Chem 142
Summer 2011
Name ________________________________________
Section ________________
Worksheet 9: Buffers, Titrations, and Solubility
1. What is the pH and percent dissociation of the acid in each of the
following solutions? What is the pH after 50.0 mL of 3.00 M NaOH is
added to the solution? What is the pH after 80.0 mL of 4.00 M HCl is
added to the original solution? Rank the buffers in order of increasing
buffer capacity based on your calculations.
a. 0.200 M HC2H3O2 (Ka = 1.8 x 10-5) in the presence of 0.500 M
NaC2H3O2 in a total volume of 1.00 L
& "#C2 H 3O2! $% )
& 0.500 )
!5
pH = pK a + log (
+ = ! log 1.8 , 10 + log ('
+
0.200 *
' [ HC2 H 3O2 ] *
pH = 4.74 + 0.3979 = 5.14
(
)
#$ H + %& #$C2 H 3O2" %& x ( 0.500 )
K a = 1.8 ! 10 =
=
0.200
[ HC2 H 3O2 ]
"5
x = !" H + #$ = 7.2 % 10 &6 M
7.2 ! 10 "6
%dissoc. =
! 100% = 3.6 ! 10 "3%
0.200
Addition of 50.00 mL of 3.00 M NaOH
HC2H3O2 (aq) + OH- (aq) ! C2H3O2- (aq) + H2O (l)
OH150.
0
HC2H3O2
200.
50.
Initial (mmol)
Final (mmol)
C2H3O2500.
650.
HC2H3O2 (aq) ! C2H3O2- (aq) + H+ (aq)
[HC2H3O2]
[C2H3O2-]
[H+]
Initial
0.04762
0.6190
0
Change
-x
+x
+x
Equilibrium
0.04762-x
0.6190+x
x
Assume x is small compared to initial concentrations
#$ H + %& #$C2 H 3O2" %& x ( 0.6190 )
"5
K a = 1.8 ! 10 =
=
0.04762
[ HC2 H 3O2 ]
x = !" H + #$ = 1.4 % 10 &6 M
(
)
(
)
pH = ! log "# H + $% = ! log 1.4 & 10 !6 = 5.86
Page 1 of 16
Chem 142
Summer 2011
Addition of 80.00 mL of 4.00 M HCl
C2H3O2- (aq) + H+ (aq) ! HC2H3O2 (aq)
C2H3O2500.
180.
Initial (mmol)
Final (mmol)
H+
320.
0
HC2H3O2
200.
520.
HC2H3O2 (aq) ! C2H3O2- (aq) + H+ (aq)
[HC2H3O2]
[C2H3O2-]
[H+]
Initial
0.482
0.167
0
Change
-x
+x
+x
Equilibrium
0.482-x
0.167+x
x
Assume x is small compared to initial concentrations
#$ H + %& #$C2 H 3O2" %& x ( 0.167 )
"5
K a = 1.8 ! 10 =
=
0.482
[ HC2 H 3O2 ]
x = !" H + #$ = 5.2 % 10 &5 M
(
)
(
)
pH = ! log "# H + $% = ! log 5.2 & 10 !5 = 4.28
b. 0.200 M HC3H5O2 (Ka = 1.3 x 10-5) in the presence of 0.400 M
NaC3H5O2 in a total volume of 1.00 L
& "C3 H 5O2! $% )
& 0.400 )
!5
pH = pK a + log ( #
+ = ! log 1.3 , 10 + log ('
+
0.200 *
' [ HC3 H 5O2 ] *
pH = 4.89 + 0.30 = 5.19
(
)
#$ H + %& #$C3 H 5O2" %& x ( 0.400 )
K a = 1.3 ! 10 =
=
0.200
[ HC3 H 5O2 ]
"5
x = !" H + #$ = 6.5 % 10 &6 M
%dissoc. =
6.5 ! 10 "6
! 100% = 3.3 ! 10 "3%
0.200
Page 2 of 16
Chem 142
Summer 2011
Addition of 50.00 mL of 3.00 M NaOH
HC3H5O2 (aq) + OH- (aq) ! C3H5O2- (aq) + H2O (l)
OH150.
0
HC3H5O2
200.
50.0
Initial (mmol)
Final (mmol)
C3H5O2400.
550.
