# What is the solubility of hydrogen (in units of grams per liter) in water at 25 °C, when the #H_2# gas over the solution has a partial pressure of .286 atm?

##### 1 Answer

#### Explanation:

The idea here is that you need to use the equation for **Henry's Law** to determine the *molar solubility* of hydrogen gas at that temperature and partial pressure, then use its *molar mass* to convert this from *moles per liter* to *grams per liter*.

So, according to **Henry's Law**, the solubility of a gas in a liquid is *proportional* to its partial pressure above the liquid. Mathematically, this is written as

#color(blue)(k_H = c_"aq"/P)" "# , where

*Henry's constant*, specific to each gas and dependent on the temperature

Now, the value of Henry's constant for hydrogen gas at

#k_H = 7.8 * 10^(-4)"mol"/("L" * "atm")#

So, plug in your values and calculate the molar solubility of hydrogen gas at that temperature

#k_H = c_"aq"/P implies c_"aq" = k_H xx P#

#c_"aq" = 7.8 * 10^(-4)"mol"/("L" * color(red)(cancel(color(black)("atm")))) * 0.286 color(red)(cancel(color(black)("atm")))#

#c_"aq" = 2.231 * 10^(-4)"mol"/"L"#

Hydrogen gas has a molar mass of **one mole** of hydrogen gas will have a mass of

In your case, the solubility of hydrogen gas in *grams per liter* will be

#2.231 * 10^(-4) color(red)(cancel(color(black)("mol")))/"L" * "2.0159 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)(4.50* 10^(-4) "g/L")#

The answer is rounded to three sig figs, the number of sig figs you have for the partial pressure of the gas.