
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
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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.
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