Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Questions on Henry's Law, Solubility Product, Dissociation Constant | CHEM 1120, Assignments of Chemistry

Material Type: Assignment; Class: General Chemistry II; Subject: Chemistry; University: Roane State Community College; Term: Unknown 1989;

Typology: Assignments

Pre 2010

Uploaded on 08/18/2009

koofers-user-oqz
koofers-user-oqz 🇺🇸

10 documents

1 / 8

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Solubility and Dissociation Equilibria Problems
Solution Equilibrium - Part I
Henry's Law, Solubility Product and Dissociation Constant
ANSWERS
Henry's Law
1. Calculate the amount of H2CO3 which is dissolved in water at 20EC if the pressure of the CO2
over the solution is 0.050 atm. The KH for CO2 in water is 79.
[]
2
2
CO
H
P
KCO
=
[]
2
0.050
79 CO
=
Ans: 6.3 × 10-4
2. What is the KH for H2S dissolving in water if 0.10 M H2S is present at equilibrium when
0.33 atm of H2S gas is in contact with the solution?
[]
2
2
HS
H
P
KHS
=0.33
0.10
H
K=
Ans: 3.3
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 O2 is 20% of the gases). The KH for O2 in water
is 2.15 x 103.
[]
2
2
O
H
P
KO
=
[]
3
2
0.20
2.15 10 O
×=
Ans: 9.3 × 10-5
Solubility and Dissociation Equilibria Problems
4. Calculate the calcium ion concentration for a solution in contact with CaF2 if the fluoride
concentration is 0.010 M and the Ksp for CaF2 is 3.0 x 10!11.
2
2
sp
KCaF
+−
⎡⎤
=⎣⎦
[
]
2
11 2
3.0 10 0.010Ca
−+
⎡⎤
×=
⎣⎦
Ans: 3.0 × 10-7
pf3
pf4
pf5
pf8

Partial preview of the text

Download Questions on Henry's Law, Solubility Product, Dissociation Constant | CHEM 1120 and more Assignments Chemistry in PDF only on Docsity!

Solubility and Dissociation Equilibria Problems

Solution Equilibrium - Part I

Henry's Law, Solubility Product and Dissociation Constant

ANSWERS

Henry's Law

  1. 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

  1. 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.

  1. 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

  1. 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

  1. 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

  1. 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 2x

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× -

  1. 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

  1. 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

  • 2OH

! 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.

  1. 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

!

  • Cl

!

. (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

  1. 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!

  • S

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

  • ] = 5.0 × 10 -

Solubility and Dissociation Equilibria Problems

Solution Equilibrium - Part I

Henry's Law, Solubility Product and Dissociation Constant

ANSWERS

Henry's Law

  1. 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

  1. 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.

  1. 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

  1. 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

  1. 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.

  1. 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 ×

  1. 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

  1. 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

  • 4NH 3

(aq) º Zn(NH

3 ) 4

2+

  • 2OH

!

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

  1. 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

  • 2OH

! 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.

  1. 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

!

  • Cl

!

. (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

  1. 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!

  • S

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

  • ] = 5.0 × 10 -