Problem Set pH and Aqueous Solutions. Aqueous Neutralization Reactions and pH Edward A. Mottel...
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![Page 1: Problem Set pH and Aqueous Solutions. Aqueous Neutralization Reactions and pH Edward A. Mottel Department of Chemistry Rose-Hulman Institute of Technology.](https://reader038.fdocuments.net/reader038/viewer/2022110323/56649d795503460f94a5cc4f/html5/thumbnails/1.jpg)
Problem Set
pH and Aqueous Solutions
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Aqueous Neutralization Reactions and pH
Edward A. Mottel
Department of Chemistry
Rose-Hulman Institute of Technology
![Page 3: Problem Set pH and Aqueous Solutions. Aqueous Neutralization Reactions and pH Edward A. Mottel Department of Chemistry Rose-Hulman Institute of Technology.](https://reader038.fdocuments.net/reader038/viewer/2022110323/56649d795503460f94a5cc4f/html5/thumbnails/3.jpg)
Aqueous Solutions and pH
Reading assignment: • Zumdahl: Chapter 7.3-7.6, 7.11, 8.1
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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)
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pH
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
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pH
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?
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pH
If the pH of an aqueous solution is 3.50 at 25 C,what is the hydroxide ion concentration?
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04/19/23
pH
7
0
14
acids
bases
neutral @ 25 °C
1 M HCl
1 M KOH
pH values can be outside 0-14
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Titration
reactant 2
reactant 1
an analytical techniqueto determinethe concentrationof a solution
buret
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Thermometric Titration
T (C)
Vacid
acid-base reactionsare usually exothermic
Equivalence Pointvolume
Moles of acid = Moles of base
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Titration
HF + NaOH NaF + H2O
Balanced equation
reactant 2
reactant 1
an analytical techniqueto determinethe concentrationof a solution
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Titration
Moles of acid = Moles of base
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
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04/19/23
Titration
Moles of acid = Moles of base
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)
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Equivalence Point Detection
Physical methods• Thermometric • Potentiometric• Visual indicator• Conductometric
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Potentiometric Titration
20.0 mL of 0.400 M HCl with Aqueous Ammonia
0 4 8 12 16 20 24 280
2
10
8
6
4
pH
Volume NH3 (mL)
Equivalence Pointvolume
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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
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SO3
CCH3
H3C
Br
Br
OH
OBrBr
O
SO3
CCH3
H3C
Br
Br
O
BrBr
Bromocresol Green
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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
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Potentiometric Titration
20.0 mL of 0.400 M HCl with Aqueous Ammonia
0 4 8 12 16 20 24 280
2
10
8
6
4
pH
Volume NH3 (mL)
phenolphthalein
bromocresol green
equivalencepoint
endpoint
endpoint
For this titration,which indicatorwould be better?
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Conductometric Titration
corr
Vacid
Equivalence Pointvolume
Depending on the acid-base system studied,the graph may have many different shapes.
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Conductometric Titration
Why does the line change slope?
corr
Vacid
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04/19/23
Conductometric Titration
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)
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04/19/23
Conductometric Titration
If the total number of ions isn’t changing before theequivalence point, why is the conductivity changing?
corr
Vacid
Equivalence Pointvolume
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Problem Set
Weak Acid – Strong Base Titration
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04/19/23
Potentiometric Titration
20.0 mL of 0.400 M HCl with Aqueous Ammonia
0 4 8 12 16 20 24 280
2
10
8
6
4
pH
Volume NH3 (mL)
Equivalence Pointvolume
Why does the pH change rapidly at the equivalence point?
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So maybe you would like to be aCPA?
(Chemistry Problem Analyzer)
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Hydrofluoric Acid Solution
What is the pH of a 20.0 mL of 0.100 M hydrofluoric acid at 25 °C?
HF HFHFH+
F-
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Substitute into theMass-Action Expression
[HF]
[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
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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
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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-
K+
HFH2O H2O
HF
K+
F-
F-
K+
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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
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EquationIDIRIM
CE
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Did you hear aboutthe newpirate movie?
it’s rated arrrrrrrr!
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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
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EquationIDIRI
MCE
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%?
0 0 0
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
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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
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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
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04/19/23
Cross-ReactionsWhich of these reactions are favorable?
