Computation of Henry's Law Constant from Solubility Data Given: Toluene (M = 92.0) is mixed in water until saturated. From Perry et al. (1984), the solubility of toluene is 0.050 g of toluene per 100. g of water at 16.0 °C and 1.0 atm. To do: Estimate Henry's law constant for toluene in water.

Question

Computation of Henry's Law Constant from Solubility Data
Given: Toluene (M = 92.0) is mixed in water until saturated. From Perry et al. (1984), the solubility of toluene is 0.050 g of toluene per 100. g of water at 16.0 °C and 1.0 atm.

To do: Estimate Henry's law constant for toluene in water.

Accepted Answer

The mol fraction of toluene in water is

[latex]x_{toluene}=\frac{c_{molar,toluene}}{c_{molar,L}} =\frac{\frac{m_{toluene}}{M_{toluene}V} }{\frac{1}{V}\sum\limits_{j}{\left(\frac{m_j}{M_j} ight) } } =\frac{\frac{m_{toluene}}{M_{toluene}} }{\frac{m_{toluene}}{M_{toluene}}+ \frac{m_{water}}{M_{water}} } =\frac{\frac{0.050g}{92.0\frac{g}{mol} } }{\frac{0.050 g}{92.0\frac{g}{mol} } +\frac{100.g}{18.0\frac{g}{mol} } } =9.78 imes 10^{-5}[/latex]

From Appendix A. 8, the vapor pressure of toluene is found to be 40.0 mm Hg at 31.8 °C and 20.0 mm Hg at 18.4 °C. The vapor pressure at 16.0 °C is not given. Since [latex]H^′[/latex] depends on vapor pressure [latex](P_{\mathrm{v},j})[/latex] which in turn is keenly dependent on temperature, the vapor pressure at 16.0 °C is
estimated using the Clausius-Claperon equation. The following notation is used:
- [latex]T_1[/latex] = 16.0 °C = 289.15 K [latex]P_{\mathrm{v},1}[/latex](at [latex]T_1[/latex]) = unknown (to be found)
- [latex]T_2[/latex] = 18.4 °C = 291.55 K [latex]P_{\mathrm{v},2}[/latex](at [latex]T_2[/latex]) = 20.0 mm Hg
- [latex]T_3[/latex] = 31.8°C = 304.95K [latex]P_{\mathrm{v},3}[/latex](at [latex]T_3[/latex]) = 40.0 mm Hg

From Eq. (4-28),

[latex]\ln \left(\frac{P_{\mathrm{v}2}}{P_{\mathrm{v}1}} ight) =\left(\frac{M\hat{h}_{\mathrm{f}g}}{R_u} ight) \left(\frac{(T_2-T_1)}{T_2T_1} ight)[/latex] (4-28)

[latex]\frac{\hat{h}_{\mathrm{f}g} }{R_u} =\ln \left(\frac{P_{\mathrm{v},3}}{P_{\mathrm{v},2}} ight) \frac{T_3T_2}{T_3-T_2} =\ln \left(\frac{40.0 mm Hg}{20.0 mm Hg} ight) \frac{(304.95 K)(291.55 K)}{304.95 K- 29 1.55 K} =4599. K[/latex]

and

[latex]\ln \left(\frac{P_{\mathrm{v},2}}{P_{\mathrm{v},1}} ight) =\frac{\hat{h}_{\mathrm{f}g} }{R_u} \frac{T_2-T_1}{T_2T_1} =\ln \left(\frac{20.0 mm Hg}{P_{\mathrm{v},1}} ight) =(4599. K)\frac{291.55 K-289.15 K}{(291.55 K)(289.15 K)}[/latex]

which can be solved for [latex]P_{\mathrm{v},1}[/latex] ,

[latex]P_{\mathrm{v},1}[/latex] = 17.55 mm Hg

Assuming that [latex]P_j[/latex] is equal to the vapor pressure [latex]P_{\mathrm{v},1}[/latex] at the solubility condition of 0.050 g of toluene per 100. g of water, Henry's law constant at 16.0 °C can then be estimated from Eq. (4-97),

[latex]P_j=Hc_{molar j}=Hx_jc_{molar,L}=H^′{x}_j[/latex] (4-97)

[latex]H^′=\frac{P_{toluene}}{x_{toluene}} =\frac{17.55 mm Hg}{9.78 imes 10^{-5}} \left(\frac{1.0 atm}{760. mm Hg} ight) \left(\frac{1.013 imes 10^5\frac{N}{m^2} }{1.0 atm} ight) =2.39 imes 10^7\frac{N}{m^2}[/latex]

Discussion: The value listed in Appendix A.9 is 3.72 × [latex]10^7[/latex] N/m². The agreement is not so good, but of the correct order of magnitude, illustrating that the (linear) Henry's law approximation is not very accurate, at least for this example. The main reason for the discrepancy is the nonlinear curve of [latex]P_j[/latex] versus [latex]X_j[/latex], as illustrated in Figure 4.16. Namely, even though the concentration of toluene in water is small (of order [latex]10^{-4}[/latex]), the slope of the [latex]P_j[/latex] versus [latex]X_j[/latex] curve is not constant as [latex]X_j[/latex] increases, and hence the estimated value of [latex]H^′[/latex] is too low.

Question 4.14: Computation of Henry's Law Constant from Solubility Data Giv...

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