HC3H5O2 (aq) ! C3H5O2- (aq) + H+ (aq)
[HC3H5O2]
[C3H5O2-]
[H+]
Initial
0.0476
0.524
0
Change
-x
+x
+x
Equilibrium
0.0476-x
0.524+x
x
Assume x is small compared to initial concentrations
#$ H + %& #$C3 H 5O2" %& x ( 0.524 )
"5
K a = 1.3 ! 10 =
=
0.0476
[ HC3 H 5O2 ]
x = !" H + #$ = 1.2 % 10 &6 M
(
)
(
)
pH = ! log "# H + $% = ! log 1.2 & 10 !6 = 5.93
Addition of 80.00 mL of 4.00 M HCl
C3H5O2-(aq) + H+ (aq) ! HC3H5O2 (aq)
C2H3O2400.
80.0
Initial (mmol)
Final (mmol)
H+
320.
0
HC2H3O2
200.
520.
HC3H5O2 (aq) ! C3H5O2- (aq) + H+ (aq)
[HC2H3O2]
[C2H3O2-]
[H+]
Initial
0.481
0.0741
0
Change
-x
+x
+x
Equilibrium
0.481-x
0.0741+x
x
Assume x is small compared to initial concentrations
# H + % #C2 H 3O2" %& x ( 0.0741)
K a = 1.3 ! 10 "5 = $ & $
=
0.481
[ HC2 H 3O2 ]
x = !" H + #$ = 8.4 % 10 &5 M
(
)
(
)
pH = ! log "# H + $% = ! log 8.4 & 10 !5 = 4.07
Page 3 of 16
Chem 142
Summer 2011
c. 0.400 M HONH2 (Kb = 1.1 x 10-8) in the presence of 0.400 M
HONH3Cl in a total volume of 1.00 L
% !" HONH 3+ #$ (
% 0.400 (
+8
pOH = pK b + log '
* = + log 1.1 , 10 + log '&
*
0.400 )
& [ HONH 2 ] )
pOH = 7.96 + 0 = 7.96
pH = 14 ! pH = 14 ! 7.96 = 6.04
(
)
#$ HONH 3+ %& #$OH " %& ( 0.400 ) x
K b = 1.1 ! 10 =
=
0.400
[ HONH 2 ]
"8
x = "#OH ! $% = 1.1 & 10 !8 M
1.1 ! 10 "8
%dissoc. =
! 100% = 2.8 ! 10 "6%
0.400
Addition of 50.00 mL of 3.00 M NaOH
HONH3+ (aq) + OH- (aq) ! HONH2 (aq) + H2O (l)
Initial (mmol)
Final (mmol)
HONH3+
400.
250.
OH150.
0
HONH2
400.
550.
HONH2 (aq) + H2O (l) ! HONH3+ (aq) + OH- (aq)
[HONH2]
[HONH3+]
[OH-]
Initial
0.524
0.238
0
Change
-x
+x
+x
Equilibrium
0.524-x
0.238+x
x
Assume x is small compared to initial concentrations
[ HONH ] #$OH
=
%& ( 0.238 ) x
=
0.524
#$ HONH %&
x = "#OH ! $% = 2.4 & 10 !8 M
K b = 1.1 ! 10
"8
"
2
+
3
(
)
(
)
pOH = ! log "#OH ! $% = ! log 2.4 & 10 !8 = 7.62
pH = 14 ! pH = 14 ! 7.62 = 6.38
Page 4 of 16
Chem 142
Summer 2011
Addition of 80.00 mL of 4.00 M HCl
HONH2 (aq) + H+ (aq) ! HONH3+ (aq) + H2O (l)
H+
320.
0
HONH2
400.
80.0
Initial (mmol)
Final (mmol)
HONH3+
400.
720.
HONH2 (aq) + H2O (l) ! HONH3+ (aq) + OH- (aq)
[HONH2]
[HONH3+]
[OH-]
Initial
0.0741
0.667
0
Change
-x
+x
+x
Equilibrium
0.0741-x
0.667+x
x
Assume x is small compared to initial concentrations
[ HONH ] #$OH
%& ( 0.667 ) x
=
0.0741
#$ HONH %&
x = "#OH ! $% = 1.2 & 10 !9 M
K b = 1.1 ! 10 "8 =
(
"
2
+
3
)
(
)
pOH = ! log "#OH ! $% = ! log 1.2 & 10 !9 = 8.91
pH = 14 ! pH = 14 ! 8.91 = 5.09
in order of increasing buffer capacity based on the range of pH
resulting from the added acid or base: b < a < c
c has equal concentrations of base and conjugate acid, so this
should have the best buffer capacity  the calculations make sense!
