11
Reactions in Aqueous
Solutions II: Calculations
1
Chapter Goals
Aqueous Acid-Base Reactions 酸鹼中和反應
•
•
•
Calculations Involving Molarity 有關體積莫耳濃度
的計算
Titrations 滴定
Calculations for Acid-Base Titrations 酸鹼滴定的計
算
Oxidation-Reduction Reactions 氧化還原反應
•
•
•
Balancing Redox Equations 氧化還原方程式平衡
Adding in H+, OH- , or H2O to Balance Oxygen or
Hydrogen
Calculations for Redox Titrations 計算氧化反原滴定
2
Concentrations of Solutions
• Second common unit of concentration:
Molarity is defined as the number of moles of solute
per literof solution. 體積莫耳濃度是指每公升溶液中所含
的溶質莫耳數
Molarity =
M=
Number of moles of solute
Number of liters of solution
moles
L
M=
mmoles
mL
M x L = moles solute
and
M x mL = mmol solute
3
Calculations Involving
Molarity
• The reaction ratio is the relative number of
moles of reactants and products shown in the
balanced equation.
H2SO4 + 2NaOH  Na2SO4 + 2H2O
1mol
2mol
1mol
2mol
Reaction ratio
1mol H2SO4 or 2mol NaOH
2mol NaOH
1mol H2SO4
4
Calculations Involving Molarity
Example 11-1: Acid-Base Reactions
If 100.0 mL of 1.00 M NaOH and 100.0 mL of 0.500 M
H2SO4 solutions are mixed, what will the concentration
of the resulting solution be?
What is the balanced reaction?
It is very important that we always use a balanced
chemical reaction when doing stoichiometric
calculations.
2NaOH+ H2SO4  Na2SO4 + 2H2O
M=mole/liter  mole= M x liter
NaOH =1.00M x 0.1L =0.1 mole
H2SO4 =0.500M x 0.1L =0.05 mole
5
Calculations Involving Molarity
2NaOH+ H2SO4  Na2SO4 + 2H2O
Reaction Ratio
2 mol
1 mol
Before Reaction
0.1 mol
0.05 mol
After Reaction
0 mol
0 mol
1 mol
2 mol
0.05 mol
0.1 mol
•What is the total volume of solution?
100.0 mL + 100.0 mL = 200.0 mL
•What is the sodium sulfate amount, in mol?
0.05 mol
• What is the molarity of the solution?
M = 0.05 mol/0.2 L = 0.250 M Na2SO4
6
Calculations Involving Molarity
Example 11-2: Acid-Base Reactions
If 130.0 mL of 1.00 M KOH and 100.0 mL of 0.500 M H2SO4
solutions are mixed, what will be the concentration of
KOH and K2SO4 in the resulting solution?
What is the balanced reaction?
2 KOH+ H2SO4  K2SO4 + 2H2O
M=mole/liter  mole= M x liter
KOH =1.00M x 0.130L =0.130 mole
H2SO4 =0.500M x 0.1L =0.05 mole
7
2KOH+ H2SO4  K2SO4 + 2H2O
Reaction Ratio
2 mol
1mol
Before Reaction
0.130 mol 0.05 mol
After Reaction
0.030 mol
0 mol
1 mol
2 mol
0.05 mol 0.1 mol
•What is the total volume of solution?
130.0 mL + 100.0 mL = 230.0 mL
•What are the potassium hydroxide and potassium sulfate
amounts?
0.03 mol & 0.05 mol
• What is the molarity of the solution?
M = 0.03 mol/0.230 L = 0.130 M KOH
M = 0.05 mol/0.230 L = 0.217 M K2SO4
8
Example 11-3: Acid-Base Reactions
What volume of 0.750 M NaOH solution would be
required to completely neutralize 100 mL of 0.250 M
H3PO4?
