.032 L
n
NaCl= 0.00320 mol / L (from (a), 1 extra digit carried)
Required: amount concentration of sodium chloride in g/L Solution: Step 1. Convert the amount of solute to mass.
mNaCl
= 0.003 20 mol !58.54 g
1 mol
mNaCl
= 0.187 g [1 extra digit carried]
Step 2. Determine concentration of the solute using the mass.
cNaCl
=m
NaCl
VNaCl
=0.187 g
0.025 L
cNaCl
= 7.5 g/L
Statement: The amount concentration of sodium chloride is 7.5 g/L. 2. (a) NiSO4(aq) + 2 NaOH(aq) → Ni(OH)2(s) + Na2SO4(aq) (b) Given:
c
NiSO4
= 0.45 mol / L
VNiSO
4
= 50.0 mL
VNaOH
= 25.0 mL
cNaOH
= 1.00 mol / L
Required: mass of nickel hydroxide precipitate, m
Ni(OH)2
Solution: Step 1. Convert all volumes of the solutions to litres.
VNiSO
4
= 50.0 mL !1 L
1000 mL
VNiSO
4
= 0.0500 L
VNaOH
= 25.0 mL !1 L
1000 mL
VNaOH
= 0.0250 L
Step 2. Write a dissociation equation listing the calculated amounts and the required value(s). NiSO4(aq) + 2 NaOH(aq) → Ni(OH)2(s) + Na2SO4(aq)
VNiSO
4
= 0.0500 L
cNiSO
4
= 0.45 mol / L
VNaOH
= 0.0250 L
cNaCl
= 1.00 mol/L
Copyright © 2011 Nelson Education Ltd. Chapter 9: Solutions and Their Reactions 9.5-9
Step 3. Determine the amounts of both substances.
c =n
V
nNiSO
4
= cNiSO
4
VNiSO
4
=0.45 mol
1L ! 0.0500 L
nNiSO
4
= 0.0225 mol [1 extra digit carried]
nNaOH
= cNaOH
VNaOH
=1.00 mol
1L ! 0.0250 L
nNaOH
= 0.0250 mol
Step 4. Determine the amount of sodium hydroxide required to react with 0.0225 mol of nickel(II) sulfate solution.
nNaOH
= 0.0225 molNiSO
4
!2 mol
NaOH
1 molNiSO
4
nNaOH
= 0.0450 mol
Since this amount is greater than the amount present, sodium hydroxide is the limiting reagent and nickel(II) sulfate is the excess reagent. Step 5. Determine the amount of nickel hydroxide precipitate.
nNi OH( )
2
= 0.0250 molNaOH
!
1 molNi OH( )
2
2 molNaOH
nNi OH( )
2
= 0.0125 molNi OH( )
2
Step 6. Determine the mass of nickel hydroxide precipitate.
mNi OH( )
2
= 0.0125 mol !92.71 g
1 mol
mNi OH( )
2
= 1.16 g
Statement: The mass of nickel(II) hydroxide precipitate produced in this reaction is 1.16 g. 3. (a) Given:
m
CaCO3
= 15.2 g
V
Na2CO
3
= 200.0 mL
Required: amount concentration of sodium carbonate solution, c
Na2CO
3
Solution: Step 1. Convert the volume of the solution to litres.
VNa
2CO
3
= 200.0 mL !1 L
1000 mL
VNa
2CO
3
= 0.2000 L
Copyright © 2011 Nelson Education Ltd. Chapter 9: Solutions and Their Reactions 9.5-10
Step 2. Determine the amount of calcium carbonate.
nCaCO
3
= 15.2 g !1 mol
100.09 g
nCaCO
3
= 0.151 86 mol [2 extra digits carried]
Step 3. Write a dissociation equation listing the calculated amounts and the required value(s). Na2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2 NaCl(aq)
V
Na2CO
3
= 0.200 L m
CaCO3
= 15.2 g
n
CaCO3
= 0.151 86 mol
Step 4. Determine the amount of sodium carbonate.
nNa
2CO
3
= 0.151 86 molCaCO
3
!
1 molNa
2CO
3
1 molCaCO
3
nNa
2CO
3
= 0.151 86 mol
Step 5. Determine the amount concentration.
cNa
2CO
3
=
nNa
2CO
3
VNa
2CO
3
=0.151 86 mol
0.2000 L
cNa
2CO
3
= 0.759 mol/L
Statement: The amount concentration of original sodium carbonate solution is 0.759 mol/L. (b) Given:
m
CaCO3
= 15.2 g
cCaCl
2
= 0.500 mol/L
nCaCO
3
= 0.151 86 mol (from (a), 2 extra digits carried)
Required: volume of calcium chloride solution, V
CaCl
Solution: Step 1. Write a dissociation equation listing the calculated amounts and the required value(s). Na2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2 NaCl(aq)
c
CaCl2
= 0.500 mol/L m
CaCO3
= 15.2 g
n
CaCO3
= 0.151 86 mol
Step 2. Determine the amount of calcium chloride.
nCaCl
2
= 0.151 86 molCaCO
3
!
