Chapter 12, Problem 12.88SP

Interpretation Introduction

Interpretation:

The solubility of sulphur hexafluoride in water have to determine at 25 ^oC in mol/L and a partial pressure of 2.00 atm.

Concept introduction:Henry’s law:

Henry made a systematic investigation of the solubility of the gasses in liquids and established that the mass of gas (m) dissolved by a unit volume of given liquid is proportional to the pressure (p) of the gas at constant temperature. Thus, $\frac{\text{m}}{\text{p}}{\text{ = constant = K}}_{\text{1}}$. Now the mass of gas (m) dissolved in unit volume is proportional to its concentration (C₂) in solution, whereas the pressure (p) of the gas is given by ${\text{P=C}}_{\text{1}}\text{RT}$ (where C₁ is the concentration in the gas phase, R = Molar gas constant, T= Temperature). Thus, ${\text{K}}_{\text{1}}=\frac{m}{P}=\frac{K^1{C}_2}{C_{1}RT}$ or, at constant temperature, $\frac{{\text{C}}_{\text{2}}}{{\text{C}}_{\text{1}}}{\text{ = K}}_{\text{2}}\text{(Constant)}$. If V be the volume of the gas dissolved per unit volume of the solution, then m = $\frac{\text{M}}{\text{RT}}$ and PV = zPV, where z is a constant. so, ${\text{K}}_{\text{1}}\text{= }\frac{\text{m}}{\text{P}}\text{ = }\frac{\text{zPV}}{\text{P}}$ or, V = $\frac{K_1}{z}$ = constant. The mass (m) of the gas dissolved in unit volume of the liquid is proportional to the concentration or mol-fraction (x₂), thus $x_{\text{2}}\text{=K}\times {\text{P}}_{\text{2}}$. The constant K= $\frac{x_{\text{2}}}{P_{\text{2}}}\text{ = }\frac{\text{mol fraction of the solute (gas) in solution}}{\text{Pressure in the gas phase}}\text{=Henry's Law constant}$. The henry’s constant has different values for different gasses.

Given:

Sulfur hexafluoride, is used as non-flammable insulator at high voltage transformer, has a Henry’s law constant of $2.4\times {10}^{-4}$ mol/(L.atm) at 25 ^oC.