Question

The Henry's law constant for the solubility of argon gas in water is \(1.0 \times 10^{-3}\) M/atm at \(30^{\circ} \mathrm{C}\). (a) Express the constant in \(\mathrm{M} / \mathrm{mm} \mathrm{Hg}\). (b) If the partial pressure of argon gas at \(30^{\circ} \mathrm{C}\) is \(693 \mathrm{~mm} \mathrm{Hg},\) what is the concentration \((\mathrm{in} M)\) of the dissolved argon gas at \(30^{\circ} \mathrm{C} ?\) (c) How many grams of argon gas can be dissolved in \(29 \mathrm{~L}\) of water at \(693 \mathrm{~mm} \mathrm{Hg}\) and \(30^{\circ} \mathrm{C}\) ? (Ignore the partial pressure of water.)

Short answer

Answer: The Henry's law constant for the solubility of argon gas in water is 1.32 x 10^-6 M/mmHg. The concentration of dissolved argon gas at 30°C is 9.14 x 10^-4 M. In 29 L of water at 693 mmHg and 30°C, 1.06 g of argon gas can be dissolved.

Step-by-step solution

Convert atm to mmHg

We are asked to convert the Henry's law constant from M/atm to M/mmHg. Recall that 1 atm = 760 mmHg. So, we will multiply the given constant by the conversion factor: \((1.0 \times 10^{-3}\ \text{M/atm}) \cdot(\frac{1}{760}\ \text{atm/mmHg})\)

Calculate the new constant

Multiply the given constant by the conversion factor: \((1.0 \times 10^{-3}\ \text{M/atm}) \cdot(\frac{1}{760}\ \text{atm/mmHg}) = 1.32 \times 10^{-6}\ \text{M/mmHg}\) The Henry's law constant for the solubility of argon gas in water expressed in M/mmHg is \(1.32 \times 10^{-6}\ \text{M/mmHg}\). #b)# Concentration of dissolved argon gas in M

Apply Henry's law

Use the Henry's law formula: \(C = kP\), where \(C\) is the concentration of the gas in solution, \(k\) is the Henry's law constant, and \(P\) is the partial pressure of the gas. Here, \(k = 1.32 \times 10^{-6}\ \text{M/mmHg}\) and \(P = 693\ \text{mmHg}\).

Calculate the concentration

Multiply the constant and partial pressure: \(C = (1.32 \times 10^{-6}\ \text{M/mmHg})(693\ \text{mmHg}) = 9.14 \times 10^{-4}\ \text{M}\) The concentration of dissolved argon gas at 30°C is \(9.14 \times 10^{-4}\ \text{M}\). #c)# Grams of argon gas dissolved in 29 L of water

Calculate the moles of argon gas

Use the concentration from part (b) and the given volume of water: \(n = CV\), where \(n\) is the number of moles, \(C = 9.14 \times 10^{-4}\ \text{M}\), and \(V = 29\ \text{L}\). \(n = (9.14 \times 10^{-4}\ \text{M})(29\ \text{L}) = 0.0265\ \text{moles}\)

Calculate the mass of argon gas

Use the molecular weight of argon gas (Ar) to convert moles to grams: \(m = nM\), where \(m\) is the mass, \(n = 0.0265\ \text{moles}\), and the molar mass \(M = 39.95\ \text{g/mol}\). \(m = (0.0265\ \text{moles})(39.95\ \text{g/mol}) = 1.06\ \text{g}\) In 29 L of water at \(693\ \text{mmHg}\) and \(30^{\circ} \mathrm{C}\), 1.06 g of argon gas can be dissolved.