2. What is the mass of salt that needs to be added to 100 mL of each of the
acid/base solutions to make a buffer of pH 4.50? pH 5.25? pH 6.00?
a. 0.500 M C5H5N (Kb = 1.7 x 10-9); salt is C5H5NHCl
For pH = 4.50
pOH = 14 ! pH = 14 ! 4.50 = 9.50
% !C5 H 5 NH + #$ (
% x (
+9
pOH = 9.50 = pK b + log ' "
* = + log 1.7 , 10 + log '&
*
0.500 )
& [ C5 H 5 N ] )
(
)
x = !"C5 H 5 NH + #$ = 2.69M
100mL !
1L
2.69mol C5 H 5 NH + 1mol C5 H 5 NH + 115.5g C5 H 5 NHCl
!
!
!
= 31.1g C5 H 5 NHCl
1000mL
1.0L
1mol C5 H 5 NHCl 1mol C5 H 5 NHCl
Page 5 of 16
Chem 142
Summer 2011
For pH = 5.25
pOH = 14 ! pH = 14 ! 5.25 = 8.75
% !"C5 H 5 NH + #$ (
% x (
+9
pOH = 8.75 = pK b + log '
* = + log 1.7 , 10 + log '&
*
0.500 )
& [ C5 H 5 N ] )
(
)
x = !"C5 H 5 NH + #$ = 0.478M
100mL !
1L
0.478mol C5 H 5 NH + 1mol C5 H 5 NH + 115.5g C5 H 5 NHCl
!
!
!
= 5.52g C5 H 5 NHCl
1000mL
1.0L
1mol C5 H 5 NHCl 1mol C5 H 5 NHCl
For pH = 6.00
pOH = 14 ! pH = 14 ! 6.00 = 8.00
% !C5 H 5 NH + #$ (
% x (
+9
pOH = 8.00 = pK b + log ' "
* = + log 1.7 , 10 + log '&
*
0.500 )
& [ C5 H 5 N ] )
(
)
x = !"C5 H 5 NH + #$ = 0.0850M
100mL !
1L
0.0850mol C5 H 5 NH + 1mol C5 H 5 NH + 115.5g C5 H 5 NHCl
!
!
!
= 0.982g C5 H 5 NHCl
1000mL
1.0L
1mol C5 H 5 NHCl 1mol C5 H 5 NHCl
b. 0.500 M HC3H5O3 (Ka = 1.38 x 10-4); salt is NaC3H5O3
For pH = 4.50
& "#C3 H 5O3! $% )
& x )
!4
pH = 4.50 = pK a + log (
+ = ! log 1.38 , 10 + log ('
+
0.500 *
' [ HC3 H 5O3 ] *
(
)
x = "#C3 H 5O3! $% = 2.18M
100mL !
1L
2.18mol C3 H 5O3"
1mol C3 H 5O3"
112g NaC3 H 5O3
!
!
!
= 24.4g NaC3 H 5O3
1000mL
1.0L
1mol NaC3 H 5O3 1mol NaC3 H 5O3
For pH = 5.25
& "#C3 H 5O3! $% )
& x )
!4
pH = 5.25 = pK a + log (
+ = ! log 1.38 , 10 + log ('
+
0.500 *
' [ HC3 H 5O3 ] *
(
)
x = "#C3 H 5O3! $% = 12.3M
100mL !
1L
12.3mol C3 H 5O3"
1mol C3 H 5O3"
112g NaC3 H 5O3
!
!
!
= 138g NaC3 H 5O3
1000mL
1.0L
1mol NaC3 H 5O3 1mol NaC3 H 5O3
Page 6 of 16
Chem 142
Summer 2011
For pH = 6.00
& "C3 H 5O3! $% )
& x )
!4
pH = 6.00 = pK a + log ( #
+ = ! log 1.38 , 10 + log ('
+
0.500 *
' [ HC3 H 5O3 ] *
(
)
x = "#C3 H 5O3! $% = 69.0M
100mL !
1L
69.0mol C3 H 5O3"
1mol C3 H 5O3"
112g NaC3 H 5O3
!
!
!
