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Solutions to: Chapter 15 Homework SetS.E. Van Bramer 1/3/2017
1. For the following reactions, write a balanced equation that describes what happens when the
following are mixed. Identify the acid, base, conjugate base, and conjugate acid.
a.Nitric acid and water: HNO3 + H2O -> H3O1+ + NO3
1-
Acid HNO3
Base H2O
Conjugate base NO31-
Conjugate acid H3O1+
b.Ammonia and water: NH3 + H2O -> OH1- + NH4
1+
Acid H2O
Base NH3
Conjugate base OH1-
Conjugate acid NH41+
c.Nitric acid and ammonia: HNO3 + NH3 -> NH41+ + NO3
1-
Acid HNO3
Base NH3
Conjugate base NO31-
Conjugate acid NH41+
d.Sulfuric acid and sodium hydroxide: H2SO4 + 2 NaOH -> 2 H2O + SO42- + 2 Na1+
Acid H2SO4
Base NaOH
Conjugate base SO42-
Conjugate acid H2O
e.Sodium hydrogen sulfate and ammonia: HNO3 + H2O -> H3O1+ + NO3
1-
Acid HNO3
Base H2O
Conjugate base NO31-
Conjugate acid H3O1+
f.Acetic acid and water: CH3COOH + H2O -> H3O1+ + CH3COO
1-
Acid CH3COOH
Base H2O
Conjugate base CH3COO1-
Conjugate acid H3O1+
g.Acetic acid and sodium hydroxide: CH3COOH + NaOH -> H2O + CH3COO1- + Na1+
Acid CH3COOH
Base NaOH
Conjugate base CH3COO1-
Conjugate acid H2O
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2.Determine the equlibrium concentration of H3O1+ and OH1-. Identify the acid and the conjugate
base (or the base and the conjugate acid). Calculate the pH.
Constants Used:
Mmole
liter:= Kw 1.0 10
14−⋅ M
2⋅:=
a.A 0.10 M solution of HNO3
Given concentraion of the acid:
Cacid 0.10 M⋅:=
The concentrations (this assumes that Kw is not significant):
CH3O Cacid:= CH3O 0.1M=
pH log Cacid M1−
⋅
−:= pH 1=
pOH 14 pH−:= pOH 13=
b.1.00 g of HCl is diluted to 500.0 mL with water
The concentration of HCl
massacid 1.00 gm⋅:=
MWacid 1.00794 35.4527+( ) gm⋅ mole1−
⋅:= MWacid 36.461 gm mole1−
⋅=
moleacid
massacid
MWacid
:= moleacid 0.027mole=
volumeacid 500 mL⋅:=
Cacid
moleacid
volumeacid:= Cacid 0.055M=
The concentrations:
CH3O Cacid:= CH3O 0.055M=
pH log Cacid M1−
⋅
−:= pH 1.261=
pOH 14 pH−:= pOH 12.739=
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c.10.0 g of HCl is diluted to 500.0 mL with water
The concentration of HCl
massacid 10.00 gm⋅:=
MWacid 1.00794 35.4527+( ) gm⋅ mole1−
⋅:= MWacid 36.461 gm mole1−
⋅=
moleacid
massacid
MWacid
:= moleacid 0.274mole=
volumeacid 500 mL⋅:=
Cacid
moleacid
volumeacid:= Cacid 0.549M=
The concentrations:
CH3O Cacid:= CH3O 0.549M=
pH log Cacid M1−
⋅
−:= pH 0.261=
pOH 14 pH−:= pOH 13.739=
d.5.00 grams of acetic acid is diluted to 250.0 mL with water
The Ka for acetic acid: Ka 1.8 105−
⋅ M⋅:=
The concentration of acetic acid
massacid 5.00 gm⋅:=
MWacid 2 12.001⋅( ) 4 1.0079⋅( )+ 2 15.999⋅( )+[ ] gm⋅ mole1−
⋅:=
moleacid
massacid
MWacid
:= moleacid 0.083mole=
volumeacid 250 mL⋅:=
Cacid
moleacid
volumeacid:= Cacid 0.333M=
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The equlibrium solution for HA + H2O <-> H3O+ + A-