3 NaOH+ H3PO4  Na3PO4 + 3H2O
M=mol/liter  mol= M x liter  liter = mol/M
? mol H3PO4 =0.250M x 0.1L =0.025 mole
complete neutralize need 0.025x3=0.075 mol NaOH
?L NaOH = 0.075 mol / 0.750M =0.10 L
9
Titrations滴定法
Acid-base Titration Terminology
•
Titration 滴定– A method of determining the
concentration of one solution by reacting it with a
solution of known concentration.
• Primary standard 一級標準– A chemical compound
which can be used to accurately determine the
concentration of another solution. Examples include
KHP and sodium carbonate 碳酸鈉.
• Standard solution 標準溶液 – A solution whose
concentration has been determined using a primary
standard.
• Standardization 標定– The process in which the
concentration of a solution is determined by
accurately measuring the volume of the solution
required to react with a known amount of a primary
standard.
一級標準：高純度物質
－做為參考物質。由一級標準滴定標
準溶液，以測定濃度的程序，稱為標定(standarddization)10
Titrations
Acid-base Titration Terminology
•
Indicator 指示劑– A substance that exists in
different forms with different colors depending on
the concentration of the H+ in solution. Examples
are phenolphthalein 酚酞 and bromothymol blue溴
百里酚藍.
• Equivalence point當量點– The point at which
stoichiometrically equivalent amounts of the acid
and base have reacted. 酸與鹼當量數相等
• End point 滴定終點– The point at which the
indicator changes color and the titration is
stopped.
bromothymol blue溴百里酚藍: Colorless in acidic
11
solution, reddish violet in basic solution
Titrations
Acid-base Titration Terminology
12
Calculations for Acid-Base
Titrations
•The properties of an ideal primary standard include
the following:
– It must not react with or absorb the components of
the atmosphere, such as water vapor, oxygen, and
carbon dioxide. 不能跟大氣中物質反應
– It must react according to one invariable reaction
– It must have a high percentage of purity (高純度)
– It should have a high formula weight to minimize the
effect of error in weighing (大分子量)
– It must be soluble in the solvent of interest
– It should be nontoxic (無毒)
– It should be readily available (inexpensive 較便宜)
– It should be environmentally friendly
13
Calculations for Acid-Base
Titrations
• Potassium hydrogen phthalate苯二甲酸鹽緩
衝溶液is a very good primary standard.
– It is often given the acronym, KHP.
– KHP has a molar mass of 204.2 g/mol.
14
標準酸或鹼的選取
在標定酸或鹼的濃度時，常選用一些分子量較大純度高的酸
或鹼來做為滴定的標準品，如鄰苯二甲酸氫鉀
(potassium hydrogen phthalate，簡記為KHP)等，稱
之為一級標準品，可用以定量我們所配製鹼 (酸) 液之精
確濃度。
而經滴定測得精確濃度的標準鹼 (酸) 溶液，稱之為二級標準
鹼 (酸)。利用鄰苯二甲酸氫鉀以標定所配製之氫氧化鈉溶
液濃度。因KHP 為單質子酸 (monoprotic acid)，與氫
氧化鈉為1：1 中和 (如式1)，故利用此化學計量關係可求
出氫氧化鈉溶液的濃度。
HOOCC6H4COOK(aq) + NaOH(aq) → C6H4(COO-)2(aq)+ K+
+
(aq) + Na (aq) + H2O(l)
15
•Example 11-4: Calculate the molarity of a
NaOH solution if 27.3 mL of it reacts with
0.4084 g of KHP.
NaOH + KHP  NaKP +H2O
?mole NaOH = 0.4084 g KHP x 1 mol KHP x1 mol NaOH
1 mol KHP
204.2gKHP
= 0.002 mol NaOH
0.002 mol NaOH
= 0.0733 M NaOH
?M NaOH =
0.0273 L NaOH
16
• Example 11-5: Calculate the molarity of a
sulfuric acid solution if 23.2 mL of it reacts with
0.212 g of Na2CO3.
