problem set ph and aqueous solutions. aqueous neutralization reactions and ph edward a. mottel...

Problem Set

pH and Aqueous Solutions

Aqueous Neutralization Reactions and pH

Edward A. Mottel

Department of Chemistry

Rose-Hulman Institute of Technology

Aqueous Solutions and pH

Reading assignment: • Zumdahl: Chapter 7.3-7.6, 7.11, 8.1

04/19/23

Autoionization of Water

Water is a very weak electrolyte

H2O(l) H+(aq) + OH–(aq)

2 H2O(l) H3O+(aq) + OH–(aq)

Dissociation constant

Kw = [H+] [OH– ] = 1.0 x 10–14 (@ 25 C)

Note: H+(aq) H3O+ (aq)

04/19/23

Base 10 logarithmic scale of hydrogen ion concentration

• pH = -log [H+ ]

If pH = -log [H+ ], what does [H+ ] equal?

[H+ ] = 10–pH

Be carefuluse log and not ln

pure waterat 25 C Kw = [H+] [OH–] = 1.0 x 10–14

1.0 x 10–7 M 1.0 x 10–7 M

pH = -log(1.0 x 10–7) = 7.0

What’s the pH of pure water?

If the pH of an aqueous solution is 3.50 at 25 C,what is the hydroxide ion concentration?

04/19/23

neutral @ 25 °C

1 M HCl

1 M KOH

pH values can be outside 0-14

04/19/23

Titration

reactant 2

reactant 1

an analytical techniqueto determinethe concentrationof a solution

04/19/23

Thermometric Titration

acid-base reactionsare usually exothermic

Equivalence Pointvolume

Moles of acid = Moles of base

04/19/23

Titration

HF + NaOH NaF + H2O

Balanced equation

reactant 2

reactant 1

an analytical techniqueto determinethe concentrationof a solution

04/19/23

Titration

nacid · Macid · Vacid = nbase · Mbase · Vbase

Most acids have one acidic proton (HCl, CH3COOH),some have two (H2SO4),some have three (H3PO4)

the purpose of a titrationis to determine theequivalence point

04/19/23

Titration

nacid · Macid · Vacid = nbase · Mbase · Vbase

Most bases accept one proton (NaOH, NH3),some accept two (Ca(OH)2),some accept three (Al(OH)3)

04/19/23

Equivalence Point Detection

Physical methods• Thermometric • Potentiometric• Visual indicator• Conductometric

04/19/23

Potentiometric Titration

20.0 mL of 0.400 M HCl with Aqueous Ammonia

0 4 8 12 16 20 24 280

Volume NH3 (mL)

04/19/23

Equivalence Point Detection

Visual indicator methods• Color change • Endpoint versus equivalence point

Endpoint: when the indicator changes colorEquivalence point:

moles of acid = moles of base• Factors to consider when selecting an

indicator

04/19/23

Bromocresol Green

Common Acid-Base Visual Indicators

Indicator Acidic Color Basic Color pH Range

thymol blue red yellow 1.2-2.8

methyl orange orange yellow 2.9-4.4

bromocresol green yellow blue 3.8-5.4

methyl red red yellow 4.4-6.0

bromothymol blue yellow blue 6.0-7.6

cresol red yellow red 7.2-8.8

phenolphthalein colorless red 8.3-10.0

thymolphthalein colorless blue 9.3-11.0

alizarin yellow yellow red 10.0-12.1

04/19/23

0 4 8 12 16 20 24 280

Volume NH3 (mL)

phenolphthalein

bromocresol green

equivalencepoint

endpoint

For this titration,which indicatorwould be better?

04/19/23

Conductometric Titration

Depending on the acid-base system studied,the graph may have many different shapes.

04/19/23

Why does the line change slope?

04/19/23

HCladded K+

OH– H+Cl–

TotalIons

0 50 50 0 0 100

20 50 30 0 20 100

40 50 10 0 40 100

60 50 0 10 60 120

80 50 0 30 80 160

100 50 0 50 100 200

H2O(l) + KCl(aq)KOH(aq) + HCl(aq)

04/19/23

If the total number of ions isn’t changing before theequivalence point, why is the conductivity changing?