Initial Concentraion Cacid
Equlibrium Concentration Cacid X− X X
Equlibrium expression:
Ka
X2
Cacid X−=
Assuming X is smaller than [acid], this reduces to:
Ka
X2
Cacid
=
Solving for X: X Ka Cacid⋅:= X 2.449 103−
× M=
Equlibrium Concentrations:
CHA Cacid X−:= CHA 0.331M=
CH3O X:= CH3O 2.449 103−
× M=
CA X:= CA 2.449 103−
× M=
pH log CH3O M1−
⋅
−:= pH 2.611=
pOH 14 pH−:= pOH 11.389=
Solving for X without assumptions:
Ka
X2
Cacid X−=
X
1−
2Ka⋅
1
2Ka
24 Ka⋅ Cacid⋅+⋅+
1−
2Ka⋅
1
2Ka
24 Ka⋅ Cacid⋅+⋅−
:= X2.44 10
3−×
2.458− 103−
×
M=
Equlibrium Concentrations (Calculated for both roots, select the reasonable
answer [top or bottom in ()]:
CHA Cacid X−:= CHA
0.331
0.336
M=
CH3O X:= CH3O
2.44 103−
×
2.458− 103−
×
M=
i 0 1, 1..:=
pHi
log CH3OiM
1−⋅
−:= pH2.613
2.609 1.364i−
=
pOH 14 pH−:= pOH11.387
11.391 1.364i+
=
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e.5.00 grams of sodium hydroxide is diluted to 500.0 mL with water
The concentration sodium hydroxide is:
massbase 5.00 gm⋅:=
MWbase 22.990 15.999+ 1.0079+( ) gm⋅ mole1−
⋅:=
molebase
massbase
MWbase
:= molebase 0.125mole=
volumebase 500 mL⋅:=
Cbase
molebase
volumebase:= Cbase 0.25M=
The concentrations (assumes Kw is not significant):
COH Cbase:= COH 0.25M=
pOH log Cbase M1−
⋅
−:= pOH 0.602=
pH 14 pOH−:= pH 13.398=
f.5.00 grams of ammonia is diluted to 500.0 mL with water
The Kb for ammonia
Kb 1.8 105−
⋅ M⋅:=
The concentration of ammonia
massbase 5.00 gm⋅:=
MWbase 14.00674 3 1.0079⋅+( ) gm⋅ mole1−
⋅:=
molebase
massbase
MWbase
:= molebase 0.294mole=
volumebase 500 mL⋅:=
Cbase
molebase
volumebase:= Cbase 0.587M=
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The equlibrium solution for B + H2O <-> HB1+ + OH1-
Initial Concentraion Cbase
Equlibrium Concentration Cbase X− X X
Equlibrium expression:
Kb
X2
Cbase X−=
Assuming X is smaller than [base], this reduces to:
Kb
X2
Cbase
=
Solving for X:
X Kb Cbase⋅:=
X 3.251 103−
× M=
Equlibrium Concentrations:
CB Cbase X−:= CB 0.584M=
CHB X:= CHB 3.251 103−
× M=
COH X:= COH 3.251 103−
× M=
pOH log COH M1−
⋅
−:= pOH 2.488=
pH 14 pOH−:= pH 11.512=
Solving for X without assumptions:
Kb
X2
Cbase X−=
X
1−
2Kb⋅
1
2Kb
24 Kb⋅ Cbase⋅+⋅+
1−
2Kb⋅
1
2Kb
24 Kb⋅ Cbase⋅+⋅−
:=
X3.242 10
3−×
3.26− 103−
×
M=
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Equlibrium Concentrations (Calculated for both roots, select the reasonable
answer [top or bottom in ()]:
CB Cbase X−:= CB
0.584
0.59
M=
CHB X:= CHB
3.242 103−
×
3.26− 103−
×
M=
COH X:= COH
3.242 103−
×
3.26− 103−
×
M=
i 0 1, 1..:=pOH
ilog COH
iM
1−⋅
−:= pOH2.489
2.487 1.364i−
=
pH 14 pOH−:= pH11.511
11.513 1.364i+
=
g. 5.00 grams of sodium chloride is diluted to 500.0 mL with water
Since sodium is the conjugate acid of a strong base (NaOH), and chloride is the
congjute base of a strong acid (HCl). NaCl will dissociate completely in water and will
not effect the pH.
