Question

Calculate the solubility in water in terms of mole fraction if partial pressure of $ C{{O}_{2}}$ is $ 2\times {{10}^{-3}}$ bar at 298K temperature, the $ {{K}_{a}}$ value for $ C{{O}_{2}}$ is $ 6.02\times {{10}^{-4}}$ bar.
A)$ 3.322\times {{10}^{-3}}$
B) $ 3.011\times {{10}^{-3}}$
C) $ 3.322\times {{10}^{-4}}$
D) $ 3.011\times {{10}^{-6}}$

Answer 1

Hint The answer to this question is obtained by applying the Henry’s law which states that the total amount of gas that is dissolved in a liquid is directly proportional to the partial pressure of that gas which is given by, $ P={{K}_{H}}\chi $

Complete step – by – step answer:
In the classes of physical chemists, we have studied the concepts of various laws associated with the gases which come under the chapter called thermodynamics and some related topics.
We shall now see the calculation of solubility of $ C{{O}_{2}}$ in water in terms of mole fraction.
- Henry’s law states that the total amount of gas that is dissolved in a liquid is directly proportional to the partial pressure of that gas.
- Henry’s law can be written in various forms and thus in terms of mole fraction we can write the Henry’s law in the form of equation as,
$ P={{K}_{H}}\chi $
where, P is the partial pressure of the gas
$ {{K}_{H}}$ is the Henry’s constant and in the above case it is $ {{K}_{a}}$
$ \chi $ is the mole fraction of the gas which is to be found.
Now, according to the data given, we have partial pressure of carbon dioxide given as P =$ 2\times {{10}^{-3}}$ bar
Also, $ {{K}_{a}}$ for carbon dioxide is$ 6.02\times {{10}^{-4}}$ bar.
Now, substituting these values in the above formula, we get
$ 2\times {{10}^{-3}}=6.02\times {{10}^{-4}}\chi $
Now, solving the above equation, we get
\[\chi =\dfrac{2\times {{10}^{-3}}bar}{6.02\times {{10}^{-4}}bar}=0.3322\times {{10}^{1}}\]
\[\Rightarrow \chi =3.322\]
Thus, mole fraction of carbon dioxide dissolved is 3.322 moles.
Solubility in moles per litre will be in 1 litre of solution that is number of moles obtained divided by 1 litre and that is
\[\dfrac{3.322}{1litre}mol/L=3.322mol/L\]
Now, solubility in moles per milliliter is obtained when we divide the above obtained answer by thousand and this will be,
\[\dfrac{3.322}{1000}mol/mL\]
Hence, the final answer would be,
\[\Rightarrow S=3.322\times {{10}^{-3}}mol/mL\]

Therefore, the correct answer is option A) in terms of milliliters.

Note: Note that the ratio of number of moles of the solute to that of one litre of the solution will be nothing but the molarity of the solution which is denoted as “M” and this can also be written in terms of milliliters where this doesn’t become molarity.