1 molCaCl
2
1 molCaCO
3
nCaCl
2
= 0.151 86 mol
Copyright © 2011 Nelson Education Ltd. Chapter 9: Solutions and Their Reactions 9.5-11
Step 3. Rearrange the equation and substitute values.
cCaCl
2
=
nCaCl
2
VCaCl
2
VCaCl
2
=
nCaCl
2
cCaCl
2
=0.151 86 mol
0.500mol
L
VCaCl
2
= 0.304 L
Statement: The volume of calcium chloride solution required to produce 15.2 g of calcium carbonate is 0.304 L or 304 mL. 4. Given:
c
HCl= 12.0 mol/L
m
Fe3O
3
= 224 g
Required: volume of hydrochloric acid, V
HCl
Solution: Step 1. Determine the amount of solute from the mass of solute.
nFe
2O
3
= 224 g !1 mol
159.7 g
nFe
2O
3
= 1.4026 mol [2 extra digits carried]
Step 2. Write a dissociation equation listing the calculated amounts and the required value(s). Fe2O3(s) + 6 HCl(aq) → 2 FeCl3(aq) + 3 H2O(l)
m
Fe2O
3
= 224 g c
HCl= 12.0 mol/L
n
Fe2O
3
= 1.4026 mol
Step 3. Determine the amount of hydrochloric acid.
nHCl
= 1.4026 molFe
2O
3
!6 mol
HCl
1molFe
2O
3
nHCl
= 8.4158 molHCl
[2 extra digits carried]
Copyright © 2011 Nelson Education Ltd. Chapter 9: Solutions and Their Reactions 9.5-12
Step 4. Rearrange the equation and substitute values to calculate the volume of hydrochloric acid.
cHCl
=n
HCl
VHCl
VHCl
=n
HCl
cHCl
=8.4158 mol
12.0 mol
L
VHCl
= 0.701 L
Statement: The volume of hydrochloric acid is 0.701 L. 5. (a) 2 Al(s) + 3 CuSO4(aq) → 3 Cu(s) + Al2(SO4)3(aq) (b) Given:
c
CuSO4
= 0.100 mol/L
V
CuSO4
= 150 mL
Required: mass of aluminum, m
Al
Solution: Step 1. Convert the volume of the solution to litres.
VCuSO
4
= 150 mL !1 L
1000 mL
VCuSO
4
= 0.15 L
Step 2. Determine the amount of copper(II) sulfate.
cCuSO
4
=
nCuSO
4
VCuSO
4
nCuSO
4
=VCuSO
4
! cCuSO
4
= 0.15 L !0.100 mol
1 L
nCuSO
4
= 0.015 mol
Step 3. Determine the amount of aluminum.
nAl= 0.015 mol
CuSO4
!2 mol
Al
3 molCuSO
4
nAl= 0.010 mol
Al
Copyright © 2011 Nelson Education Ltd. Chapter 9: Solutions and Their Reactions 9.5-13
Step 4. Determine the mass of aluminum.
mAl= 0.010 mol !
26.98 g
1 mol
mAl= 0.27 g
Statement: The mass of aluminum required to react with the copper(II) sulfate solution is 0.27 g. 6. Given:
m
NaOH= 45!10
6 t
c
NaCl= 6.0 mol/L
Required: volume of sodium chloride, V
NaCl
Solution: Step 1. Convert the mass of sodium hydroxide to grams.
mNaOH
= 45!106
t !10
6 g
1 t
mNaOH
= 45!1012
g
Step 2. Determine the amount of sodium hydroxide.
nNaOH
= 45!1012
g !1 mol
40.00 g
nNaOH
= 1.125!1012
mol [2 extra digits carried]
Step 3. Write a dissociation equation listing the calculated amounts and the required value(s). 2 NaCl(aq) + 2 H2O(l) → 2 NaOH(aq) + Cl2(g) + H2(g)
c
NaCl= 6.0 mol/L
n
NaOH= 1.125!10
12mol
Step 4. Determine the amount of sodium chloride.
nNaCl
= 1.125 !1012
molNaOH
!1 mol
NaCl
1 molNaOH
nNaCl
= 1.125 !1012
molNaCl
Step 5. Rearrange the equation and substitute values to calculate the volume of hydrochloric acid.
cNaCl
=n
NaCl
VNaCl
VNaCl
=n
NaCl
cNaCl
=1.125!10
12mol
6.0 mol
L
VNaCl
= 1.9 !1011
L
Statement: The volume of sodium chloride solution required is 1.9 × 1011 L.
Copyright © 2011 Nelson Education Ltd. Chapter 9: Solutions and Their Reactions 9.5-14
7. Given: c
AgNO3
= 0.50 mol / L
VAgNO
3
= 15.0 mL
msolute
= 0.42 g
Vsolution
= 100.0 mL
Required: identify the solute Solution: Step 1. Convert the volumes of the solutions to litres.