pH = 7
pOH = 7
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h. 5.00 grams of potassium acetate is diluted to 500.0 mL with water
This salt containes acetate ions, the conjugate base for acetic acid. The equlibrium constant for
acetic acid is:
Ka 1.8 105−
⋅ M⋅:=
Kb
Kw
Ka
:= Kb 5.556 1010−
× M=
The concentration of the potassium acetate:
masssalt 5 gm⋅:=
MWsalt 39.0983( ) 2 12.001⋅( )+ 3 1.0079⋅( )+[ ] 2 15.999⋅( )+[ ] gm⋅ mole1−
⋅:=
molesalt
masssalt
MWsalt
:= molesalt 0.051mole=
volumesalt 500 mL⋅:=
Csalt
molesalt
volumesalt:= Csalt 0.102M=
The equlibrium solution for A- + H2O <-> OH- + HA
Initial Concentraion Csalt
Equlibrium Concentration Csalt X− X X
Equlibrium expression:
Kb
X2
Csalt X−=
Assuming X is smaller than [acid], this reduces to:
Kb
X2
Csalt
=
Solving for X:
X Kb Csalt⋅:=
X 7.525 106−
× M=
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Equlibrium Concentrations:
CHA X:= CHA 7.525 106−
× M=
COH X:= COH 7.525 106−
× M=
CA Csalt X−:= CA 0.102M=
pOH log COH M1−
⋅
−:= pOH 5.124=
pH 14 pOH−:= pH 8.876=
Solving for X without assumptions:
Kb
X2
Csalt X−=
X
1−
2Kb⋅
1
2Kb
24 Kb⋅ Csalt⋅+⋅+
1−
2Kb⋅
1
2Kb
24 Kb⋅ Csalt⋅+⋅−
:=
X7.524 10
6−×
7.525− 106−
×
M=
Equlibrium Concentrations (Calculated for both roots, select the reasonable
answer [top or bottom in ( ) ]:
CHA X:= CHA
7.524 106−
×
7.525− 106−
×
M=
COH X:= COH
7.524 106−
×
7.525− 106−
×
M=
CA Csalt X−:= CA
0.102
0.102
M=
pOHi
log COHiM
1−⋅
−:= pOH5.124
5.124 1.364i−
=
pH 14 pOH−:= pH8.876
8.876 1.364i+
=
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i.5.00 grams of sulfuric acid is diluted to 500 mL with water
Sulfuric acid is a diprotic acid. K1 is very large (since sulfuric acid is a strong acid).
Since K1 is much large than K2, we will ignore the contribution of K2 to the reaction and
simply treat this as a monoprotic strong acid.
The concentration of sulfuric acid is:
massacid 5.00 gm⋅:=
MWacid 2 1.0079⋅ 32.066+ 4 15.9994⋅+( ) gm⋅ mole1−
⋅:=
moleacid
massacid
MWacid
:=
volumeacid 500 mL⋅:=
Cacid
moleacid
volumeacid:=
Cacid 0.102M=
The concentrations:
CH3O Cacid:= CH3O 0.102M=
pH log Cacid M1−
⋅
−:= pH 0.992=
pOH 14 pH−:= pOH 13.008=
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j.5.00 grams of sodium hydrogen sulfate is diluted to 500.0 mL with water
This is a weak acid problem. Sodium hydrogen sulfate is a salt. When it dissolves in
water it will dissociate into sodium ions (conjugate acid of a strong base, NaOH, so it
will have no effect on the pH) and hydrogen sulfate.
Hydrogen sulfate is the conjugate base of a strong acid (sulfuric acid), so it will not act
as a base. However, it is also a weak base. This is the equlibrium that we must pay
attention to in this problem.
