
At what partial pressure, oxygen will have solubility of $2.6 \times {10^{ - 4}}$ $mol{\text{ d}}{{\text{m}}^{ - 3}}$ in water of ${25^ \circ }C$ ? Henry constant for ${O_2}$ in water at ${25^ \circ }C$ is $1.3 \times {10^{ - 3}}$ $mol{\text{ d}}{{\text{m}}^{ - 3}}{\text{ at}}{{\text{m}}^{ - 1}}$
Answer
488.1k+ views
Hint: Henry’s law constant is a proportionality constant for the relationship between the amount of gas that is dissolved in a liquid and the partial pressure of the gas above the liquid when the temperature is kept constant. Moreover, according to Henry’s law at constant temperature, the amount of gas dissolved in a liquid is directly proportional to the partial pressure of the gas above the liquid. Henry’s law constant is inversely proportional to the solubility of the gas in solution.
Complete answer:
Step 1: Identifying the values given in the question:
So, according to the question, Solubility of the oxygen at ${25^ \circ }C$ is \[2.6 \times {10^{ - 4}}mol{\text{ }}d{m^{ - 3}}\] and Henry’s constant for oxygen at ${25^ \circ }C$ is \[1.3 \times {10^{ - 3}}\]\[mol{\text{ }}d{m^{ - 3}}at{m^{ - 1}}.\]
Step 2: Substituting the values in the mathematical formula of the Henry’s law:
Mathematically, Henry’s law is given by the equation:
$p = {K_H}.S$
Where p is the partial pressure of the gas in the atmosphere
S is the solubility of the dissolved gas
${K_H}$ is the Henry’s constant
Now, $p = 2.6 \times {10^{ - 4}} \times 1.3 \times {10^{ - 3}}$
$p = 3.38 \times {10^{ - 7}}atm$
Therefore, at $3.38 \times {10^{ - 7}}$ atm partial pressure, oxygen will have a solubility $2.6 \times {10^{ - 4}}mol{\text{ d}}{{\text{m}}^{ - 3}}$ in water at ${25^ \circ }C$.
Note:
The nature of gas, nature of solvent, temperature, and pressure are the factors on which the value of Henry’s law constant is dependent. According to Henry's law as there is an increase in temperature the value of Henry’s law constantly increases and the solubility of the gas in liquid decreases. The solubility of carbon dioxide in an unopened carbonated drink is the result of Henry’s law.
Complete answer:
Step 1: Identifying the values given in the question:
So, according to the question, Solubility of the oxygen at ${25^ \circ }C$ is \[2.6 \times {10^{ - 4}}mol{\text{ }}d{m^{ - 3}}\] and Henry’s constant for oxygen at ${25^ \circ }C$ is \[1.3 \times {10^{ - 3}}\]\[mol{\text{ }}d{m^{ - 3}}at{m^{ - 1}}.\]
Step 2: Substituting the values in the mathematical formula of the Henry’s law:
Mathematically, Henry’s law is given by the equation:
$p = {K_H}.S$
Where p is the partial pressure of the gas in the atmosphere
S is the solubility of the dissolved gas
${K_H}$ is the Henry’s constant
Now, $p = 2.6 \times {10^{ - 4}} \times 1.3 \times {10^{ - 3}}$
$p = 3.38 \times {10^{ - 7}}atm$
Therefore, at $3.38 \times {10^{ - 7}}$ atm partial pressure, oxygen will have a solubility $2.6 \times {10^{ - 4}}mol{\text{ d}}{{\text{m}}^{ - 3}}$ in water at ${25^ \circ }C$.
Note:
The nature of gas, nature of solvent, temperature, and pressure are the factors on which the value of Henry’s law constant is dependent. According to Henry's law as there is an increase in temperature the value of Henry’s law constantly increases and the solubility of the gas in liquid decreases. The solubility of carbon dioxide in an unopened carbonated drink is the result of Henry’s law.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