K = 1HF + HF
K > 1 K < 1
HF + K+
HF + OH
HF + H2O
K+ + K+
K+ + OH
K+ + H2OOH+ OH
H2O + H2O
OH+ H2O
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EquationIDIRI
MCE
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
-1.00
How many millimoles of HF reactant?
-1.00 +1.00 +1.00
1.00 1.00 1.00 0.00 lots+1
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04/19/23
Assume volumes are additive.What is the total volume of solution?
EquationIDIRI
MCE
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
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
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04/19/23
Cross-ReactionsWhich of these reactions are favorable?
HF + HF HF + F
HF + K+
HF + H2O
F+ F
F+ K+
F+ H2O
K+ + K+
K+ + H2O
H2O + H2O
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04/19/23
HF + HFHF + F
Equilibrium ReactionWhich of these reactions are favorable?
K = 1K > 1 K < 1
HF + K+
HF + H2O
F+ F
F+ K+
F + H2OK+ + K+
K+ + H2OH2O + H2O
This is the most favorable“unfavorable”
reaction
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EquationIDIRI
MCE
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
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
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04/19/23
Why is this anEquilibrium Reaction?
What was the goal of the calculation?
HF(aq) + F¯(aq) F¯(aq) + HF(aq)
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EquationIDIRI
MCE
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
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
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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
x is thehydrogen ionconcentration
A negativeconcentrationis not possible
pH = - log [H+] = - log (6.9 x 10-4)
pH = 3.16
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04/19/23
Work out the pHof the solution for
your first data point
return toIDIRICE
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04/19/23
Titration of 20.0 mL of 0.100 M HF with 0.100 M KOH
pH
2.00
4.00
6.00
8.00
10.00
12.00
14.00
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
pH
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04/19/23
pH After the Equivalence Point
20.0 mL of 0.100 M HF 30.0 mL of 0.100 M KOH
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04/19/23
EquationIDIRI
MCE
HF H+ F¯ H2OKOH K+ OH¯
Start with the initial conditions
2.00 0 0 3.00 0 0 lots
Is HF a strong acid? Does HF dissociate 100%?
0 0 0
Is KOH a strong base? Does KOH dissociate 100%?
Write the strong electrolyte,dissociation reaction here.
KOH K+ + OH¯-3.00 +3.00 +3.00 02.00 0 0 0 3.00 3.00 lots
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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
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04/19/23
Cross-ReactionsWhich of these reactions are favorable?
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
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04/19/23
EquationIDIRI
MCE
Write the favorable reaction here
HF + OH¯ F¯ + H2O
What is the limiting reagent?How many millimoles of HF reactant?
-2.00
How many millimoles of hydroxide ion reactant?
-2.00+2.00 +2.00
0 2.00 3.00 1.00 lots+2
HF H+ F¯ H2OKOH K+ OH¯2.00 0 0 3.00 0 0 lots
0 0 0 KOH K+ + OH¯-3.00 +3.00 +3.00 02.00 0 0 0 3.00 3.00 lots
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04/19/23
Assume volumes are additive.What is the total volume of solution?
EquationIDIRI
MCE
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
HF H+ F¯ H2OKOH K+ OH¯2.00 0 0 3.00 0 0 lots
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
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04/19/23
Cross-ReactionsWhich of these reactions are favorable?
F+ F
F+ K+
F+ OH F+ H2O K+ + K+
K+ + OH K+ + H2O
OH + OH OH + H2O
H2O + H2O
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04/19/23
F+ F
Equilibrium ReactionWhich of these reactions are favorable?
K = 1K > 1 K < 1
F+ K+
F + OH
F + H2OK+ + K+
K+ + OH
K+ + H2O
OH+ OH
OH+ H2O
H2O + H2O
What is the best available acid?
F, K+ , OH, H2O
F + H2O
K+ + H2O
OH+ H2O
H2O + H2O
What is the best available base?
F + OH
K+ + OH
OH+ OH
OH+ H2O
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04/19/23
EquationIDIRI
MCE
Write the equilibrium reaction here.
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
HF H+ F¯ H2OKOH K+ OH¯2.00 0 0 3.00 0 0 lots
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¯
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04/19/23
Why is this anEquilibrium Reaction?