Problem Set

Weak Acid – Strong Base Titration

04/19/23

0 4 8 12 16 20 24 280

Volume NH3 (mL)

Why does the pH change rapidly at the equivalence point?

04/19/23

So maybe you would like to be aCPA?

(Chemistry Problem Analyzer)

04/19/23

Hydrofluoric Acid Solution

What is the pH of a 20.0 mL of 0.100 M hydrofluoric acid at 25 °C?

HF HFHFH+

04/19/23

Substitute into theMass-Action Expression

[H+] [F¯]= 7.2 x 10–4

HF(aq) + H2O(l) H3O+(aq) + F¯(aq) K = 7.2 x 10–4

[0.100-x]

[+x] [+x]= 7.2 x 10–4

04/19/23

Solve for [H+]

x = 0.00813, -0.00885

[0.100-x]

[+x] [+x]= 7.2 x 10–4

x is thehydrogen ionconcentration

A negativeconcentrationis not possible

pH = - log [H+] = - log (0.00813)

pH = 2.09

04/19/23

pH Before the Equivalence Point

What is the pH of a solutioncontaining 20.0 mL of0.100 M hydrofluoric acidand 10.0 mL of0.100 M potassium hydroxideat 25 °C?

HF HFHF

KOHKOH HFHF

HF HFHFK+

OH-OH-

HFH2O H2O

04/19/23

IDIRICE

Initial 1 Dissociation Initial 2 Reaction Initial 3 M Change Equilibrium

Convert to moles

Strong electrolytes dissociate 100%

Sum of previous two steps

Favorable (K>1) reactions occur 100%

Sum of previous two steps

Convert to concentration

Most favorable unfavorable reaction

Use in mass-action expression

04/19/23

EquationIDIRIM

Did you hear aboutthe newpirate movie?

it’s rated arrrrrrrr!

04/19/23

pH Before the Equivalence Point

20.0 mL of 0.100 M HF 10.0 mL of 0.100 M KOH

How many millimoles of each reactant are there?

20.0 mL x 0.100 M = 2.00 mmoles of HF

10.0 mL x 0.100 M = 1.00 mmoles of KOH

04/19/23

EquationIDIRI

HF H+ F¯ H2OKOH K+ OH¯

Start with the initial conditions

2.00 0 0 1.00 0 0 lots

Is HF a strong acid? Does HF dissociate 100%?

Is KOH a strong base? Does KOH dissociate 100%?

Write the strong electrolyte,dissociation reaction here.

KOH K+ + OH¯-1.00 +1.00 +1.00 02.00 0 0 0 1.00 1.00 lots

04/19/23

Consider the Cross Reactions

What chemical species are available to react?

HF, K+, OH¯, H2O

Are the cross-reactions of any of these speciesfavorable (K>1)?

Hint: acid-base reactions are usually favorableidentify any acids and bases

04/19/23

Cross-ReactionsWhich of these reactions are favorable?

HF + HF HF + K+

HF + OH

HF + H2O

K+ + K+

K+ + OH

K+ + H2O

OH+ OH

OH+ H2O

H2O + H2O

04/19/23

K = 1HF + HF

K > 1 K < 1

HF + K+

HF + OH

HF + H2O

K+ + K+

K+ + OH

K+ + H2OOH+ OH

H2O + H2O

OH+ H2O

04/19/23

EquationIDIRI

Write the favorable reaction here

HF + OH¯ H2O + F¯

What is the limiting reagent?

How many millimoles of hydroxide ion reactant?

HF H+ F¯ H2OKOH K+ OH¯2.00 0 0 1.00 0 0 lots

0 0 0 KOH K+ + OH¯-1.00 +1.00 +1.00 02.00 0 0 0 1.00 1.00 lots

How many millimoles of HF reactant?

-1.00 +1.00 +1.00

1.00 1.00 1.00 0.00 lots+1

04/19/23

Assume volumes are additive.What is the total volume of solution?

