Question

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Answer 1

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

`N_2`
`7.81* 10^(-1)`
`6.70* 10^(-4)`

`O_2`
`2.10* 10^(-1)`
`1.30* 10^(-3)`

Ar
`9.34* 10^(-3)`
`1.40* 10^(-3)`

`CO_2`
`3.33* 10^(-4)`
`3.50* 10^(-2)`

`CH_4`
`2.00* 10^(-6)`
`1.40* 10^(-3)`

`H_2`
`5.00* 10^(-7)`
`7.80* 10^(-4)`

Answer: The solubility of hydrogen gas in water at given atmospheric pressure is
`1.48* 10^(-10)M`

Step-by-step explanation:

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

`p_{\text{hydrogen gas}}=p_T* \chi_{\text{hydrogen gas}}`

where,

`p_A` = partial pressure of hydrogen gas = ?

`p_T` = total pressure = 0.380 atm

`\chi_A` = mole fraction of hydrogen gas =
`5.00* 10^(-7)`

Putting values in above equation, we get:

`p_{\text{hydrogen gas}}=0.380* 5.00* 10^(-7)\\\\p_{\text{hydrogen gas}}=1.9* 10^(-7)atm`

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

`C_(H_2)=K_H* p_(H_2)`

where,

`K_H` = Henry's constant =
`7.80* 10^(-4)mol/L.atm`

`p_(H_2)` = partial pressure of hydrogen gas =
`1.9* 10^(-7)atm`

Putting values in above equation, we get:

`C_(H_2)=7.80* 10^(-4)mol/L.atm* 1.9* 10^(-7)atm\\\\C_(CO_2)=1.48* 10^(-10)M`

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is
`1.48* 10^(-10)M`