Question

For equilibrium at 25 oC with the atmosphere, what are the aqueous concentrations of N2 (mg/L), O2 (mg/L) and CO2 (mol/L) in natural water? Example: N2 : 0.79 atm O2 : 0.21 atm CO2 : 10-3.38 atm

Answer 1

Answer:-

Given:-

temperature (T) = 25 ⁰C = 25 + 273 = 298 K

partial pressure of N₂ gas = 0.79 atm

partial pressure of O₂ gas = 0.21 atm

partial pressure of CO₂ gas = 10^-3.38 atm

concentration of N₂ (mg/L) = ?

concentration of O₂ (mg/L) = ?

concentration of CO₂ (mol/L) = ?

As we know that

gas constant (R) = 0.08206 L.atm.K^-1.mol^-1

So according to the ideal gas equation

PV = nRT

or

P = nRT / V

or

P = CRT (since C = n /V )

where

C = concentration of gas in mol / L

therefore

partial pressure of CO₂ gas (P_CO2) = C_CO2RT

concentration of CO₂ (C_CO2) = partial pressure of CO₂ gas (P_CO2) / RT

concentration of CO₂ (C_CO2) = 10^-3.38 atm / 0.08206 L.atm.K^-1.mol^-1$\times$ 298 K

concentration of CO₂ (C_CO2) = 10^-3.38 atm / 24.45388‬ L.atm..mol^-1

concentration of CO₂ (C_CO2) = 0.041 $\times$ 10^-3.38 mol / L (i.e the answer)

similarly

concentration of N₂ (C_N2) = partial pressure of N₂ gas (P_N2) / RT

concentration of N₂ (C_N2) =  0.79 atm / 0.08206 L.atm.K^-1.mol^-1$\times$ 298 K

concentration of N₂ (C_N2) =  0.79 atm / 24.45388‬ L.atm..mol^-1

concentration of N₂ (C_N2) = 0.032 mol / L

above concentration of N₂ (C_N2) shows that 0.032 mol N₂ gas present in 1 L natural water.

Since we know that

1 mol of N₂ = 28 g N₂ = 28000 mg N₂

So

1 mol of N₂ = 28000 mg N₂

then

0.032 mol of N₂ = 0.032 $\times$ 28000 mg N₂​​​​​​​

0.032 mol of N₂ = 896.0 mg N₂​​​​​​​

therefore

concentration of N₂ (C_N2) = 0.032 mol / L = 896.0 mg / L (i.e the answer)

similarly

concentration of O₂ (C_O2) = partial pressure of O₂ gas (P_O2) / RT

concentration of O₂ (C_O2) =  0.21 atm / 0.08206 L.atm.K^-1.mol^-1$\times$ 298 K

concentration of O₂ (C_O2) =  0.21 atm / 24.45388‬ L.atm..mol^-1

concentration of O₂ (C_O2) = 0.009 mol / L

above concentration of O₂ (C_O2) shows that 0.009 mol O₂ gas present in 1 L natural water.

Since we know that

1 mol of O₂ = 32 g O₂ = 32000 mg O₂

So

1 mol of O₂ = 32000 mg O₂

then

0.009 mol of O₂ = 0.009 $\times$ 32000 mg O₂

0.009 mol of O₂ =   288.0 mg O₂

therefore

concentration of O₂ (C_O2) = 0.009 mol / L = 288.0 mg / L (i.e the answer)