Na2CO3 + H2SO4  Na2SO4 + CO2 + H2O
?mole H2SO4 = 0.212 g Na2CO3 x
1 mol Na2CO3 x 1 mol H2SO4
1 mol Na2CO3
106g Na2CO3
= 0.002 mol H2SO4
?M H2SO4 =
0.002 mol H2SO4 = 0.0862 M H SO
2
4
0.0232 L H2SO4
17
•Example 11-6: An impure sample of potassium
hydrogen phthalate, KHP, had a mass of 0.884 g. It
was dissolved in water and titrated with 31.5 mL of
0.100 M NaOH solution. Calculate the percent purity
of the KHP sample. Molar mass of KHP is 204.2 g/mol.
NaOH + KHP  NaKP + H2O
?mole NaOH = 31.5x10-3L solution x 0.1 mol NaOH
1L solution
= 0.00315 mol NaOH
1 mol KHP x 204.2g KHP
?g KHP = 0.00315mol NaOH x
1 mol KHP
1mol NaOH
= 0.643g KHP
g KHP
0.643g
% KHP = g sample x 100% = 0.884g x 100%
= 72.7%
18
Example 11-5: Standardization of an Acid Solution
Calculate the molarity of a solution of H2SO4 if 40.0ml
of the solution neutralizes 0.364g of Na2CO3?
Na2CO3 + H2SO4  Na2SO4 + CO2 + H2O
1 mol Na2CO3 1 mol H2SO4
?mole H2SO4= 0.364 g Na2CO3 x
106.0g Na2CO3 x 1 mol Na2CO3
= 0.00343 mol H2SO4
?M H2SO4= 0.00343mol H2SO4
0.04 L
= 0.0858 M H2SO4
Exercise 26
19
Example 11-6: Standardization of a Base Solution
A 20.00-ml sample of a solution of NaOH reacts with
0.3641g of KHP. Calculate the molarity of the NaOH
solution.
NaOH + KHP  NaKP +H2O
?mole NaOH = 0.3641 g KHP x
1 mol KHP x 1 mol NaOH
204.2g KHP 1 mol KHP
= 0.001783 mol NaOH
?M NaOH=
0.001783mol NaOH
0.02 L
= 0.08915 M NaOH
Exercise 28
20
Example 11-7: Titration
The titration of 36.7ml of a hydrochloric acid solution
requires 43.2 ml of standard 0.236M sodium
hydroxide solution for complete neutralization. What
is the molarity of the HCl solution?
HCl + NaOH  NaCl +H2O
1 mol HCl
?mmole HCl = 43.2ml NaOH x0.236mmol NaOH x
1ml NaOH
1 mol NaOH
= 10.2 mmol HCl
?M HCl= 10.2mmol HCl
36.7 mL
= 0.278 M HCl
21
Example 11-8: Titration
The titration of 36.7ml of a sulfuric acid solution requires
43.2 ml of standard 0.236M sodium hydroxide solution
for complete neutralization. What is the molarity of the
H2SO4 solution?
H2SO4 +2 NaOH  Na2SO4 +2 H2O
1 mol H2SO4
?mmole H2SO4 = 43.2ml NaOH x0.236mmol NaOH x
1ml NaOH
2 mol NaOH
= 5.1 mmol H2SO4
?M H2SO4 = 5.1 mmol H2SO4
36.7 mL
= 0.139 M H2SO4
Exercise 38
22
Example 11-9: Dtermination of Percent Acid
Oxalic acid, (COOH)2, is used to remove rust stains and
some ink stains from fabrics. A 0.1743-g sample of impure
oxalic acid required 39.82 ml of 0.08915 M NaOH solution for
complete neutralization. No acidic impurities were present.
Calculate the percentage purity of the (COOH)2.