VAgNO
3
= 15.0 mL !1 L
1000 mL
VAgNO
3
= 0.0150 L
Vsolution
= 100.0 mL !1 L
1000 mL
Vsolution
= 0.1000 L
Step 2. Determine the amount of silver nitrate used.
nAgNO
3
= 0.0150 L !0.50 mol
1 L
nAgNO
3
= 7.5!10"3
mol
Step 3. Using sodium chloride as the solute, determine the amount of sodium chloride.
nNaCl
= 0.42 g !1 mol
58.44 g
nNaCl
= 7.2 !10"3
mol
Step 4. Using sodium chloride as the solute, write the dissociation equation. NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3 (aq)
n
NaCl= 7.2 !10
"3mol
n
AgNO3
Step 5. Using sodium chloride as the solute, determine the amount of silver nitrate required.
nAgNO
3
= 7.2 !10"3
molNaCl
!
1 molAgNO
3
1 molNaCl
nAgNO
3
= 7.2 !10"3
mol
Step 6. Using calcium chloride as the solute, determine the amount of calcium chloride.
nCaCl
2
= 0.42 g !1 mol
110.98 g
nCaCl
2
= 3.78 !10"3
mol [1 extra digit carried]
Step 7. Using calcium chloride as the solute, write the dissociation equation. CaCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Ca(NO3)2(aq)
n
CaCl2
= 3.78 !10"3
mol n
AgNO3
Copyright © 2011 Nelson Education Ltd. Chapter 9: Solutions and Their Reactions 9.5-15
Step 8. Using calcium chloride as the solute, determine the amount of silver nitrate required.
nAgNO
3
= 3.78 !10"3
molCaCl
2
!
2 molAgNO
3
1 molCaCl
2
nAgNO
3
= 7.56 !10"3
mol
Statement: The actual amount of silver nitrate available is 7.5 ×10-3 mol. If the unknown solution contains sodium chloride, it would take 7.2 × 10-3 mol of silver nitrate to precipitate all of the chloride ions. If the unknown solution contains calcium chloride, it would take 7.56 × 10-3 mol of silver nitrate to precipitate all of the chloride ions. The unknown solution is sodium chloride because there is not enough silver nitrate to precipitate all of the chloride ions if the solution was calcium chloride. 8. (a) Given:
c
Na2CO
3
= 0.5 mol / L
Step 1: Write the dissociation reaction. Na2CO3(aq) → 2 Na+(aq) + CO3
2-(aq) Step 2. Convert the concentration of the compound into concentration of cations.
cNa
+= 0.5
molNa
2CO
3
L!
2 molNa
+
1molNa
2CO
3
cNa
+= 1 mol/L
Statement: The amount concentration of cations is 1 mol/L. (b) Given:
c
(NH4
)2
SO3
= 0.2 mol / L
Step 1: Write the dissociation reaction. (NH4)2SO3(s) → 2 NH4
+ + SO32–
Step 2. Convert the concentration of the compound into concentration of cations.
cNH
4
+= 0.2
mol(NH
4)2
SO3
L!
2 molNH
4
+
1mol(NH
4)2
SO3
cNH
4
+= 0.4 mol/L
Statement: The amount concentration of cations is 0.4 mol/L. (c) Given:
c
Fe2
(SO4
)3
= 1.5 mol / L
Step 1: Write the dissociation reaction. Fe2(SO4)3(s) → 2 Fe3+ + 3 SO4
2–
Step 2. Convert the concentration of the compound into concentration of cations.
cFe
3+= 1.5
molFe
2(SO
4)3
L!
2 molFe
3+
1molFe
2(SO
4)3
cFe
3+= 3.0 mol/L
Statement: The amount concentration of cations is 3.0 mol/L.
Copyright © 2011 Nelson Education Ltd. Chapter 9: Solutions and Their Reactions 9.5-16
9. Given: c
Na+= 0.85 mol/L
V
Na2CO
3
= 200.0 mL
Required: mass of sodium carbonate, m
Na2CO
3
Solution: Step 1. Convert the volume of the solution to litres.
VNa
2CO
3
= 200.0 mL !1 L
1000 mL
VNa
2CO
3
= 0.2000 L
Step 2. Write the dissociation reaction. Na2CO3(aq) → 2 Na+(aq) + CO3
2-(aq) Step 3. Determine the concentration of sodium carbonate.
cNa
2CO
3
= 0.85 mol
Na+
L!
1 molNa
2CO
3
2 molNa
+
cNa
2CO
3
= 0.425 mol/L [1 extra digit carried]
Step 4. Determine the amount of sodium carbonate.
cNa
2CO
3
=
nNa
2CO
3
VNa
2CO
3
nNa
2CO
3
= cNa
2CO
3
!VNa
2CO
3
= 0.425 mol
L! 0.2000 L
nNa
2CO
3
= 0.085 mol
Step 5. Determine the mass of sodium carbonate.
mNa
2CO
3
= 0.085 mol !105.99 g
1 mol
mNa
2CO
3
= 9.0 g
Statement: The mass of sodium carbonate is 9.0 g.