Liquids and Solutions Crib Sheet

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Kelley Nathaniel Rich
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1 Liquids and Solutions Crib Sheet Determining the melting point of a substance from its solubility Consider a saturated solution of B in a solvent, A. Since the solution is saturated, pure solid B is in equilibrium with the B in solution: µ B(s) = µ B(l) and so if B is present in solution at a mole fraction x B : i.e. µ B(s) = µ B(l) + T ln x B ln x B = µ B(s) µ B(l) = fusg B T T since the chemical potential of a pure substance is simply the molar Gibbs free energy. ecall that fus G B is the Gibbs free energy change for the process B (s) B (l). Using G = H T S, we have ln x B = fush B T + fuss B Now, at T = T, the melting point of B, fus G B = 0 and so we can add fus G B T to the right hand side to get ln x B = fush B T = fush B T + fuss B fuss B + ( fus H B T = 0 fuss B ) Assuming that fus H B and fuss B are constant across the temperature range of interest, we have ln x B = fush ( B 1 T 1 ) T Therefore, a plot of ln x B, the saturation mole fraction of B in A, against 1/T yields a straight line with slope fus H B / and intercept fush B /T. That is, we can deduce the melting point of B and its enthalpy of fusion from measurements of its solubility at different temperatures.

2 Determining the molar mass of a solute Consider a solution of B in a solvent, A. At the melting point of the solvent, µ A(s) = µ A(l) = µ A(l) + T ln x A where x A = 1 x B. As before, ln(1 x B ) = µ A(s) µ A(l) T = fusg A T = fush A T + fuss A Now, if x B = 0 then T = T, the melting point of pure A, and fus G A = 0, so as before fus G A T = fush A T fuss A = 0 And if, as before, we assume that fus H A and fuss A are constant with temperature we can add this equation to the right hand side to get ln(1 x B ) = fush A ( 1 T 1 ) T Finally, if B is only sparingly soluble in A, x B 1 and we can use the approximation ln(1 + y) y for small y; that is ln(1 x B ) x B. Also, note that 1 T 1 T = T T T T T T 2 when T T. Thus, x B = fush A T T 2 So from a measurement of how the freezing point of the solvent changes ( T ) when a known mass of B is dissolved in it, x B can be found (if fus H A is known). But n B = x B (n A + n B ) x B n A So for a known amount of solvent, n A, we know how many moles of B have been dissolved. If this corresponds to a known (measured) mass of B, m, the molar mass of B is simply m B = m/n B.

3 Mixing of ideal solutions (or gases) Consider two pure liquids, A and B, separated by a partition. Their total Gibbs free energy is = n A µ A + n B µ B G i where there are n A moles of A and n B moles of B - recall that the chemical potential of a pure substance is the molar Gibbs free energy, µ = G m. Suppose now that the partition is removed and the liquids mix. If the solution is ideal, there is no enthalpy change on mixing: mix H ideal = 0 In an ideal solution the A-B interaction energy is the same as the average A-A and B-B interaction energy and the driving force for mixing is purely entropic: A and B molecules are randomly distributed about one another. The total Gibbs free energy of the mixture is G f = n A (µ A + T ln x A ) + n B (µ B + T ln x B ) where x A is the mole fraction of A in the mixture and x B is the mole fraction of B. Therefore, the Gibbs free energy change on mixing is mix G ideal = G f G i = n A T ln x A + n B T ln x B Now, n A = x A n and n B = x B n, where n = n A + n B. Thus, Furthermore, since S = ( G/ T ) p, mix G ideal = nt (x A ln x A + x B ln x B ) mix S ideal = n(x A ln x A + x B ln x B ) egular solutions eal solutions have values for mix G, mix H, and mix S that differ from their ideal values. The difference is called the excess: mix G = mix G ideal + G excess mix H = H excess mix S = mix S ideal + S excess A regular solution is one for which there is a non-zero enthalpy of mixing, but no excess entropy of mixing (S excess = 0): the A-B interaction is different from the average A-A and B-B interaction, but not so much so that the A and B molecules are no longer randomly distributed. This suggests that the enthalpy of mixing should be defined as H excess = mix H = βx A x B where β = zn A ɛ, z is the co-ordination number of both A and B, and ɛ = ɛ AB 1 2 ɛ AA 1 2 ɛ BB is the energy change when one A-B bond is formed in the mixture.

4 Osmotic pressure and polymer solution Consider a semi-permeable membrane separating pure solvent, A, and a solvent containing a solute, B, which cannot pass through the membrane. The pure solvent is subject to a pressure, p (usually just the ambient atmospheric pressure), and the solution subject to a greater pressure, p + Π which prevents the influx of solvent across the membrane. At equilibrium, µ A(p) = µ A (p + Π, x A ) where x A is the mole fraction of solvent on the solution side of the membrane. Therefore, µ A(p) = µ A(p + Π) + T ln x A Now, for a pure substance, µ = G m, and dg = V dp SdT, so at constant temperature dµ = V m dp. Thus, µ A(p + Π) = µ A(p) + p+π p V m dp If V m is constant with p (the solvent is incompressible), this becomes and so substituting in above, µ A(p + Π) = µ A(p) + V m Π 0 = V m Π + T ln x A If x B = 1 x A 1 (the solute is only sparingly soluble), and we have ln x A = ln(1 x B ) x B Π = T x B V m But x B = n B /(n A + n B ) n B /n A when n A n B, so Π = T n B, since V = V m n A V The concentration of the solute, [B] = n B /V, so Π = T [B] This equation is valid for ideal solutions. Solutions of polymers (even dilute solutions) are not ideal, and it is usual to treat them with a virial expansion: Π [B] = T (1 + B [B] + ) where B is the second virial coefficient. This enables the molar mass of a solute to be obtained because [B] = c B /M B, where c B is the molality of B in

5 the solution (the known mass of B dissolved e.g. in g dm 3 ), and M B is the molar mass of B (e.g. in g mol 1 ), so Π c B = T M B (1 + Bc B ) A plot of Π/c B against c B should yield a straight line with slope T B/M B and intercept T/M B. Note that B = B /M B is temperature- and solvent-dependent. The so-called θ-condition occurs when B = 0 and the solution behaves ideally (cf. the Boyle temperature for real gases). Debye-Hückel theory Debye-Hückel theory describes the behaviour of dilute ionic solutions. Each ion is thought of as being a point charge surrounded by a diffuse ionic atmosphere of opposite charge (when the thermal motions of all the surrounding ions has been averaged) present in a neutral, inert solvent which is considered to be a continuous medium. Each ion therefore gives rise to a shielded Coulomb potential which decreases exponentially as a function of distance from the ion (of charge z i ): φ i (r) = z i r e r/rd where r D is the Debye length, and depends on the ionic strength, I: r D I 1/2, where I = 1 2 z 2 i c i c Here, c i is the concentration of ions of charge z i. A larger ionic strength gives rise to greater shielding. The chemical potential of the ions is given in terms of their activities, a i µ i = µ i + T ln a i where a i = γ i c i c and γ i is the activity coefficient; Debye-Hückel theory provides a way of estimating γ i for weak ionic solutions: log 10 γ ± = z + z A I where A is a constant which depends on the solvent and temperature (A = for water at 298 K).

Lecture 6. NONELECTROLYTE SOLUTONS

Lecture 6. NONELECTROLYTE SOLUTONS NONELECTROLYTE SOLUTIONS SOLUTIONS single phase homogeneous mixture of two or more components NONELECTROLYTES do not contain ionic species. CONCENTRATION UNITS percent