(COOH)2 +2 NaOH  Na2(COO)2 + 2H2O
1 mol (COOH)2
?mole (COOH)2 = 0.03982L NaOH x 0.08915mol NaOH x
1L NaOH
2 mol NaOH
= 0.001775 mol (COOH)2
?g (COOH)2 = 0.001775 mole x
90.04g (COOH)2
1mol (COOH)2
= 0.1598g (COOH)2
0.1598g (COOH)
% purity = 0.1743g sample 2 x 100% = 91.68%
Exercise 38
23
Oxidation-Reduction
Reactions
• We have previously gone over the basic concepts
of oxidation & reduction in Chapter 4.
• Rules for assigning oxidation numbers were also
introduced in Chapter 4.
– Refresh your memory as necessary.
• We shall learn to balance redox reactions using the
half-reaction method.
24
Oxidation-Reduction
Reactions: An Introduction
• Oxidation is an increase in the oxidation number.
氧化反應為增加氧化數
– Corresponds to the loss of electrons (失去電子).
• Reduction is a decrease in the oxidation number.
還原反應為氧化數減少
– Good mnemonic – reduction reduces the
oxidation number.
– Corresponds to the gain of electrons (得到電子)
25
Balancing Redox Reactions
Half reaction method rules:
1. Write the unbalanced reaction.
2. Break the reaction into 2 half reactions:
– One oxidation half-reaction and
– One reduction half-reaction
Each reaction must have complete formulas for
molecules and ions.
3. Mass balance each half reaction by adding
appropriate stoichiometric coefficients. To
balance H and O we can add:
H+ or H2O in acidic solutions.
OH- or H2O in basic solutions.
26
Balancing Redox Reactions
4. Charge balance the half reactions by adding
appropriate numbers of electrons.
 Electrons will be products in the oxidation halfreaction.
 Electrons will be reactants in the reduction
half-reaction.
5. Multiply each half reaction by a number to
make the number of electrons in the oxidation
half-reaction equal to the number of electrons
reduction half-reaction.
6. Add the two half reactions.
7. Eliminate any common terms and reduce
coefficients to smallest whole numbers.
27
Example 11-12
Tin (II) ions are oxidized to tin (IV) by bromine. Use the half
reaction method to write and balance the net ionic
equation.
Starting Reaction Sn2+ +Br2  Sn4+ + BrMass balance the half-reaction
Sn2+  Sn4+
Charge balance the half reaction Add the two half-reaction
Sn2+  Sn4+ + 2eSn2+  Sn4+ + 2eElectrons are products thus this
Br2 + 2e- 2Bris the oxidation half-reaction
Sn2+ +Br2  Sn4+ + 2BrMass balance the half-reaction
Br2  2BrCharge balance the half reaction
Br2 + 2e- 2BrThis is the reduction half-reaction
28
Example 11-13
Dichromate ions oxidize iron (II) ions to iron (III) ions and are
reduced to chromium (III) ions in acidic solution. Write and
balance the net ionic equation for the reaction.
Starting Reaction
2- + Fe2+  Cr3+ + Fe3+
Cr2O+6
7
Mass balance the half-reaction
Fe2+  Fe3+
Charge balance the half reaction
Add the two half-reaction
2+
3+
Fe  Fe + e
6(Fe2+  Fe3+ + e-)
This is the oxidation half-reaction
14H+ + Cr2O72- + 6e- 
Mass balance the half-reaction
3+ + 7H O
2Cr
2
Cr2O72-  2Cr3+
2+
+
26Fe 14H + Cr2O7 
Cr2O72-  2Cr3+ + 7H2O
3+ + 2Cr3+ + 7H O
6Fe
+
23+
2
14H + Cr2O7  2Cr + 7H2O
Charge balance the half reaction
14H+ + Cr2O72-+ 6e-  2Cr3+ + 7H2O
This is the reduction half-reaction
29
Example 11-14
In basic solution hydrogen peroxide oxidizes chromite ions,
Cr(OH)4-, to chromate ions, CrO42-. The hydrogen
peroxide is reduced to hydroxide ions. Write and
balance the net ionic equation for this reaction.
