The following table provides some information on carbon dioxide solubility in wa

Question

The following table provides some information on carbon dioxide solubility in water (mol/L) (mol L 1 atm 1) (atm) 11.00 3.80x10-2 7.20x10-2 1.00 3.40x10-2 Part A What is the Henry's law constant for CO2 at 20 C? Express your answer to three significant figures and include the appropriate units. Value Units Submit Hints My Answers Give Up Review Part Part B What pressure is required to achieve a CO2 concentration of 7.20x10 2 Mat 20 C? Express your answer to three significant figures and include the appropriate units. TT 20.0 20.0 25.0

Explanation / Answer

Henry’s Law Constant

The amount of dissolved gas is proportional to its partial pressure in the gas phase. The proportionality factor is called the Henry's law constant.

Part A

S=kP where S is the solubility in

where

S is the solubility of a gas at a fixed temperature in a particular solvent (in units of M or mL gas/L)

k is Henry's law constant (often in units of M/atm)

P is the partial pressure of the gas (often in units of Atm)

Now we can rearrange our equation from above to solve for the constant:

k=S/P= 3.80x10-2/1=3.80x10-2mol.L-1 atm-1

Part B

S=kP or P=S/k

S=7.20x10-2 mol.L-1

K=3.80x10-2mol.L-1 atm-1

Substituting the values we have

P=7.20x10-2 mol.L-1/ 3.80x10-2mol.L-1 atm-1=1.8947atm

Part C

Solubility at 25°C

S=kP

K at 25°C =3.40x10-2 mol.L-1

P=1 atm

S=kP=3.40x10-2 mol.L-1 x 1atm=3.40x10-2 mol.L-1

Completing the table we have

S mol/L

P (atm)

K (mol. 3.40x10-2 mol.L-1.atm-1

T °C

3.80x10-2

1

3.80x10-2

20

7.20x10-2

1.8947

3.80x10-2(as calculated before at 20°C)

20

3.40x10-2

1

3.40x10-2

25

Henry’s Law Constant

The amount of dissolved gas is proportional to its partial pressure in the gas phase. The proportionality factor is called the Henry's law constant.

Part A

S=kP where S is the solubility in

where

S is the solubility of a gas at a fixed temperature in a particular solvent (in units of M or mL gas/L)

k is Henry's law constant (often in units of M/atm)

P is the partial pressure of the gas (often in units of Atm)

Now we can rearrange our equation from above to solve for the constant:

k=S/P= 3.80x10-2/1=3.80x10-2mol.L-1 atm-1

Part B

S=kP or P=S/k

S=7.20x10-2 mol.L-1

K=3.80x10-2mol.L-1 atm-1

Substituting the values we have

P=7.20x10-2 mol.L-1/ 3.80x10-2mol.L-1 atm-1=1.8947atm

Part C

Solubility at 25°C

S=kP

K at 25°C =3.40x10-2 mol.L-1

P=1 atm

S=kP=3.40x10-2 mol.L-1 x 1atm=3.40x10-2 mol.L-1

Completing the table we have

S mol/L

P (atm)

K (mol. 3.40x10-2 mol.L-1.atm-1

T °C

3.80x10-2

1

3.80x10-2

20

7.20x10-2

1.8947

3.80x10-2(as calculated before at 20°C)

20

3.40x10-2

1

3.40x10-2

25

S mol/L

P (atm)

K (mol. 3.40x10-2 mol.L-1.atm-1

T °C

3.80x10-2

1

3.80x10-2

20

7.20x10-2

1.8947

3.80x10-2(as calculated before at 20°C)

20

3.40x10-2

1

3.40x10-2

25

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