1 11 Reactions in Aqueous Solutions II: Calculations

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11Reactions in Aqueous

Solutions II: Calculations

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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

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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.

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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


• 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

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• 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?

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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


After

Reaction: 30 mmol


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• 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

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• 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!

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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

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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.

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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.

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Titrations

Acid-base Titration Terminology

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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

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• 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

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• 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

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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


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% 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 ?

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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.

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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.

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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.

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• 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

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reaction. halfreduction theis This

Br 2e2Br

reaction.-halfother thebalance Charge

Br 2Br

reaction.-halfother thebalance Mass

e2Sn Sn

--2

-2

-+4+2

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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

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• 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

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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

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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

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• 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.


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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.


2

-2

-2

nd

2

reaction-halfreduction Cl 2e 2Cl


reaction.-half 2 thebalance Charge 4.


-2

-2

-2

nd

2

OH 2 Cl 2ClO 2 OH 4Cl 2

Cl 2e 2Cl


reactions.-half two theAdd 5.


2---

2

-2

-2

-2

2

OH ClClO OH 2Cl

Cl 2e 2Cl


numbers. holesmallest w toReduce 6.


2---

2

-2

-2

-2

2

OH 2 ClO 2OH 4Cl

reaction.-half 1 thebalance Mass 1.


2-

2

st

2

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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


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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

. .

. .

. .

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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


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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

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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?

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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

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11Reactions in Aqueous

Solutions II: Calculations

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1 11 Reactions in Aqueous Solutions II: Calculations