EquationIDIRI

0 0 0 KOH K+ + OH¯-1.00 +1.00 +1.00 02.00 0 0 0 1.00 1.00 lots

HF + OH¯ H2O + F¯-1.00 1.00 -1.00 +1.00

1.00 1.00 1.00 0.00 lots+1

1.00 mmol / 30.0 mL

0.033 0.033 0.033 lots

04/19/23

HF + HF HF + F

HF + K+

HF + H2O

F+ H2O

K+ + K+

K+ + H2O

H2O + H2O

04/19/23

HF + HFHF + F

Equilibrium ReactionWhich of these reactions are favorable?

K = 1K > 1 K < 1

HF + K+

HF + H2O

F + H2OK+ + K+

K+ + H2OH2O + H2O

This is the most favorable“unfavorable”

reaction

04/19/23

EquationIDIRI

0 0 0 KOH K+ + OH¯-1.00 +1.00 +1.00 02.00 0 0 0 1.00 1.00 lots

HF + OH¯ H2O + F¯-1.00 1.00 -1.00 +1.00

1.00 1.00 1.00 0.00 lots+1

0.033 0.033 0.033 lots

Write the equilibrium reaction here.

HF + F¯ F¯+ HF

04/19/23

Why is this anEquilibrium Reaction?

What was the goal of the calculation?

HF(aq) + F¯(aq) F¯(aq) + HF(aq)

04/19/23

EquationIDIRI

0 0 0 KOH K+ + OH¯-1.00 +1.00 +1.00 02.00 0 0 0 1.00 1.00 lots

HF + OH¯ H2O + F¯-1.00 1.00 -1.00 +1.00

1.00 1.00 1.00 0.00 lots+1

0.033 0.033 0.033 lots

Write an equation that includes the equilibrium terms and the goal of the calculation.

HF + F¯ F¯+ HFHF +H2O H3O+ +F¯

Let -x equal the amount of HF which reacts.

-x +x +x -x

0.033-x x 0.033+x 0.033 lots-x

04/19/23

Solve for [H+]

x = 6.9 x 10–4, -0.0344

[0.033 - x]

[+x] [0.033+x]= 7.2 x 10–4

pH = - log [H+] = - log (6.9 x 10-4)

pH = 3.16

04/19/23

Work out the pHof the solution for

your first data point

return toIDIRICE

04/19/23

Titration of 20.0 mL of 0.100 M HF with 0.100 M KOH

0.0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0

Vol KOH (mL)

Vol KOH pH0.0 2.094.0 2.628.0 2.98

10.0 3.1612.0 3.3316.0 3.7518.0 4.1019.5 4.7420.020.522.025.028.030.032.036.0

04/19/23

pH After the Equivalence Point

20.0 mL of 0.100 M HF 30.0 mL of 0.100 M KOH

04/19/23

EquationIDIRI

2.00 0 0 3.00 0 0 lots

KOH K+ + OH¯-3.00 +3.00 +3.00 02.00 0 0 0 3.00 3.00 lots

04/19/23

HF + HF HF + K+

HF + OH

HF + H2O K+ + K+

K+ + OH

K+ + H2O

OH+ OH

OH+ H2O

H2O + H2O

04/19/23

K = 1HF + HF

K > 1 K < 1

HF + K+

HF + OH

HF + H2OK+ + K+

K+ + OH

K+ + H2OOH+ OH

H2O + H2O

OH+ H2O

What is the best available acid?

HF, K+ , OH, H2O

HF + HFHF + K+

HF + OH

HF + H2O

What is the best available base?

HF + OH

K+ + OH

OH+ OH

OH+ H2O

04/19/23

EquationIDIRI

HF + OH¯ F¯ + H2O

What is the limiting reagent?How many millimoles of HF reactant?