+3
Cr(OH)4- + H2O2  CrO42-+6
Basic solution
Mass balance the half-reaction
Cr(OH)4- + OH-  CrO42Add the two half-reaction
2Cr(OH)4 + OH  CrO4 + H2O 2 (Cr(OH) - + 4OH-  CrO 2- + 4H O +3e-)
4
4
2
Cr(OH)4- + 4OH-  CrO42- + 4H2O 3 (H O + 2e-  2OH-)
2 2
Charge balance the half reaction
Cr(OH)4- + 4OH-  CrO42- + 4H2O +3e-
2Cr(OH)4- +8OH- + 3H2O2 
2CrO42- + 8H2O+ 6OH-
oxidation half-reaction
Mass balance the half-reaction 2Cr(OH) - +2OH- + 3H O 
4
2 2
H2O2  2OH
Charge balance the half reaction
2CrO42- + 8H2O
H2O2 + 2e-  2OH-
reduction half-reaction
30
Example 11-15
When chlorine is bubbled into basic solution, it forms
hypochlorite ions and chloride ions. Write and balance
the net ionic equation.
This is a disproportionation redox reaction. The same
species, in this case Cl2, is both reduced and oxidized.
- +ClCl+0

ClO
2
+1
Basic solution
Mass balance the half-reaction
Cl2 + 4OH-  2ClO- +2H2O
Add the two half-reaction
Charge balance the half reaction Cl + 4OH-  2ClO- + 2H O +2e2
2
Cl2 + 4OH  2ClO +2H2O+2e
oxidation half-reaction
Mass balance the half-reaction
Cl2  2Cl-
Cl2 + 2e-  2Cl-
2Cl2 + 4OH-  2ClO- + 2Cl-+ 2H2O
Cl2 + 2OH-  ClO- + Cl-+ H2O
Charge balance the half reaction
Cl2 + 2e-  2Cl-
reduction half-reaction
Disproportionation Reaction 不均化反應 is a redox reaction in
which the same element is oxidized and reduced.
31
Concentrations of Solutions
• Second common unit of concentration:
Molarity is defined as the number of moles of
solute per literof solution.
Number of moles of solute
Molarity =
Number of liters of solution
moles
M=
L
mmoles
M=
mL
M x L = moles solute
and
M x mL = mmol solute
32
Example 11-16
What volume of 0.200 M KMnO4 is required to oxidize
35.0 mL of 0.150 M HCl? The balanced reaction is:
2KMnO4 + 16HCl  2KCl + 2MnCl2+ 5Cl2+ 8H2O
?mmole HCl = 35ml solution x 0.150M
= 5.25 mmol HCl
2 mmol KMnO
?mmol KMnO4 = 5.25 mmol HCl x 16mmol HCl 4
= 0.656 mmol KMnO4
1mL
? ml KMnO4 = 0.656 mmol KMnO4 x 0.20mmol KMnO
4
= 3.28ml KMnO4
33
Example 11-17
A volume of 40.0 mL of iron (II) sulfate is oxidized to
iron (III) by 20.0 mL of 0.100 M potassium dichromate
solution. What is the concentration of the iron (II)
sulfate solution? From Example 11-19 the balanced
equation is:
6Fe2+ 14H+ + Cr2O72-  6Fe3+ + 2Cr3+ + 7H2O
?mmole Cr2O72- = 20.0ml solution x 0.100M
= 2.00 mmol Cr2O726 mmol Fe2+
2+
2?mmol Fe = 2.00 mmol Cr2O7 x 1mmol Cr O 22 7
2+
= 12.0 mmol Fe
1M
2+
? M Fe = 12.0 mmol KMnO4 x 40.0mmol KMnO
4
= 0.30 M Fe2+
34
Example 11-10: Balancing Redox Equations
A useful analytical procedure involves the oxidation of
iodide to free iodine. The free iodine is then titrated with a
standard solution of sodium thiosulfate, Na2S2O3. Iodine
oxidizes S2O32- ions to tetrathionate ions, S4O62- and is
reduced to I- ions. Write the balanced net ionic equation for
this reaction.
