Question

A compound has a molecular weight of 153.2 and a Henry’s constant of 85 atm. What is
the concentration of the compound in air in ppmv which is in equilibrium with a 35.0
mg/L aqueous solution?

Answer 1

Answer on Question #48060 – Chemistry – Inorganic Chemistry

Question:

A compound has a molecular weight of 153.2 and a Henry's constant of 85 atm. What is the concentration of the compound in air in ppmv which is in equilibrium with a 35.0 mg/L aqueous solution?

Solution:

According to Henry's law partial pressure of dissolved gas is: $p_1 = K \times x$, where $K$ – Henry constant, $x$ – mole fraction of gas in solution.

$x = \mu_1 / (\mu_1 + \mu_2)$, where $\mu_1$ – the mole number of the gas, $\mu_2$ – the mole number of the solvent.

In 1 L of the solution there are 35 mg of the gas ($\mu_1 = \mathrm{m} / \mathrm{M_w} = 0.035\mathrm{g} / 153.2\mathrm{g} / \mathrm{mol} = 0.2285$ mmol) and 1000 g of water ($\mu_2 = 1000\mathrm{g} / 18\mathrm{g} / \mathrm{mol} = 55.56$ mol).

Thus, the mole fraction is: $x = 0.2285 \times 10^{-3} \mathrm{mol} / (0.2285 \times 10^{-3} \mathrm{mol} + 55.56 \mathrm{mol}) = 4.1127 \times 10^{-6}$

The partial fraction is: $p_1 = 85$ atm $\times 4.1127 \times 10^{-6} = 0.3496 \times 10^{-3}$ atm

As known the ratio of the partial pressure of two gases to the total pressure is the same as the ratio of the number of moles of gas to the total number of moles.

Taking into account that the total pressure is 1 atm ($p_{\text{total}}$), the concentration of the compound in ppmv can be calculated as: $C(\text{ppmv}) = p_1 / p_{\text{total}} \times 1000000 = 349.6$ ppmv

Answer: 349.6 ppmv

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