-2.00+2.00 +2.00

0 2.00 3.00 1.00 lots+2

0 0 0 KOH K+ + OH¯-3.00 +3.00 +3.00 02.00 0 0 0 3.00 3.00 lots

04/19/23

EquationIDIRI

2.00 mmol / 50.0 mL

0.040 0.060 0.020 lots

HF + OH¯ F¯ + H2O-2.00 -2.00+2.00 +2.00

0 2.00 3.00 1.00 lots+2

0 0 0 KOH K+ + OH¯-3.00 +3.00 +3.00 02.00 0 0 0 3.00 3.00 lots

20.0 mL of 0.100 M HF is titrated with 30.0 mL of 0.100 M KOH

04/19/23

F+ OH F+ H2O K+ + K+

K+ + OH K+ + H2O

OH + OH OH + H2O

H2O + H2O

04/19/23

K = 1K > 1 K < 1

F + OH

F + H2OK+ + K+

K+ + OH

K+ + H2O

OH+ OH

OH+ H2O

H2O + H2O

F, K+ , OH, H2O

F + H2O

K+ + H2O

OH+ H2O

H2O + H2O

F + OH

K+ + OH

OH+ OH

OH+ H2O

04/19/23

EquationIDIRI

0.040 0.060 0.020 lots

HF + OH¯ F¯ + H2O-2.00 -2.00+2.00 +2.00

0 2.00 3.00 1.00 lots+2

0 0 0 KOH K+ + OH¯-3.00 +3.00 +3.00 02.00 0 0 0 3.00 3.00 lots

H2O + OH¯ H2O + OH¯

04/19/23

Why is this anEquilibrium Reaction?

What was the goal of the calculation?

H2O(l) + OH¯(aq) OH¯(aq) + H2O(l)

04/19/23

EquationIDIRI

H2O H+ + OH¯

Let +x equal the amount of H+ which forms.

+x +x -x

x 0.040 0.060 0.020+x lots-x

0.040 0.060 0.020 lots

HF + OH¯ F¯ + H2O-2.00 -2.00+2.00 +2.00

0 2.00 3.00 1.00 lots+2

0 0 0 KOH K+ + OH¯-3.00 +3.00 +3.00 02.00 0 0 0 3.00 3.00 lots

H2O + OH¯ H2O + OH¯

04/19/23

Solve for [H+]

[x] [0.020+x] = 1.0 x 10–14

x = 5.0 x 10–13, -0.020

[H+] [OH¯] = Kw

pH = - log [H+] = - log (5.0 x 10-13)

pH = 12.30

04/19/23

Vol KOH pH0.0 2.094.0 2.628.0 2.98

10.0 3.1612.0 3.3316.0 3.7518.0 4.1019.5 4.7420.020.522.024.028.030.032.036.0

?11.0011.6811.9512.2012.3012.4012.50

0.0 10.0 20.0 30.0 40.0

Vol KOH (mL)

04/19/23

pH at the Equivalence Point

20.0 mL of 0.100 M HF?? mL of 0.100 M KOH20.0 mL of 0.100 M KOH

04/19/23

EquationIDIRI

2.00 0 0 2.00 0 0 lots

KOH K+ + OH¯-2.00 +2.00 +2.00 02.00 0 0 0 2.00 2.00 lots

04/19/23

HF + HF HF + K+

HF + OH

HF + H2O K+ + K+

K+ + OH

K+ + H2O

OH+ OH

OH+ H2O

H2O + H2O

04/19/23

K = 1HF + HF

K > 1 K < 1

HF + K+

HF + OH

HF + H2OK+ + K+

K+ + OH

K+ + H2OOH+ OH

H2O + H2O

OH+ H2O

HF, K+ , OH, H2O

HF + HFHF + K+

HF + OH

HF + H2O

HF + OH

K+ + OH

OH+ OH

OH+ H2O

04/19/23

EquationIDIRI

HF + OH¯ F¯ + H2O

What is the limiting reagent?How many millimoles of HF reactant?

-2.00+2.00 +2.00

0 2.00 2.00 0 lots+2

0 0 0 KOH K+ + OH¯-2.00 +2.00 +2.00 02.00 0 0 0 2.00 2.00 lots

04/19/23

EquationIDIRI

2.00 mmol / 40.0 mL

0.050 0.050 lots

HF + OH¯ F¯ + H2O-2.00 -2.00+2.00 +2.00

0 2.00 2.00 0 lots+2

0 0 0 KOH K+ + OH¯-2.00 +2.00 +2.00 02.00 0 0 0 2.00 2.00 lots

04/19/23

F+ H2O K+ + K+

K+ + H2O

H2O + H2O

04/19/23

K = 1K > 1 K < 1

F + H2O

K+ + K+

K+ + H2O

H2O + H2OWhat is the best available acid?

F, K+ , H2O

F + H2O

K+ + H2O

H2O + H2O

F + H2O

What is the best acid - base reaction?