Starting Reaction I2 + S2O32-  I-+ S4O62balance the ox. half-reaction
Add the two half-reaction
S2O32-  S4O622 S2O32-  S4O62-+ 2e2 S2O32-  S4O62I2 + 2e- 2 I2 S2O32-  S4O62-+ 2ebalance the red. half-reaction I2 + 2S2O32-  2I-+ S4O62I2  II2  2II2 + 2e- 2 IExercise 38
35
Example 11-11: Net Ionic Equations
Permanganate ions oxidize ion(II) to iron(III) in sulfuric acid
solution. Permanganate ions are reduced to manganese(II)
ions. Write the balanced net ionic equation for this reaction.
Starting Reaction Fe2+ + MnO4-  Fe3++ Mn2+
balance the ox. half-reaction Add the two half-reaction
5 (Fe2+  Fe3++ e- )
Fe2+  Fe3+
- + 8H++ 5e-  Mn2++ 4H O
2+
3+
MnO
4
2
Fe  Fe + e
5Fe2+ +MnO4- + 8H+ 
balance the red. half-reaction
5Fe3+ +Mn2++ 4H2O
MnO4-  Mn2+
MnO4- + 8H+  Mn2++ 4H2O
MnO4- + 8H++ 5e-  Mn2++ 4H2O
Exercise 38
36
Example 11-13: Balancing Redox equations
In basic solution, hypochlorite ions, ClO-, oxidize chromite
ions, CrO2-, to chromate ions, CrO42-, and are reduced to
chloride ions. Write the balanced net ionic equation for this
reaction.
Starting Reaction ClO- + CrO2-  Cl-+ CrO2-
balance the ox. half-reaction
CrO2-  CrO2CrO2- + 4OH- CrO42- + 2H2O
CrO2- + 4OH- CrO42- + 2H2O + 3e-
Add the two half-reaction
2(CrO2- + 4OH- CrO42- + 2H2O + 3e-)
3(ClO- + H2O+ 2e-  Cl-+ 2OH-)
2CrO2- + 8OH- + 3ClO- + 3H2O 
2CrO42- + 4H2O + 3Cl-+ 6OH-
balance the red. half-reaction
ClO-  Cl- + 2OH- + 3ClO- 
2CrO
2
ClO + H2O  Cl + 2OH
2- + H O + 3Cl2CrO
4
2
ClO + H2O+ 2e  Cl + 2OH
Exercise 54
37
Example 11-14: Redox Titration
What volume of 0.0200 M KMnO4 solution is required to
oxidize 40.0ml of 0.100M FeSO4 in sulfuric acid solution?
MnO4- + 8H+ + 5Fe2+  5Fe3+ + Mn2+ + 4H2O
?mmol Fe2+ = 40.0ml solution x 0.100M
= 4.00 mmol Fe2+
1 mmol MnO42+
?mmol MnO4 = 4.00 mmol Fe x 5mmol Fe2+
= 0.8 mmol MnO41ml
? ml KMnO4 = 0.8 mmol KMnO4 x 0.02mmol KMnO
4
= 40.0 ml KMnO4
Exercise 54
38
Example 11-15: Redox Titration
A 20.00 ml sample of Na2SO3 was titrated with 36.30 ml
of 0.5130 M K2Cr2O7 solution in the presence of H2SO4
solution. Calculate the molarity of the Na2SO3 solution.
3SO32- + Cr2O72- + 8H+  3SO42- + 2Cr3+ + 4H2O
?mmol Cr2O72- = 36.30ml solution x 0.5130M
= 1.862 mmol Cr2O7223
mmol
SO
?mmol SO32- = 1.862 mmol Cr2O72- x 1mmol Cr O3 22 7
2= 5.586 mmol SO3
? M Na2SO3 = 5.586 mmol Na2SO3 / 20ml Na2SO3
=0.2793 M Na2SO3
Exercise 66
39