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Solubility and Dissociation Equilibria Problems
Solution Equilibrium - Part I
Henry's Law, Solubility Product and Dissociation Constant
ANSWERS
Henry's Law
- Calculate the amount of H 2 CO 3 which is dissolved in water at 20EC if the pressure of the CO 2
over the solution is 0.050 atm. The KH for CO 2 in water is 79.
[ ]
2
2
CO
H
P
K
CO
[ 2 ]
CO
Ans: 6.3 × 10
- What is the K H for H 2 S dissolving in water if 0.10 M H 2 S is present at equilibrium when
0.33 atm of H 2 S gas is in contact with the solution?
[ ]
2
2
H S
H
P
K
H S
H
K =
Ans: 3.
- What is the expected oxygen content, in molarity, of water that one would find in a pure
stream at sea level (1 atm total pressure and O 2 is 20% of the gases). The K H for O 2 in water
is 2.15 x 10
3 .
[ ]
2
2
O
H
P
K
O
[ ]
3
2
O
× =
Ans: 9.3 × 10
Solubility and Dissociation Equilibria Problems
- Calculate the calcium ion concentration for a solution in contact with CaF 2 if the fluoride
concentration is 0.010 M and the K sp for CaF 2 is 3.0 x 10
! 11 .
2 2
sp K Ca F
[ ]
11 2 2 3.0 10 Ca 0.
− + × = ⎡^ ⎤ ⎣ ⎦
Ans: 3.0 × 10
Solubility and Dissociation Equilibria Problems
- What is the molar solubility of CaF 2 ? The K sp for CaF 2 is 3.0 x 10
! 11 .
mf CaF 2 [Ca
2+ ] [F
! ]
before equilibrium 1 0 0
at equilibrium 1 1x 2x
2 2
sp
K Ca F
[ ][ ]
11 2 3.0 10 x 2 x
− × =
11 3 3.0 10 4 x
− × =
Ans: 3.9 × 10
- What is the resultant concentration of both Ca
2+ and F
! ions if solid CaF 2 is added to a
solution which is originally 7.5 x 10
! 4 M in Ca(NO 3 ) 2 ? The K sp for CaF 2 is 3.0 x 10
! 11 .
[Ca
2+ ] [F
! ]
before equilibrium 7.5 x 10
! 4
at equilibrium 7.5 x 10
! 4
x = 9.4 ×
[ ]
11 4 2 3.0 10 7.5 10 x 2 x
− − × = ⎡^ × + ⎤ ⎣ ⎦
Ans: [Ca
2+ ] = 8.4 × 10
Ans: [F
- ] = 1.8× -
- To a 0.10 M solution of Ca(NO 3 ) 2 is added enough Na 2 CO 3 to make the solution 0.10 M in
CO 3
2! ion. What is the resulting Ca
2+ concentration? The K sp for CaCO 3 is 4.8 x 10
! 9 .
Type I since [CO 3 ] = 0.10 when finished.
2 2
sp 3 K Ca CO
9 2 4.8 10 Ca 0.
− + × = ⎡ ⎤ ⎣ ⎦
Ans: 4.8 × 10
Introduction to Solution Equilibria -
Henry's Law, Solubility Product and Dissociation Constant
- Calculate the concentration of the OH
! ion at which 0.10 mol of the precipitate Zn(OH) 2
will dissolve into 1.0 L of pure water. The K sp for Zn(OH) 2 is 4.5 x 10
! 17 and K d for
Zn(OH) 4
2! is 3.5 x 10
! 16 .
The reaction: Zn(OH) 2
! º Zn(OH) 4
2!
Sub:
2
4
2
sp
RXN
d
K Zn OH
K
K OH
−
17
16 2
OH
−
− −
×
×
ANS [OH
! ] = 0.
- What concentration of cyanide ion, CN
! , is needed to dissolve AuCl solid into a water
solution to yield a solution which is 1.0 M in Au(CN) (^2)
! ions. The Ksp AuCl(s) is 2.0 x 10
! 13
and the Kd for the Au(CN) (^2)
! ion is 5.0 x 10
! 39
. The relevant reaction is:
AuCl(s) + 2CN
!
º Au(CN)
2
!
!
. (Notice that as much Cl
! goes into solution as does
gold in the form Au(CN) (^2)
! .)
Sub:
2
2
sp
RXN
d
K Au CN^ Cl
K
K CN
− −
−
13 25
39 2
2.0 10 1.0^ 1.
CN
−
− −
×
= × =
×
ANS [CN
! ] = 1.6 × 10
- The most insoluble form of nickel sulfide is the γ (gamma) form. The K sp for the NiS(γ) is
2.0 x 10
! 28
. What concentration of CN
! ion would be required to dissolve this sulfide to
the extent of obtaining 0.10 M Ni(CN) 4
2! ions in solution? The K d for Ni(CN) 4
2! is
1.0 x 10
! 31 .
NiS(γ) + 4 CN
! º Ni(CN) 4
2!
2-
Sub:
2 2
4
4
sp
RNX
d
K Ni CN^ S
K
K CN
− −
−
28 3
31 4
2.0 10 0.10^ 0.
CN
−
− −
×
= × =
×
ANS [CN
Solubility and Dissociation Equilibria Problems
Solution Equilibrium - Part I
Henry's Law, Solubility Product and Dissociation Constant
ANSWERS
Henry's Law
- Calculate the amount of H 2 CO 3 which is dissolved in water at 20EC if the pressure of the CO 2
over the solution is 0.050 atm. The KH for CO 2 in water is 79.
