Question

The solubility of nitrogen, N2, in water is 4.52 ✕ 10−4 mol/L at 0°C when the nitrogen pressure above water is 0.741 atm. Calculate the solubility of nitrogen in water when the partial pressure of nitrogen above water is 1.086 atm at 0°C?

Answer 1

Answer: The concentration of nitrogen gas when the partial pressure is 1.086 atm above water is $6.62\times 10^{-4}mol/L$

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{N_2}=K_H\times p_{N_2}$

where,

$K_H$ = Henry's constant = ?

$C_{N_2}$ = molar solubility of nitrogen gas = $4.52\times 10^{-4}mol/L$

$p_{N_2}$ = partial pressure of nitrogen gas = 0.741 atm

Putting values in above equation, we get:

$4.52\times 10^{-4}mol/L=K_H\times 0.741atm\\\\K_H=\frac{4.52\times 10^{-4}mol/L}{0.741atm}=6.10\times 10^{-4}mol/L.atm$

Now, calculating the concentration when pressure is changed by using above equation:

$p_{N_2}$ = 1.086 atm

$K_H=6.10\times 10^{-4}mol/L.atm$

Putting values in above equation, we get:

$C_{N_2}=6.10\times 10^{-4}mol/L.atm\times 1.086atm\\\\C_{N_2}=6.62\times 10^{-4}mol/L$

Hence, the concentration of nitrogen gas when the partial pressure is 1.086 atm above water is $6.62\times 10^{-4}mol/L$