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Experiment 3 - Solubility of Solids in Liquids

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Experiment 3 - Solubility of Solids in Liquids 1 EXPERIMENT 3 - SOLUBILITY OF SOLIDS IN LIQUIDS THEORY When a sufficiently large amount of solute is in contact with the limited amount of solvent, dissolving of the solute starts taking place. After a certain period of time, the reverse process, precipitation, starts to occur. When dissolving and precipitation of the solute takes place at equal rates (i.e. equilibrium has been established), the solution is said to be saturated. After this point the concentration of the solute in the solution remains unchanged. The concentration of the saturated solution is defined as the solubility of the solute in the given solvent. Solubility is generally a function of temperature. For a pure solid A to be in equilibrium with a solution containing A, the chemical l s potential of A, μA, must be the same in the solid (s) and liquid (l) phases; μA = μA . The chemical potential of component A in an ideal solution is l 0l μA = μ A + RTln X A (1) 0l where μA is the chemical potential of pure liquid A, and X is the solubility. Hence, the equilibrium condition can be written s 0l μA = μ A + RTln X A (2) 0l s where μA and μA are the molar free energies of pure liquid and pure solid A. Remember that : ⎛ ∂G ⎞ μ = ⎜ ⎟ (3) i ⎜ ∂n ⎟ ⎝ i ⎠ P,, T nJ Hence GG0s − 0l A A = ln X (4) RT A and since ∂(GT/ ) H = − (5) ∂T T 2 differentiation of equation (4) with respect to T yields (with ΔHf, the latent heat of fusion) HH0l − 0s ΔH dln X A A = f = A (6) RT 2 RT 2 dT It is a good approximation to take ΔHf independent of T over moderate ranges of temperature. Alternately, equation 6 can be written in the following form (see equation 7 below). As a result, the linear variation of the solubility, XA, of pure solid in an ideal solution with temperature is − ΔH ⎛ 1 1 ⎞ f ⎜ ⎟ ln X A = ⎜ − ⎟ (7) RTT⎝ 2 1 ⎠ In determining the solubility of a solute, it is practically more correct to start off at a high temperature, cool the solution down to the required temperature and wait for thermal equilibrium to be reached. This approach is known as “reaching equilibrium from the molecular rather than the crystal side”. 2 The solubility of the solute (benzoic acid in this case) can be determined as a function of temperature. As shown in equation (7), a plot of lnXA versus 1/T should give a straight line with the slope being equal to heat of solution (ΔHf). Experimentally, the heat of solution measured is the heat absorbed when 1 mole of solid is dissolved in an almost saturated solution. This value is different from the heat of solution measured at “infinite dilution”, and the difference is equivalent to the heat of dilution. PROCEDURE 1. Weigh 2 grams of benzoic acid and add it to approximately 250 ml of distilled water that has been heated up to 50-600C in a beaker. Caution, above 600C benzoic acid may decompose. 2. After all benzoic acid dissolves, distribute this solution into six separate boiling tubes. There should be 30-40 ml sample in each tube. 3. Keep two of the tubes at room temperature, and stir them well with a stirrer. 4. Place two of the tubes in a thermostat bath at 300C and place the other two at 400C. Stir them well with a stirrer while maintaining the temperature constant for 15-20 minutes. 5. Check for the formation of crystals. 6. Use glass wool to filter 10 ml samples from each tube into a weighing bottle, then weigh it on an analytical balance. 7. Transfer the “weighed” solution to an Erlenmeyer flask, titrate it with the standard NaOH solution, using phenolphthalein as the indicator. 8. Repeat your “solubility” determinations for each sample. You should have duplicate results for each solution that were kept either at room temperature, at 300C or at 400C. Please remember to record the room temperature! CALCULATIONS 1. Use titration results to determine the solubility, X, of benzoic acid. Define the solubility as moles per 1000 g of solvent. 2. Plot lnX versus 1/T. 3. Draw tangents to the curve at least four temperatures, and determine ΔHf from each slope. 4. Determine the deviations in each value obtained. Why do you think it differs for each temperature? DATA SHEET T (0C) Weight (g) NaOH (ml) Weight (g) NaOH (ml) 25 30 40 .
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