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I was trying to calculate amount of salt needed to lower the melting point of water by 5 °C. According to my solutions, I need to make sure $$n\mu(\ce{H2O(s)}) =n\mu(\ce{H2O(l)})+ RT\log(x)$$where $ x=\frac{n}{n+2m}$

I, however, thought that I need to make sure that

$$n\mu(\ce{H2O(s)})+m\mu(\ce{NaCl(s)}) =n\mu(\ce{H2O(l)})+m\mu(\ce{NaCl(aq)})$$

After all, the solid ice still has a certain contribution. And I would have assumed that water can be seen as the solved according to the first convention? So why don't we need to consider the solid ice, and what is wrong with my way of thinking?

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You should write independent equilibrium expressions for the chemical potential of each species.

In a freezing point depression problem the solvent in frozen form is at equilibrium with the solvent in liquid form, so you can write that $$\mu(\ce{H2O(s))}=\mu(\ce{H2O(sol))}$$ Next, by Raoult's law (assuming it holds), the chemical potential of solvent in the solution (the right-hand term) differs from that of the pure liquid by the term $RT\log(\chi_1)$ where $\chi_1$ is the mole fraction of solvent, so that
$$\mu(\ce{H2O(s))}=\mu(\ce{H2O(l))} + RT\log(\chi_1)$$ This is just the beginning. Next you need to recognize that the standard Gibbs free energy change for melting is just $\Delta_m G^\circ=\mu(\ce{H2O(l))}-\mu(\ce{H2O(s))}$ (assuming the solution is at $\pu{1 bar}$ or approx normal pressure) and invoke the Gibbs-Helmholtz equation in differential form:

$$\left(\frac{\partial(\Delta G /T)}{\partial (1/T)}\right)_p=\Delta H$$

so that

$$\left(\frac{\partial\log(\chi_1)}{\partial (1/T)}\right)_p=-\frac{\Delta_m H^\circ}{R}$$

Now you integrate the van't Hoff equation from $\chi_1 = 1$ (pure solvent) to $\chi_1 = 1-\chi_2<1$, while assuming $\Delta_m H^\circ$ to be constant over this range of $T$. Finally, assuming $T\approx T_m$, expressing $\chi_1$ in terms of solute molality, and assuming the solute concentration is low, results in the freezing point equation $$\Delta T = K_m m_2 $$ where $K_m$ is the cryoscopic constant $$K_m=\frac{RT_m^2M_2}{1000 \Delta_m H^\circ}$$ Here $M_2$ is the molar mass of the solute. You can use this last expression to compute the molality of solute $m_2$ required to depress the freezing temperature by $\Delta T$.

The process of dissolving the solute is not important (whatever happens before it is fully dissolved doesn't matter as long as it is soluble). Also, it is assumed an ideal solution forms, which means interactions of the solvent with solute, and between solvent molecules are equally strong. And the salt remains in solution (none is incorporated into the solid phase). Note also that liquid solvent remains in equilibrium with solid solvent at the melting point $T_m$, and that we assume that the solute concentration is dilute (this is important to derive the last expression, but Raoult's law may hold even for substantial solute concentrations).

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  • $\begingroup$ Thank you! Two questions: 1. So the salt we use doesn't matter, we might as well have used KI if the amount is the same? 2. Why doesn't the enthalpy and entropy of dissolving our salt matter at all? After all, by freezing our water, we force the salt to become solid again, right? So why don't we have to consider that in our equations? $\endgroup$
    – DottyPhone
    Jul 17 '19 at 16:32
  • $\begingroup$ The process of dissolving the solute is not important (whatever happens before it is fully dissolved doesn't matter as long as it is soluble). Also, it is assumed an ideal solution forms, which means interactions of the solvent with solute, and between solvent molecules are equally strong. And no, the salt remains in solution. And remember that the liquid solvent remains in equilibrium with solid (at a constant T) and we assume the solute concentration is dilute. $\endgroup$
    – Buck Thorn
    Jul 17 '19 at 16:40
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    $\begingroup$ @MathewMahindaratne thanks for your edit - that was a weird oversight of mine. $\endgroup$
    – Buck Thorn
    Jul 17 '19 at 19:57

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