[ ]
2
2
CO
H
P
K
CO
[ 2 ]
CO
Ans: 6.3 × 10
- What is the K H for H 2 S dissolving in water if 0.10 M H 2 S is present at equilibrium when
0.33 atm of H 2 S gas is in contact with the solution?
[ ]
2
2
H S
H
P
K
H S
H
K =
Ans: 3.
- What is the expected oxygen content, in molarity, of water that one would find in a pure
stream at sea level (1 atm total pressure and O 2 is 20% of the gases). The K H for O 2 in water
is 2.15 x 10
3 .
[ ]
2
2
O
H
P
K
O
[ ]
3
2
O
× =
Ans: 9.3 × 10
Solubility and Dissociation Equilibria Problems
- Calculate the calcium ion concentration for a solution in contact with CaF 2 if the fluoride
concentration is 0.010 M and the K sp for CaF 2 is 3.0 x 10
! 11 .
2 2
sp K Ca F
[ ]
11 2 2 3.0 10 Ca 0.
− + × = ⎡^ ⎤ ⎣ ⎦
Ans: 3.0 × 10
Introduction to Solution Equilibria -
Henry's Law, Solubility Product and Dissociation Constant
- The K d for CdCl 4
2! is 1.0 x 10
! 4
. What is the concentration of Cd
2+ in a solution of chloride
which has 0.010 M of the CdCl 4
2! ion present. The Cl
! concentration in this solution was
measured and found to be 0.10 M.
4 2
2
4
d
Cd Cl
K
CdCl
−
4 2
4
Cd
−
× =
Ans: [Cd
2+ ] = 0.
- A solution is 0.010 M in total silver and 0.010 M in NH 3. What is the Ag
ion concentration?
The Kd for Ag(NH 3 ) (^2)
is 6.3 x 10
! 8
. (Note - the total silver concentration is [Ag
+ ] +
[Ag(NH 3 ) 2
+ ] .)
Let x = [Ag
]
[ ]
2
3
3 2
d
Ag NH
K
Ag NH
2 9
x
x
− × =
Ans: x = 6.3 ×
- At what [OH
! ] will a 0.10 M solution of Zn(NO 3 ) 2 develop a precipitate of Zn(OH) 2 (s)? The
K sp for Zn(OH) 2 is 4.5 x 10
! 17 .
2 2
sp
K Zn OH
2 17 4.5 10 (0.10) OH
− − × = ⎡^ ⎤ ⎣ ⎦
Ans: [OH- ] = 2.1 × 10
- Calculate the [NH 3 ] required to redissolve 0.10 mol of Zn(OH) 2 in 1.0 L of pure water if
the concentration of OH
! is kept constant at 1.8 x 10
! 5 M. The Ksp for Zn(OH) 2 is
4.5 x 10
! 17 and the Kd for Zn(NH 3 ) (^4)
2+ is 3.4 x 10
! 10 .
The relevant reaction is: Zn(OH) 2
(aq) º Zn(NH
3 ) 4
2+
!
Sub:
[ ]
2 2
3 4
4
3
sp
RXN
d
K Zn NH^ OH
K
K (^) NH
[ ]
2 5
7
4
3
NH
−
−
×
× =
ANS [NH 3 ] = 0.
Introduction to Solution Equilibria -
Henry's Law, Solubility Product and Dissociation Constant
- Calculate the concentration of the OH
! ion at which 0.10 mol of the precipitate Zn(OH) 2
will dissolve into 1.0 L of pure water. The K sp for Zn(OH) 2 is 4.5 x 10
! 17 and K d for
Zn(OH) 4
2! is 3.5 x 10
! 16 .
The reaction: Zn(OH) 2
! º Zn(OH) 4
2!
Sub:
2
4
2
sp
RXN
d
K Zn OH
K
K OH
−
17
16 2
OH
−
− −
×
×
ANS [OH
! ] = 0.
- What concentration of cyanide ion, CN
! , is needed to dissolve AuCl solid into a water
solution to yield a solution which is 1.0 M in Au(CN) (^2)
! ions. The Ksp AuCl(s) is 2.0 x 10
! 13
and the Kd for the Au(CN) (^2)
! ion is 5.0 x 10
! 39
. The relevant reaction is:
AuCl(s) + 2CN
!
º Au(CN)
2
!
!
. (Notice that as much Cl
! goes into solution as does
gold in the form Au(CN) (^2)
! .)
Sub:
2
2
sp
RXN
d
K Au CN^ Cl
K
K CN
− −
−
13 25
39 2
2.0 10 1.0^ 1.
CN
−
− −
×
= × =
×
ANS [CN
! ] = 1.6 × 10
- The most insoluble form of nickel sulfide is the γ (gamma) form. The K sp for the NiS(γ) is
2.0 x 10
! 28
. What concentration of CN
! ion would be required to dissolve this sulfide to
the extent of obtaining 0.10 M Ni(CN) 4
2! ions in solution? The K d for Ni(CN) 4
2! is
1.0 x 10
! 31 .
NiS(γ) + 4 CN
! º Ni(CN) 4
2!
2-
Sub:
2 2
4
4
sp
RNX
d
K Ni CN^ S
K
K CN
− −
−
28 3
31 4
2.0 10 0.10^ 0.
CN
−
− −
×
= × =
×
ANS [CN
