# 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?

Dec 18, 2015

$4.50 \cdot {10}^{- 4} \text{g/L}$

#### 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

${k}_{H}$ - Henry's constant, specific to each gas and dependent on the temperature
${c}_{\text{aq}}$ - the molar concentration of the dissolved gas
$P$ - the partial pressure of the gas above the liquid

Now, the value of Henry's constant for hydrogen gas at ${25}^{\circ} \text{C}$ is equal to

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

${k}_{H} = {c}_{\text{aq"/P implies c_"aq}} = {k}_{H} \times P$

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

${c}_{\text{aq" = 2.231 * 10^(-4)"mol"/"L}}$

Hydrogen gas has a molar mass of $\text{2.0159 g/mol}$, which means that one mole of hydrogen gas will have a mass of $\text{2.0159 g}$.

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.