04/19/23

EquationIDIRI

H2O + F¯ HF + OH¯

0.050 0.050 lots

HF + OH¯ F¯ + H2O-2.00 -2.00+2.00 +2.00

0 2.00 2.00 0 lots+2

0 0 0 KOH K+ + OH¯-2.00 +2.00 +2.00 02.00 0 0 0 2.00 2.00 lots

04/19/23

Strategy

How to get at the pH

Can’t use the equationHF(aq) + H2O(l) F–(aq) + H3O+(aq)

So use the equationH2O(l) + F–(aq) HF(aq) + OH–(aq)

And solve for OH–

Then useH2O(l) H+(aq) + OH–(aq)

04/19/23

EquationIDIRI

H2O + F¯ HF + OH¯

Let +x equal the amount of OH¯ which forms.

+x+x -x -x

x 0.050-x 0.050 x lots-x H2O + F¯ HF + OH¯

0.050 0.050 lots

HF + OH¯ F¯ + H2O-2.00 -2.00+2.00 +2.00

0 2.00 2.00 0 lots+2

0 0 0 KOH K+ + OH¯-2.00 +2.00 +2.00 02.00 0 0 0 2.00 2.00 lots

04/19/23

What is the Equilibrium Constant for

H2O(l) + F–(aq) HF(aq) + OH–(aq)

H+(aq) + F–(aq) HF(aq)

H2O(l) H+(aq) + OH–(aq)

1 / 7.2 x 10–4 M

1.0 x 10–14 M2

1.4 x 10–11 M

04/19/23

Solve for [OH–]

x = 8.3 x 10–7 , - 8.3 x 10–7

[F–]

[HF] [OH–]= 1.4 x 10–11 M

x is thehydroxide ionconcentration

[OH–] = 8.3 x 10–7 M

=0.050 - x

(x) (x)

04/19/23

Solve for [H+] then pH

[H+] [8.3 x 10–7] = 1.0 x 10–14 M2

[H+] = 1.2 x 10–8 M

pH = - log [H+] = - log (1.2 x 10–8)

pH = 7.92

[H+] [OH¯] = Kw

04/19/23

Vol KOH pH0.0 2.104.0 2.628.0 3.03

10.0 3.1612.0 3.1916.0 3.7818.0 4.1519.5 4.9020.0 7.9220.5 11.0022.0 11.6824.0 11.9528.0 12.2230.0 12.3032.0 12.3636.0 12.50

0.0 10.0 20.0 30.0 40.0

Vol KOH (mL)

04/19/23

Group Activity

What species are predominately present when• 20.0 mL of 0.100 M HF reacts with• 20.0 mL of 0.100 M KOH?

Give another way in which a solution containing• 2.00 mmol of K+ and 2.00 mmol of F¯ in

40.0 mL of solution could be prepared.

04/19/23

pH of a Salt

What is the pH of a salt solution containing• 2.00 mmol of KF in 40.0 mL of solution?

Why is the pH of the solution not neutral? (not equal to 7)

04/19/23

pH of a SaltpH at the Equivalence Point

StrongStrong 7 HCl + NaOHNaCl

WeakStrong below 7 HNO3 + NH3NH4NO3

StrongWeak above 7 H2CO3 + KOHKHCO3

WeakWeak near 7 HF + N2H4N2H5F

BaseAcidEquivalence

Point pHExample

04/19/23

Predict whether aqueous solutions of the following salts should be acidic, basic or neutral.

KBr HBr + KOHHBr + KOHHBr + KOH neutral

Na2CO3 NaHCO3 + NaOHNaHCO3 + NaOHNaHCO3 + NaOH basic

NaHCO3 H2CO3 + NaOHH2CO3 + NaOHH2CO3 + NaOH

Acid + Base pHSalt

basicNH4I HI + NH3HI + NH3HI + NH3 acidic

CH3NH3F HF + CH3NH2HF + CH3NH2HF + CH3NH2 neutral

04/19/23

problem set ph and aqueous solutions. aqueous neutralization reactions and ph edward a. mottel...

ph base

moles of base slide

h aq h

equivalence point slide

solution slide

o aq slide

acid v acid

base v base

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