= 773g NaC3 H 5O3
1000mL
1.0L
1mol NaC3 H 5O3 1mol NaC3 H 5O3
c. 0.500 M HCO2H (Ka = 1.8 x 10-4); salt is NaCO2H
For pH = 4.50
& "#CO2 H ! $% )
& x )
!4
pH = 4.50 = pK a + log (
+ = ! log 1.8 , 10 + log ('
+
0.500 *
' [ HCO2 H ] *
(
)
x = "#CO2 H ! $% = 2.85M
100mL !
1L
2.85mol CO2 H "
1mol CO2 H "
68g NaCO2 H
!
!
!
= 19.4g NaCO2 H
1000mL
1.0L
1mol NaCO2 H 1mol NaCO2 H
For pH = 5.25
& "#CO2 H ! $% )
& x )
!4
pH = 5.25 = pK a + log (
+ = ! log 1.8 , 10 + log ('
+
0.500 *
' [ HCO2 H ] *
(
)
x = "#CO2 H ! $% = 16.0M
100mL !
1L
16.0mol CO2 H "
1mol CO2 H "
68g NaCO2 H
!
!
!
= 109g NaCO2 H
1000mL
1.0L
1mol NaCO2 H 1mol NaCO2 H
For pH = 6.00
& "#CO2 H ! $% )
& x )
!4
pH = 6.00 = pK a + log (
+ = ! log 1.8 , 10 + log ('
+
0.500 *
' [ HCO2 H ] *
(
)
x = "#CO2 H ! $% = 90.0M
1L
90.0mol CO2 H "
1mol CO2 H "
68g NaCO2 H
100mL !
!
!
!
= 612g NaCO2 H
1000mL
1.0L
1mol NaCO2 H 1mol NaCO2 H
Page 7 of 16
Chem 142
Summer 2011
3. Calculate the pH after 0.000 mL, 10.00 mL, 30.00 mL and 40.00 mL of
titrant have been added to 50.00 mL of the solution
a. 0.1500 M NaOH titrated with 0.2500 M HCl
at 0.000 mL, only NaOH, a strong base, in solution
"#OH ! $% = 0.1500M
(
)
pOH = ! log "#OH ! $% = ! log ( 0.1500 ) = 0.8239
pH = 14 ! pOH = 14 ! 0.8239 = 13.1761
at 10.00 mL, pH determined by excess OH- in solution
OH- (aq) + H+ (aq) ! H2O (l)
H+
Initial (mmol)
2.500
Final (mmol)
0
OH7.500
5.000
5.000mmol
"#OH ! $% =
= 0.08333M
60.00mL
pOH = ! log "#OH ! $% = ! log ( 0.08333) = 1.0792
pH = 14 ! pOH = 14 ! 1.0792 = 12.9218
(
)
at 30.00 mL, titration is at equivalence point, pH = 7.0000
OH- (aq) + H+ (aq) ! H2O (l)
H+
7.500
0
Initial (mmol)
Final (mmol)
OH7.500
0
at 40.00 mL, pH determined by excess H+ in solution
OH- (aq) + H+ (aq) ! H2O (l)
H+
10.00
2.500
Initial (mmol)
Final (mmol)
OH7.500
0
2.500mmol
= 0.02778M
90.0mL
pH = ! log "# H + $% = ! log ( 0.02778 ) = 1.5563
!" H + #$ =
(
)
Page 8 of 16
Chem 142
Summer 2011
b. 0.1500 M HCOOH titrated with 0.2500 M NaOH
at 0.000 mL, only HCOOH, a weak acid, in solution
# H + % #COOH " %&
x2
K a = 1.8 ! 10 "4 = $ & $
=
0.1500
[ HCOOH ]
x = "#COOH ! $% = "# H + $% = 0.0059M
(
)
pH = ! log "# H + $% = ! log ( 0.0059 ) = 2.28
at 10.00 mL, pH determined by equilibrium of excess HCOOH
in solution
HCOOH (aq) + OH- (aq) ! COOH- (aq) + H2O (l)
Initial (mmol)
Final (mmol)
OH2.500
0
HCOOH
7.500
5.000
COOH0
2.500
HCOOH (aq) ! COOH- (aq) + H+ (aq)
[HCOOH]
[COOH-]
[H+]
Initial
0.08333
0.04167
0
Change
-x
+x
+x
Equilibrium
0.08333-x
0.04167+x
x
Assume x is small compared to initial concentrations of acid and
conjugate base.