What was the goal of the calculation?
H2O(l) + OH¯(aq) OH¯(aq) + H2O(l)
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04/19/23
EquationIDIRI
MCE
Write an equation that includes the equilibrium terms and the goal of the calculation.
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
HF H+ F¯ H2OKOH K+ OH¯2.00 0 0 3.00 0 0 lots
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¯
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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
x is thehydrogen ionconcentration
A negativeconcentrationis not possible
pH = - log [H+] = - log (5.0 x 10-13)
pH = 12.30
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04/19/23
Titration of 20.0 mL of 0.100 M HF with 0.100 M KOH
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
Titration of 20.0 mL of 0.100 M HF with 0.100 M KOH
2.00
4.00
6.00
8.00
10.00
12.00
14.00
0.0 10.0 20.0 30.0 40.0
Vol KOH (mL)
pH
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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
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04/19/23
EquationIDIRI
MCE
HF H+ F¯ H2OKOH K+ OH¯
Start with the initial conditions
2.00 0 0 2.00 0 0 lots
Is HF a strong acid? Does HF dissociate 100%?
0 0 0
Is KOH a strong base? Does KOH dissociate 100%?
Write the strong electrolyte,dissociation reaction here.
KOH K+ + OH¯-2.00 +2.00 +2.00 02.00 0 0 0 2.00 2.00 lots
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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
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04/19/23
Cross-ReactionsWhich of these reactions are favorable?
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
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04/19/23
EquationIDIRI
MCE
Write the favorable reaction here
HF + OH¯ F¯ + H2O
What is the limiting reagent?How many millimoles of HF reactant?
-2.00
How many millimoles of hydroxide ion reactant?
-2.00+2.00 +2.00
0 2.00 2.00 0 lots+2
HF H+ F¯ H2OKOH K+ OH¯2.00 0 0 2.00 0 0 lots
0 0 0 KOH K+ + OH¯-2.00 +2.00 +2.00 02.00 0 0 0 2.00 2.00 lots
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04/19/23
Assume volumes are additive.What is the total volume of solution?
EquationIDIRI
MCE
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
HF H+ F¯ H2OKOH K+ OH¯2.00 0 0 2.00 0 0 lots
0 0 0 KOH K+ + OH¯-2.00 +2.00 +2.00 02.00 0 0 0 2.00 2.00 lots
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04/19/23
Cross-ReactionsWhich of these reactions are favorable?
F+ F
F+ K+
F+ H2O K+ + K+
K+ + H2O
H2O + H2O
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04/19/23
F+ F
Equilibrium ReactionWhich of these reactions are favorable?
K = 1K > 1 K < 1
F+ K+
F + H2O
K+ + K+
K+ + H2O
H2O + H2OWhat is the best available acid?
F, K+ , H2O
F + H2O
K+ + H2O
H2O + H2O
What is the best available base?
F+ F
F+ K+
F + H2O
What is the best acid - base reaction?
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04/19/23
EquationIDIRI
MCE
Write the equilibrium reaction here.
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
HF H+ F¯ H2OKOH K+ OH¯2.00 0 0 2.00 0 0 lots
0 0 0 KOH K+ + OH¯-2.00 +2.00 +2.00 02.00 0 0 0 2.00 2.00 lots
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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)
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04/19/23
EquationIDIRI
MCE
Write an equation that includes the equilibrium terms and the goal of the calculation.
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
HF H+ F¯ H2OKOH K+ OH¯2.00 0 0 2.00 0 0 lots
0 0 0 KOH K+ + OH¯-2.00 +2.00 +2.00 02.00 0 0 0 2.00 2.00 lots
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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
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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
A negativeconcentrationis not possible
[OH–] = 8.3 x 10–7 M
=0.050 - x
(x) (x)
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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
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04/19/23
Titration of 20.0 mL of 0.100 M HF with 0.100 M KOH
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
Titration of 20.0 mL of 0.100 M HF with 0.100 M KOH
2.00
4.00
6.00
8.00
10.00
12.00
14.00
0.0 10.0 20.0 30.0 40.0
Vol KOH (mL)
pH
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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.
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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)
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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
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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
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