# 16.8: When the Solute Obeys Henry's Law, the Solvent Obeys Raoult's Law

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In Section 16.4, we conclude that any sufficiently dilute solute obeys Henry’s law, and define a hypothetical, pure-liquid standard state that makes the solute activity equal to its mole fraction, \(\tilde{a}_A\left(P,y_A,y_B\right)=y_A\). In Section 16.7, we find that the mole fractions and activities of the components of any binary solution are related by

\[y_Ad \ln \tilde{a}_A + y_B d \ln \tilde{a}_B =0.\]

For a solute that obeys Henry’s law, we have

\[\begin{align*} d{ \ln \tilde{a}_B\ } &=-\left(\frac{y_A}{y_B}\right)d{ \ln y_A\ } \\[4pt] &=-\left(\frac{y_A}{y_B}\right)\left(d{ \ln y_B\ }\right)\left(\frac{d \ln y_A}{d \ln y_B}\right) \\[4pt] &=-\left(\frac{y_A}{y_B}\right)\left(d \ln y_B\right)\left(\frac{dy_A/y_A}{dy_B/y_B}\right) \\[4pt] &= -\left(\frac{y_A}{y_B}\right)\left(d \ln y_B\right)\left(\frac{-dy_B/y_A}{dy_B/y_B}\right) \\[4pt] &= d \ln y_B \end{align*}\]

This result follows for any choice of standard state for the activity of solvent \(B\). It is satisfied by \(\tilde{a}_B=ky_B\), where \(k\) is a constant. It is valid even if \(A\) is completely nonvolatile. When gas-phase \(B\) behaves as an ideal gas, and we choose the ideal gas at \(P^o\) as the standard state for both gas- and solution-phase B, we have

\[\tilde{a}_B\left(\mathrm{gas}\right)=f_B={P_B}/{P^o}={x_BP}/{P^o}\]

Since the standard states are the same, the fugacity and activity of \(B\) in solution are the same as they are in the gas phase above it. We have \(\tilde{a}_B\left(\text{solution}\right)=ky_B={x_BP}/{P^o}\). To find \(k\), we consider the system comprised of pure \(B\), for which \(y_B=x_B=1\) and \(P=P^{\textrm{⦁}}_B\). Substituting, we find \(k={P^{\textrm{⦁}}_B}/{P^o}\). With this value for \(k\),

\[\tilde{a}_A\left(\mathrm{solution}\right)={P^{\textrm{⦁}}_By_B}/{P^o}={x_BP}/{P^o}\]

so that

\[y_BP^{\textrm{⦁}}_B=x_BP.\]

This is **Raoult’s law**.

Thus when the solute obeys Henry’s law and the solvent behaves as an ideal gas in the gas phase above its solution, the solvent obeys Raoult’s law.

Evidently the converse is also true. If the solvent obeys Raoult’s law, \(y_BP^{\textrm{⦁}}_B=x_BP\). With pure ideal gas \(B\) as the standard state for \(B\) in both the gas phase and the solution phase, we have

\[\tilde{a}_B\left(\text{solution}\right)=\tilde{a}_B\left(\mathrm{gas}\right)=f_B={P_B}/{P^o}={x_BP}/{P^o}=y_B\left({P^{\textrm{⦁}}_B}/{P^o}\right)\] so that \[d{ \ln \tilde{a}_B\left(\mathrm{solution}\right)\ }=d{ \ln y_B\ }\]

From \(y_Ad{ \ln \tilde{a}_A\ }+y_Bd{ \ln \tilde{a}_B\ }=0\) and \(d{ \ln \tilde{a}_B\ }=d{ \ln y_B\ }\), we have

\[d{ \ln \tilde{a}_A\ }=-\left(\frac{y_B}{y_A}\right)d{ \ln y_B\ }=-\left(\frac{y_B}{y_A}\right)\left(\frac{dy_B}{y_B}\right)=\frac{dy_A}{y_A}=d{ \ln y_A\ }\]

so that \(\tilde{a}_A\left(\mathrm{solution}\right)=ky_A\), where \(k\) is a constant. When we choose the standard state such that \(\tilde{a}_A\left(\mathrm{ss,\ }\mathrm{solution}\right)=1\) when \(y_A=1\), we find \(k=1\) and \(\tilde{a}_A\left(\mathrm{solution}\right)=y_A\). The activity of the solute is related to its fugacity and the fugacity of its standard state by

\[\tilde{a}_A\left(\mathrm{solution}\right)=y_A=\frac{f_A\left(\mathrm{solution}\right)}{f_A\left(ss,\ \mathrm{solution}\right)}\]

When \(y_A=1\), the fugacity is that of the standard state, which is a system of the hypothetical pure liquid in equilibrium with its own ideal gas. Letting the pressure of this ideal gas be \({\textrm{ĸ}}_A\), we have \(f_A\left(ss,\ \mathrm{solution}\right)={\textrm{ĸ}}_A\), so that \(f_A\left(\mathrm{solution}\right)={\textrm{ĸ}}_Ay_A\), which is equal to the fugacity of the gas with which it is at equilibrium. The fugacity of the ideal gas is \(x_AP\), so that

\[x_AP={\textrm{ĸ}}_Ay_A.\]

This is Henry’s law. Thus, if solvent \(B\) obeys Raoult’s law, solute \(A\) obeys Henry’s law.