Question

Henry's constant for oxygen dissolved in water is $4.34 \times {10^4}\;{\rm{atm}}$ at ${\rm{25^\circ C}}$. If the partial pressure of oxygen in air is ${\rm{0}}{\rm{.2}}\;{\rm{atm}}$, under atmospheric conditions, calculate the concentration (in moles per litre) of dissolved oxygen in water in equilibrium with air at ${\rm{25^\circ C}}$.

Answer 1

We know that Henry's law is used to determine the unknown pressure quantity and unknown mole fraction at respective temperatures.

Complete step by step answer:
Henry's constant for oxygen dissolved in water is $4.34 \times {10^4}\;{\rm{atm}}$.
The temperature is ${\rm{25^\circ C}}$.
The partial pressure of oxygen in air is ${\rm{0}}{\rm{.2}}\;{\rm{atm}}$.
The henry law can be defined by with the help of vapour pressure and number of mole fraction of gas in respective solutions. The mathematical form of henry law is depicted below.
${\rm{P}} = {{\rm{K}}_{\rm{H}}} \times {\rm{X}}$
Where, ${{\rm{K}}_{\rm{H}}}$ is the henry constant, P is the vapour pressure, and X is the mole fraction.
The mole fraction of oxygen can be calculated with the help of henry haw.
${{\rm{X}}_{{{\rm{o}}_{\rm{2}}}}} = \dfrac{{\rm{P}}}{{{{\rm{K}}_{\rm{H}}}}}$
Substitute the value of henry constant and pressure in the above equation.
$\begin{array}{c} {{\rm{X}}_{{{\rm{o}}_{\rm{2}}}}} = \dfrac{{0.2\;{\rm{atm}}}}{{4.34 \times {{10}^4}\;{\rm{atm}}}}\\\ = 4.6 \times {10^{ - 6}} \end{array}$
That means $4.6 \times {10^{ - 6}}$ moles of oxygen gas present in one mole of water.
And $4.6 \times {10^{ - 6}}$ moles oxygen in ${\rm{18}}\;{\rm{g}}$ of water.
$\left( {\dfrac{{18}}{{18}} = 1\;{\rm{mole}}} \right)\left( {{\rm{18}}\;{\rm{g/mL}}} \right)$
The density of water is ${\rm{1}}\;{\rm{g/mL}}$.
So, it means ${\rm{18}}\;{\rm{g}} = {\rm{18}}\;{\rm{mL}}$
The conversion of milliliters to liters is done as follows.
$\begin{array}{c} {\rm{1}}\;{\rm{mL}} = 0.001\;{\rm{L}}\\\ {\rm{18}}\;{\rm{mL}} = 18 \times 0.001\;{\rm{L}}\\\ = 18 \times {10^{ - 3}}\;{\rm{L}} \end{array}$
Now, calculate the concentration by using the formula given below.
${\rm{Concentration}} = \dfrac{{{\rm{Moles}}}}{{{\rm{Volume}}}}$
Substitute the respective values of moles and volume in the above equation. We get,

{\rm{Concentration}} = \dfrac{{4.6 \times {{10}^{ - 6}}\;{\rm{mol}}}}{{18 \times {{10}^{ - 3}}\;{\rm{L}}}}\\\ = 2.55 \times {10^{ - 4}}\;{\rm{M}} \end{array}$$ Therefore, the concentration (in moles per litre) of dissolved oxygen in water in equilibrium with air at ${\rm{25^\circ C}}$ is $$2.55 \times {10^{ - 4}}\;{\rm{M}}$$. **Note:** The concentration or molarity of the aqueous solution usually be affected by the increasing or decreasing of temperature. The unit representation of the molarity is mol/L.