1 11 Reactions in Aqueous Solutions II: Calculations.
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Transcript of 1 11 Reactions in Aqueous Solutions II: Calculations.
1
11Reactions in Aqueous
Solutions II: Calculations
2
Chapter Goals
Aqueous Acid-Base Reactions1. Calculations Involving Molarity2. Titrations3. Calculations for Acid-Base Titrations
Oxidation-Reduction ReactionsOxidation-Reduction Reactions 4. Balancing Redox Equations
5. Adding in H+, OH- , or H2O to Balance Oxygen or Hydrogen
6. Calculations for Redox Titrations
3
Calculations Involving Molarity
• Example 11-1: 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.
4
Calculations Involving Molarity
Reaction
Ratio: 2 mmol 1 mmol 1 mmol 2 mmol
Before
Reaction: 100 mmol
50 mmol 0 mmol 0 mmol
After
Reaction:0 mmol 0 mmol 50 mmol
100 mmol
OH 2 SONa SOH + NaOH 2 24242
5
Calculations Involving Molarity
• What is the total volume of solution?
100.0 mL + 100.0 mL = 200.0 mL
• What is the sodium sulfate amount, in mmol?
50.0 mmol
• What is the molarity of the solution?
M = 50 mmol/200 mL = 0.250 M Na2SO4
6
Calculations Involving Molarity
• Example 11-2: 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?
7
Calculations Involving Molarity
2 KOH + H SO K SO + 2 H O2 4 2 4 2Reaction
Ratio: 2 mmol 1 mmol 1 mmol 2 mmol
Before
Reaction: 130 mmol
50 mmol 0 mmol 0 mmol
After
Reaction: 30 mmol
0 mmol 50 mmol 100 mmol
8
Calculations Involving Molarity
• 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?
30.0 mmol & 50.0 mmol
• What is the molarity of the solution?
M = 30.0 mmol/230.0 mL = 0.130 M KOH
M = 50.0 mmol/230.0 mL = 0.217 M K2SO4
9
Calculations Involving Molarity
• Example 11-3: What volume of 0.750 M NaOH solution would be required to completely neutralize 100 mL of 0.250 M H3PO4?
You do it!You do it!
10
Calculations Involving Molarity
NaOH L 0.100NaOH mol 0.750
NaOH L 1
POH mol 1
NaOH mol 3
POH L 1
POH mol 0.250POH L 0.100 = NaOH L ?
OH 3 + PONa POH + NaOH 3
43
43
4343
24343
11
Titrations
Acid-base Titration Terminology Titration – A method of determining the concentration
of one solution by reacting it with a solution of known concentration.
Primary standard – A chemical compound which can be used to accurately determine the concentration of another solution. Examples include KHP and sodium carbonate.
Standard solution – A solution whose concentration has been determined using a primary standard.
Standardization – 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.
12
Titrations
Acid-base Titration Terminology Indicator – A substance that exists in different forms
with different colors depending on the concentration of the H+ in solution. Examples are phenolphthalein and bromothymol blue.
Equivalence point – The point at which stoichiometrically equivalent amounts of the acid and base have reacted.
End point – The point at which the indicator changes color and the titration is stopped.
13
Titrations
Acid-base Titration Terminology
14
Calculations for Acid-Base Titrations
• Potassium hydrogen phthalate is a very good primary standard.– It is often given the acronym, KHP.– KHP has a molar mass of 204.2 g/mol.
CH
CHCH
C
CCH C
OH
O
C
OH
O
CH
CHCH
C
CCH C
O
O
C
OH
O+ KOH
-K+ + H2O
KHP
acidic H
15
Calculations for Acid-Base Titrations
• 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 + H O2
NaOH 0.0733NaOH L 0.0273
NaOH mol 0.00200 = NaOH ?
NaOH mol 0.00200KHP mol 1
NaOH mol 1
KHP g 204.2
KHP mol 1KHP g 0.4084 = NaOH mol ?
MM
16
Calculations for Acid-Base Titrations
• Example 11-5: Calculate the molarity of a sulfuric acid solution if 23.2 mL of it reacts with 0.212 g of Na2CO3.
Na CO + H SO Na SO + CO + H O2 3 2 4 2 4 2 2
4242
4242
4232
42
32
323242
SOH 0.0862 SOH L 0232.0
SOH mol 0.00200 SOH ?
SOH mol 0.00200 CONa mol 1
SOH mol 1
CONa g 106
CONa mol 1CONa g 0.212 = SOH mol ?
MM
17
Calculations for Acid-Base Titrations• 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
You do it!You do it!
18
Calculations for Acid-Base Titrations
% 7.72%100g 0.884
g 0.643%100
sample g
KHP g=KHP %
KHP g 643.0KHP mmol 1
KHP g2042.0
NaOH mmol
KHP mmol 1NaOH mmol 3.15 = KHP g ?
NaOH mmol 15.3 nsol' mL 1
NaOH mmol 0.100solution mL 31.5 = NaOH mmol ?
19
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.
20
Balancing Redox Reactions
Half reaction method rules:1. Write the unbalanced reaction.2. Break the reaction into 2 half reactions:
One oxidation half-reaction andOne 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.
21
Balancing Redox Reactions
4. Charge balance the half reactions by adding appropriate numbers of electrons.
Electrons will be products in the oxidation half-reaction.
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.
22
Balancing Redox Reactions
• 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.