& "#COOH ! $% )
& 0.04167 )
!4
pH = pK a + log (
+ = ! log 1.8 , 10 + log ('
+
0.08333 *
' [ HCOOH ] *
pH = 3.74 ! 0.30 = 3.44
(
)
at 30.00 mL, titration is at equivalence point, pH dominated by
conjugate base of the weak acid
HCOOH (aq) + OH- (aq) ! COOH- (aq) + H2O (l)
Initial (mmol)
Final (mmol)
HCOOH
7.500
0
OH7.500
0
COOH0
7.500
Page 9 of 16
Chem 142
Summer 2011
COOH- (aq) + H2O (l) ! HCOOH (aq) + OH- (aq)
[COOH-]
[HCOOH]
[OH-]
Initial
0.09375
0
0
Change
-x
+x
+x
Equilibrium
0.09375-x
x
x
Assume x is small compared to initial concentration of conjugate
base.
#$OH ! %& [ HCOOH ]
K
10 !14
x2
!11
Kb = w =
=
5.56
"
10
=
=
K a 1.8 " 10 !4
0.09375
#$COOH ! %&
x = [ HCOOH ] = "#OH ! $% = 2.3 & 10 !6 M
(
)
(
)
pOH = ! log "#OH ! $% = ! log 2.3 & 10 !6 = 5.64
pH = 14 ! pOH = 14 ! 5.64 = 8.36
at 40.00 mL, pH determined by equilibrium of excess OH- in
solution
HCOOH (aq) + OH- (aq) ! COOH- (aq) + H2O (l)
Initial (mmol)
Final (mmol)
OH10.00
2.500
HCOOH
7.500
0
COOH0
7.500
2.500mmol
"#OH ! $% =
= 0.02778M
90.00mL
pOH = ! log "#OH ! $% = ! log ( 0.02778 ) = 1.5563
(
)
pH = 14 ! pOH = 14 ! 1.5563 = 12.4437
c. 0.1500 M C5H5N titrated with 0.2500 M HCl
at 0.000 mL, only C5H5N, a weak base, in solution
#$OH " %& #$C5 H 5 NH + %&
x2
K b = 1.7 ! 10 =
=
0.1500
[ C5 H 5 N ]
"9
x = !"C5 H 5 NH + #$ = !"OH % #$ = 1.6 & 10 %5 M
(
)
(
)
pOH = ! log "#OH ! $% = ! log 1.6 & 10 !5 = 4.80
pH = 14 ! pOH = 14 ! 4.80 = 9.20
Page 10 of 16
Chem 142
Summer 2011
at 10.00 mL, pH determined by equilibrium of excess C5H5N in
solution
C5H5N (aq) + H+ (aq) ! C5H5NH+ (aq)
Initial (mmol)
Final (mmol)
C5H5N
7.500
5.000
H+
2.500
0
C5H5NH+
0
2.500
C5H5N (aq) +H2O (l) ! C5H5NH+ (aq) + OH- (aq)
[C5H5N]
[C5H5NH+]
[OH-]
Initial
0.08333
0.04167
0
Change
+x
-x
+x
Equilibrium
0.08333+x
0.04167-x
x
Assume x is small compared to initial concentrations of acid and
conjugate base.
% !"C5 H 5 NH + #$ (
% 0.04167 (
+9
pOH = pK b + log '
* = + log 1.7 , 10 + log '&
*
0.08333 )
& [ C5 H 5 N ] )
pOH = 8.77 ! 0.30 = 8.47
pH = 14 ! pOH = 14 ! 8.47 = 5.53
(
)
at 30.00 mL, titration is at equivalence point, pH dominated by
conjugate acid of the weak base
C5H5N (aq) + H+ (aq) ! C5H5NH+ (aq)
Initial (mmol)
Final (mmol)
C5H5N
7.500
0
H+
7.500
0
C5H5NH+
0
7.500
C5H5NH+ (aq) + H2O (l) ! C5H5N (aq) +H3O+ (aq)
[C5H5NH+]
[C5H5N]
[H3O+]
Initial
0.09375
0
0
Change
-x
+x
+x
Equilibrium
0.09375-x
x
x
Assume x is small compared to initial concentration of conjugate
base.
