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Question: For a gas, the Henry’s law constant is \(1.25\times {{10}^{-3}}mol.d{{m}^{-3}}at{{m}^{-1}}\) at \(25...

For a gas, the Henry’s law constant is 1.25×10−3mol.dm−3atm−11.25\times {{10}^{-3}}mol.d{{m}^{-3}}at{{m}^{-1}} at 25∘C25{}^\circ C. Calculate the solubility (In terms of molarity) of the given gas at 2.5 atm2.5\,atm and 25∘C25{}^\circ C.

Explanation

Solution

Molarity is the parameter to express concentration of the solution and is defined as the amounts of moles of solute that are dissolved in per liter of solution.

Formula used: Henry’s law formula-
C=KHPC={{K}_{H}}P

Complete step by step solution:
For a gas, given in this question is, Henry’s law constant =1.25×10−3mol.dm−3atm−1=1.25\times {{10}^{-3}}mol.d{{m}^{-3}}at{{m}^{-1}}
Temperature =25∘C=25{}^\circ C
Pressure =2.5 atm=2.5\,atm
According to Henry’s law, the amount of a gas dissolved in a liquid is directly proportional to its partial pressure.
So, solubility of the gas will be obtained by getting its concentration through Henry’s formula
C=KHPC={{K}_{H}}P
Where,
Henry’s constant KH=1.25×10−3mol.dm−3atm−1~{{K}_{H}}=1.25\times {{10}^{-3}}mol.d{{m}^{-3}}at{{m}^{-1}}
Pressure of the gas P  = 2.5 atmP~~=\,2.5\,atm
Concentration of the gas in the solution (solubility) CC = 1.25×10−3mol.dm−3atm−1×2.5 atm=~1.25\times {{10}^{-3}}mol.d{{m}^{-3}}at{{m}^{-1}}\times 2.5\,atm
Concentration = 3.125×10−3mol.dm−3=~3.125\times {{10}^{-3}}mol.d{{m}^{-3}}
We know that molarity is expressed in terms of moles per liter, so conversion of mass unit into volume is as:
1 dm3= 1L1\,d{{m}^{3}}=\,1L
So, solubility is expressed in terms of molarity as:
Molarity = 3.125×10−3 mol L−1=\,3.125\times {{10}^{-3}}\,mol\,{{L}^{-1}}

Additional information: Henry’s law’s application can be seen in manufacture of soft drinks where carbon dioxide is sealed under high pressure in soft drink bottles. Also at high altitude where pressure decreases, deficiency of oxygen in the human body arises which is called anoxia and tells us about Henry’s Law.

Note: Henry’s formula will give the concentration in the unit which involves mass; this has to be converted to moles per liter when solubility is asked in terms of molarity.
Henry’s formula that involves concentration should be used, the other formula involves moles fraction that is
P=KHXP={{K}_{H}}X
Where, P=P= partial pressure, KH={{K}_{H}}= Henry’s constant, and X=X= mole fraction