Question 2 (15 pts): A 2L container has 1 L water and 1 L air headspace. The hea

Question

Question 2 (15 pts): A 2L container has 1 L water and 1 L air headspace. The headspace has a total pressure of 1 atm. The temperature of both water and air is 298K. 1 mg of formaldehyde (HCHO) was added into the water. Since formaldehyde is a volatile compound, some formaldehyde will escape into the air. The Henry's constant (KH) for formaldehyde is 6300 M-atm What is the mass concentration of formaldehyde in the water after the equilibrium between the air and the water has been reached? What is the molar fraction of formaldehyde (YHcHo) in the air headspace at equilibrium? ??

Explanation / Answer

Using Henry's law ,you can write,

According to henry's law, at a given temperature the concentration of a gas dissolved in a solvent is directly proportional to the partial pressure of the gas,

C(formaldehyde)=KH*p(formaldehyde) ............................(1)

where C(for)=concentration of formaldehyde=Mol of formaldehyde in 1Lwater

total mol of formaldehyde in the container=ntotal=mass/molar mass=0.001g/30.031g/mol=3.330*10^-5 mol in 1L

and p=partial pressure of formaldehyde

As the Volume of air space =1L and Temperature ,T=298K and the mol of formaldehyde =n(gas)

Applying ideal gas equation, V=volume of gas,T=temperature,R=ideal gas constant=0.0821Latm/Kmol

p(formaldehyde)=p(gas)=n(gas)*RT/V=n(gas)(0.0821Latm/Kmol)(298K)/1L=n(gas)*24.465 atm/mol

or ,p(gas)=n(gas)*24.465 atm/mol........(A)

Also ,C(formaldehyde)=KH*p(formaldehyde) ............................(1)

or, n(water)=(6300M/atm)*p(gas)..........(B)

eqn (A) and (B) are related throgh the equation,

n(total)=n(gas)+n(water)=3.330*10^-5 mol

So , eqn (B) becomes,n(water)=3.330*10^-5 mol-n(gas)=(6300M/atm)*p(gas)

Also ,plug in the value of n(gas ) from eqn (A),

n(gas)=p(gas)/24.465 atm/mol

Thus,3.330*10^-5 mol-n(gas)=(6300M/atm)*p(gas)

or,3.330*10^-5 mol-(p(gas)/24.465 atm/mol)=(6300M/atm)*p(gas)

or,3.330*10^-5 mol=(6300M/atm)*p(gas)+(p(gas)/24.465 atm/mol)=p(gas)[(6300M/atm)+0.0409 mol/atm]

or,3.330*10^-5 mol=p(gas)[(6300mol/atm)+0.0409 mol/atm]

or, p(gas)=(3.330*10^-5 mol)/(6300mol/atm)+0.0409 mol/atm)=5.285*10^-9 atm

p(gas)=5.285*10^-9 atm

So,n(gas)=p(gas)/24.465 atm/mol=(5.285*10^-9 atm)/(24.465 atm/mol)=2.160*10^-10 mol

mol fraction in gas phase or air head space=n(gs)/n(total)=(2.160*10^-10 mol)/(3.330*10^-5 mol)=6.486*10^-6

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