Page 11 of 16
Chem 142
Summer 2011
C5 H 5 N ] #$ H 3O + %&
[
Kw
10 !14
x2
!6
Kb =
=
= 5.9 " 10 =
=
K a 1.7 " 10 !9
0.09375
#$C5 H 5 NH + %&
x = [ C5 H 5 N ] = !" H 3O + #$ = 7.4 % 10 &4 M
(
)
(
)
pH = ! log "# H 3O + $% = ! log 7.4 & 10 !4 = 3.13
at 40.00 mL, pH determined by equilibrium of excess H+ in
solution
C5H5N (aq) + H+ (aq) ! C5H5NH+ (aq)
H+
10.00
2.500
C5H5N
7.500
0
Initial (mmol)
Final (mmol)
C5H5NH+
0
7.500
2.500mmol
!" H + #$ =
= 0.02778M
90.00mL
pH = ! log "# H + $% = ! log ( 0.02778 ) = 1.5563
(
)
4. Silver sulfide has a solubility of 3.4 x 10-17 M at 25 °C. What is the Ksp for
silver sulfide?
Ag2S (s) ! 2Ag+ (aq) + S2- (aq)
3.4 & 10 !17 mol Ag2 S 1mol S 2 !
2!
"# S $% =
&
= 3.4 & 10 !17 mol S 2 !
1L
1mol Ag2 S
3.4 % 10 &17 mol Ag2 S 2mol Ag +
!" Ag #$ =
%
= 6.8 % 10 &17 mol Ag +
1L
1mol Ag2 S
+
2
(
K sp = !" Ag + #$ !" S 2 % #$ = 6.8 & 10 %17 M
) ( 3.4 & 10
2
%17
)
M = 1.6 & 10 %49 M 3
Page 12 of 16
Chem 142
Summer 2011
5. What is the solubility of the following salt solutions in moles per liter and
grams per liter?
a. nickel (II) carbonate (Ksp=1.4 x 10-7)
NiCO3 (s) ! Ni2+ (aq) + CO32- (aq)
x = [ NiCO3 ]
K sp = 1.4 ! 10 "7 = #$ Ni 2 + %& #$CO32 " %& = x 2
x = [ NiCO3 ] = 3.7 ! 10 "4 M
3.7 ! 10 "4 mol NiCO3 118.7g NiCO3
!
= 0.44 g L NiCO3
1L
1mol NiCO3
b. barium phosphate (Ksp=6 x 10-39)
Ba3(PO4)2 (s) ! 3Ba2+ (aq) + 2PO43- (aq)
x = !" Ba3 ( PO4 )2 #$
K sp = 6 ! 10 "39 = #$ Ba 2 + %& #$ PO43" %& = ( 3x ) ( 2x ) = 27x 3 ' 4x 2 = 108x 5
x = !" Ba3 ( PO4 )2 #$ = 9 % 10 &9 M
3
9 ! 10 "9 mol Ba3 ( PO4 )2
1L
!
2
3
601.8g Ba3 ( PO4 )2
1mol Ba3 ( PO4 )2
2
= 5 ! 10 "6 g L Ba3 ( PO4 )2
c. lead (II) bromide (Ksp=4.6 x 10-6)
PbBr2 (s) ! Pb2+ (aq) + 2Br- (aq)
x = [ PbBr2 ]
K sp = 4.6 ! 10 "6 = #$ Pb 2 + %& #$ Br " %& = x ( 2x ) = x ' 4x 2 = 4x 3
x = [ PbBr2 ] = 1.0 ! 10 "2 M
2
2
1.0 ! 10 "2 mol PbBr2 367g PbBr2
!