Given a weak acid where: Ka 1.2 102−
⋅ M⋅:=
The concentration of hydrogen sulfide (the acid) is :
massacid 5 gm⋅:=
MWacid 22.9897768( ) 1.0079( )+ 32.066( )+ 4 15.9994⋅( )+[ ] gm⋅ mole1−
⋅:=
moleacid
massacid
MWacid
:=
volumeacid 500 mL⋅:=
Cacid
moleacid
volumeacid:= Cacid 0.083M=
The equlibrium solution for HA + H2O <-> H3O+ + A-
Initial Concentraion Cacid
Equlibrium Concentration Cacid X− X X
Equlibrium expression:
Ka
X2
Cacid X−=
Assuming X is smaller than [acid], this reduces to:
Ka
X2
Cacid
=
Solving for X:
X Ka Cacid⋅:= X 0.032M=
Equlibrium Concentrations:
CHA Cacid X−:= CHA 0.052M=
CH3O X:= CH3O 0.032M=
CA X:= CA 0.032M=
pH log CH3O M1−
⋅
−:= pH 1.5=
pOH 14 pH−:= pOH 12.5=
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Solving for X without assumptions:
Ka
X2
Cacid X−=
X
1−
2Ka⋅
1
2Ka
24 Ka⋅ Cacid⋅+⋅+
1−
2Ka⋅
1
2Ka
24 Ka⋅ Cacid⋅+⋅−
:=
X0.026
0.038−
M=
Equlibrium Concentrations (Calculated for both roots, select the reasonable
answer [top or bottom in ()]:
CHA Cacid X−:= CHA
0.057
0.121
M=
CH3O X:= CH3O
0.026
0.038−
M=
CA X:= CA
0.026
0.038−
M=
i 0 1, 1..:=
pHi
log CH3OiM
1−⋅
−:= pH1.582
1.418 1.364i−
=
pOH 14 pH−:= pOH12.418
12.582 1.364i+
=
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k.5.00 grams of sodium sulfate is diluted to 500.0 mL with water
Sodium sulfate is a salt. It dissolves in water to produce sodium ions and sulfate ions.
Sodium Ions are the conjugate acid of a strong base (NaOH) so they do not effect on the pH.
Sulfate ions are the conjugate base of a weak acid (hydrogen sulfate). So it will behave as a
weak base. Hydrogen sulfate in turn is the conjugate base of a strong acid (sulfuric acid) so
it will have no additional effect on the pH.
So we will treat this problem as the conjugate base of a weak acid.
The equlibrium constant for the weak acid (Input appropriate value here):
Ka 1.2 102−
⋅ M⋅:=
Kb
Kw
Ka
:= Kb 8.333 1013−
× M=
The concentration of the sodium sulfate is
masssalt 5 gm⋅:=
MWsalt 2 22.990⋅( ) 32.066( )+[ ] 4 15.999⋅( )+[ ] gm⋅ mole1−
⋅:=
molesalt
masssalt
MWsalt
:=
volumesalt 500 mL⋅:=
Csalt
molesalt
volumesalt:= Csalt 0.07M=
The equlibrium solution for A- + H2O <-> OH- + HA
Initial Concentraion Csalt
Equlibrium Concentration Csalt X− X X
Equlibrium expression:
Kb
X2
Csalt X−=
Assuming X is smaller than [acid], this reduces to:
Kb
X2
Csalt
=
Solving for X:
X Kb Csalt⋅:=
X 2.422 107−
× M=
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Equlibrium Concentrations:
CHA X:= CHA 2.422 107−
× M=
COH X:= COH 2.422 107−
× M=
CA Csalt X−:= CA 0.07M=
pOH log COH M1−
⋅
−:= pOH 6.616=
pH 14 pOH−:= pH 7.384=
Solving for X without assumptions:
Kb
X2
Csalt X−=
X
1−
2Kb⋅
1
2Kb
24 Kb⋅ Csalt⋅+⋅+
1−
2Kb⋅
1
2Kb
24 Kb⋅ Csalt⋅+⋅−
:=
X2.422 10
7−×
2.422− 107−
×
M=
Equlibrium Concentrations (Calculated for both roots, select the reasonable
answer [top or bottom in ( ) ]:
CHA X:= CHA
2.422 107−
×
2.422− 107−
×
M=
COH X:= COH
2.422 107−
×
2.422− 107−
×
M=
CA Csalt X−:= CA
0.07
0.07
M=
pOHi
log COHiM
1−⋅
−:= pOH6.616
6.616 1.364i−
=
pH 14 pOH−:= pH7.384
7.384 1.364i+
=
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Sodium Ions are the conjugate acid of a strong base (NaOH) so they do not effect on the pH.