-+42
+2 BrSnBrSn
reaction-halfoxidation theis
this thusproducts are Electrons
e2Sn Sn
reaction.-half thebalance Charge
Sn Sn
reaction.-half thebalance Mass
-+4+2
+4+2
23
Balancing Redox Reactions
reaction. halfreduction theis This
Br 2e2Br
reaction.-halfother thebalance Charge
Br 2Br
reaction.-halfother thebalance Mass
e2Sn Sn
--2
-2
-+4+2
24
Balancing Redox Reactions
reaction balanced Br 2SnBrSn
reaction half red. Br 2e2Br
reaction half ox. e2Sn Sn
reactions. half two theAdd
reaction starting BrSnBrSn
-+42
+2
--2
-+4+2
-+42
+2
25
Balancing Redox Reactions
• 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.
reaction?-halfreduction or oxidation an thisIs
1FeFe
reaction.-half thebalance Charge
FeFe
reaction.-half thebalance Mass
reaction starting FeCrFeOCr
+3+2
+3+2
+3+3+2272
e
26
Balancing Redox Reactions
OH 7Cr 2 e 6OCrH 14
reaction.-half 2 thebalance Charge
OH 7Cr 2 OCrH 14
reaction-half 2 thebalance Mass
oxidation 1e+ FeFe
rxn starting FeCrFeOCr
2+3-2
72+
nd
2+32
72+
nd
-+3+2
+3+3+2272
27
Balancing Redox Reactions
red. OH 7Cr 2 e 6H 14OCr
ox. 1e+ FeFe 6
electrons. thebalance toreactions-half heMultiply t
rxnstart FeCrFeOCr
2+3-+2
72
-+3+2
+3+3+2272
OH 7Cr 2Fe 6H 14OCrFe 6
red. OH 7Cr 2 e 6H 14OCr
ox. 1e+ FeFe 6
reactions.-half two theAdd
rxnstart FeCrFeOCr
2+3+3+2
72+2
2+3-+2
72
-+3+2
+3+3+2272
28
Balancing Redox Reactions
• Example 11-15: When chlorine is bubbled into basic solution, it forms hypochlorite ions and chloride ions. Write and balance the net ionic equation.
You do it!You do it!
• This is a disproportionation redox reaction. The same species, in this case Cl2, is both reduced and oxidized.
Balancing Redox Reactions
29
reactionoxidation e 2 OH 2 ClO 2OH 4Cl
reaction.-half 1 thebalance Charge 2.
solution) (basic ClClOCl
-2
-2
st
2
Cl 2Cl
e 2 OH 2 ClO 2OH 4Cl
reaction.-half 2 thebalance Mass 3.
solution) (basic ClClOCl
2
-2
-2
nd
2
reaction-halfreduction Cl 2e 2Cl
e 2 OH 2 ClO 2OH 4Cl
reaction.-half 2 thebalance Charge 4.
solution) (basic ClClOCl
-2
-2
-2
nd
2
OH 2 Cl 2ClO 2 OH 4Cl 2
Cl 2e 2Cl
e 2 OH 2 ClO 2OH 4Cl
reactions.-half two theAdd 5.
solution) (basic ClClOCl
2---
2
-2
-2
-2
2
OH ClClO OH 2Cl
Cl 2e 2Cl
e 2 OH 2 ClO 2OH 4Cl
numbers. holesmallest w toReduce 6.
solution) (basic ClClOCl
2---
2
-2
-2
-2
2
OH 2 ClO 2OH 4Cl
reaction.-half 1 thebalance Mass 1.
solution) (basic ClClOCl
2-
2
st
2
30
Calculations for Redox Titrations• Just as we have done stoichiometry with
acid-base reactions, it can also be done with redox reactions.
• 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:
OH 8Cl 5MnCl 2+KCl 2HCl 16 KMnO 2 2224
You do it!You do it!
31
Calculations for Redox Titrations
35 0150 5 25
5 252
0 656
0 6561
328
mL HCl M HCl mmol HCl
mmol HCl mmol KMnO
16 mmol HCl mmol KMnO
mmol KMnO mL
0.200 mmol KMnO mL
44
44
. .
. .
. .
32
Calculations for Redox Titrations• 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:
OH 7Cr 2Fe 6H 14OCrFe 6 2+3+3+2
72+2
You do it!You do it!
33
Calculations for Redox Titrations
+2+2
+22
72
+22
72
272
272
Fe M 300.0mL 0.40
Fe mmol 0.12
Fe mmol 0.12OCr mmol 1
Fe mmol 6OCr mmol 00.2
OCr mmol 00.2OCr M 100.0mL 0.20
34
Synthesis Question
• A 0.7500 g sample of an impure FeSO4 sample is titrated with 26.25 mL of 0.0200 M KMnO4 to an endpoint. What is the % purity of the FeSO4 sample?
35
Synthesis Question
%13.53%100g 0.7500
g 0.3985 purity %
g 3985.0mg 1000
g 1mg 5.398
mg 5.398mmol
mg 8.151
Femmol 1
FeSOmmol 1 Femmol 625.2
Femmol 625.2 MnO1
Fe5mmol 525.0
mmol 525.0 KMnO0200.0mL 26.25
O H4 Mn Fe5 Fe5 H8MnO
:is reactionredox balanced
8.1518.151 FeSOof massmolar
242
2-4
2
4
2232-
4
mmolmg
molg
4
M
36
11Reactions in Aqueous
Solutions II: Calculations