= 3.67 g L PbBr2
1L
1mol PbBr2
Page 13 of 16
Chem 142
Summer 2011
6. Which of the following compounds is the most soluble?
AgCl
Ksp= 1.5 x 10-10
Ag2CrO4
Ksp= 6 x 10-39
Ag3PO4
Ksp=1.8 x 10-18
AgCl (s) ! Ag+ (aq) + Cl- (aq)
x = [ AgCl ]
K sp = 1.5 ! 10 "10 = #$ Ag + %& #$Cl " %& = x 2
x = [ AgCl ] = 1.2 ! 10 "5 M
Ag2CrO4 (s) ! 2Ag+ (aq) + CrO42- (aq)
x = [ Ag2CrO4 ]
MOST SOLUBLE
K sp = 6 ! 10 "39 = #$ Ag + %& #$CrO42 " %& = ( 2x ) x = 4x 3
x = [ Ag2CrO4 ] = 1.1 ! 10 "13 M
2
2
Ag3PO4 (s) ! 3Ag+ (aq) + PO43- (aq)
x = [ Ag3 PO4 ]
K sp = 1.8 ! 10 "18 = #$ Ag + %& #$ PO43" %& = ( 3x ) x = 27x 4
x = [ Ag3 PO4 ] = 1.6 ! 10 "5 M
3
3
7. Calculate the solubility of SrF2 (Ksp=1.9 x 10-10) in
a. Pure water
SrF2 (s) ! Sr2+ (aq) + 2F- (aq)
x = [ SrF2 ]
K sp = 1.9 ! 10 "10 = #$ Sr 2 + %& #$ F " %& = x ( 2x ) = 4x 3
x = [ SrF2 ] = 3.6 ! 10 "4 M
2
2
b. 0.100 M Sr(NO3)2
K sp = 1.9 ! 10 "10 = #$ Sr 2 + %& #$ F " %& = ( 0.100 + x ) ( 2x ) = ( 0.100 ) 4x 2
x = [ SrF2 ] = 2.2 ! 10 "5 M
The solubility is decreased by the addition of strontium ions
2
2
c. 0.400 M NaF
K sp = 1.9 ! 10 "10 = #$ Sr 2 + %& #$ F " %& = x ( 0.400 + 2x ) = x ( 0.400 )
x = [ SrF2 ] = 1.0 ! 10 "9 M
The solubility is decreased by the addition of fluoride ions
2
2
2
Page 14 of 16
Chem 142
Summer 2011
8. Calculate the equilibrium concentrations of each ion in solution
obtained by mixing the following solutions
a. 200.0 mL of 1.3 x 10-3 M AgNO3 and 100.0 mL of 4.5 x 10-5 M
Na2S. (Ksp for Ag2S = 1.6 x 10-49)
First, verify that precipitation occurs:
200.0mL
!" Ag + #$ 0 = 1.3 % 10 &3 M %
= 8.7 % 10 &4 M
( 200.0mL + 100.0mL )
100.0mL
"# S 2 ! $% 0 = 4.5 & 10 !5 M &
= 1.5 & 10 !5 M
( 200.0mL + 100.0mL )
+
2Ag (aq) + S2- (aq) ! Ag2S (s)
2
(
) ( 4.5 & 10
Q = !" Ag + #$ !" S 2 % #$ = 8.7 & 10 %4 M
Q > Ksp so precipitation occurs.
2
%5
)
M = 1.1 & 10 %11 M 3
Assume stoichiometric reaction between Ag+ and S2- to form
Ag2S
Ag+
S2Initial (mmol)
0.26
0.0045
Final (mmol)
0.2555
≈0
Initial
Change
Equilibrium
[Ag+]
8.5 x 10-4
+2x
8.5 x 10-4 + 2x
[S2-]
0
+x
x
Assume x is small compared to initial concentration of F2
(
K sp = 1.6 ! 10 "49 = #$ Ag + %& #$ S 2 " %& = 8.5 ! 10 "4
x = "# S 2 ! $% = 2.2 & 10 !43 M
)
2
x
Page 15 of 16
Chem 142
Summer 2011
b. 50.0 mL of 6.0 x 10-3 CaCl2 and 30.0 mL of 0.040 M NaF. (Ksp for
CaF2 = 4.0 x 10-11)
First, verify that precipitation occurs:
50.0mL
!"Ca 2 + #$ 0 = 6.0 % 10 &3 M %
= 3.75 % 10 &3 M
( 50.0mL + 30.0mL )
30.0mL
"# F ! $% 0 = 0.040M &
= 0.015M
( 50.0mL + 30.0mL )
Ca2+ (aq) + 2F- (aq) ! CaF2 (s)
(
)
Q = !"Ca 2 + #$ !" F % #$ = 3.75 & 10 %3 M ( 0.015M ) = 8.4 & 10 %7 M 3
2
2
Q > Ksp so precipitation occurs.
Assume stoichiometric reaction between Ca2+ and F- to form
CaF2
Ca2+
FInitial (mmol)
0.30
1.2
Final (mmol)
≈0
0.60
Initial
Change
Equilibrium
[Ca2+]
0
+x
x
[F-]
7.5 x 10-3
+2x
7.5 x 10-3 + 2x
Assume x is small compared to initial concentration of F2
(
K sp = 4.0 ! 10 "11 = #$Ca 2 + %& #$ F " %& = x 7.5 ! 10 "3
)
2
x = !"Ca 2 + #$ = 7.1 % 10 &7 M
Page 16 of 16