(ii) nature of the solvent (intermolecular forces) (iii) temperature (Solubility & Le Chatelier's Principle) Points above the solubility curve give concentrations for supersaturated solutions. (i) determine how much solute will dissolve in the solvent at a given temperature (ii) compare the solubilities of different solutes in the same solvent (iii) determine the mass of solid precipitated, or crystallised, out of solution when the temperature of the solution changes Mass is chosen primarily because the mass of a substance does not change with temperature. The density of water changes with temperature: As water is heated, its density decreases. This means that if you measure out 100 mL (100 cm3) of water at 20°C in a 100 mL volumetric flask, it will have a mass of:
100 mL of water at 20°C has a mass of 99.82 g If we then heat this 99.82 g of water in the 100 mL volumetric flask to 40°C we could see the water level rise above the mark as the volume increases (as a result of the decrease in density at higher temperature), we can calculate the volume of this mass of water at 40°C:
At 20°C, 99.82 g of water has a volume of 100 cm3, but if you increase the temperature to 40°C, the same mass of water has a greater volume, that is, 100.6 cm3. This means that you can't use the volume of water to determine concentration if you are going to the change the temperature because the volume of the water also changes with temperature! Solubility data are therefore given as a ratio of the mass of solute to a fixed mass of solvent. Typically, the units are grams of solute per 100 g of solvent (g/100 g). For aqueous solutions, the units are grams of solute per 100 g of water (g/100 g).
Do you know this? Join AUS-e-TUTE! Play the game now! Many salts, such as sodium chloride (NaCl(s)), are soluble in water at 25°C. The table below shows experimentally determined values for the solubility of sodium chloride in water at various temperatures. The points have been plotted on a graph to produce a solubility curve.
Remember that solubility refers to the maximum mass of solute that can be dissolved in a given mass of solvent at a specified temperature. Each point on the curve in the graph above tells how much solute we can add to 100 g of water at that temperature in order to form a saturated solution. The first thing we notice is that more NaCl(s) can be dissolved in 100 g of water if we increase the temperature.
What does this tell us about the region in the graph below the solubility curve? It tells us that at any point below the solubility curve the solution is unsaturated, that is, more solute could be added to the water at that temperature until you reach the line at which point the solution will be saturated.
But what happens if we could instantaneously cool the solution made up of 37 g of NaCl(s) dissolved in 100 g of water from 80°C down to 20°C?
Different substances will have different solubility curves. For lots of salts, like sodium chloride, the solubility of the salt in water increases with increasing temperature. But for gases and some other salts, like lithium sulfate, the solubility decreases as the temperature increases.
As shown above, the mass of lithium sulfate that can be dissolved in 100 g of water actually decreases as you increase the temperature of the solution.
Do you understand this? Join AUS-e-TUTE! Take the test now! The solubility curves for sodium chloride (NaCl) and lithium sulfate (Li2SO4) are plotted in the graph below:
The graph shows that as temperature increases
From the graph we see that:
From the graph we can see that if we have a solution containing 35 g of solute in 100 g of water at 40°C, then
Do you understand this? Join AUS-e-TUTE! Take the test now! We can use the solubility curve to predict the mass of solid that should precipitate (or crystallise) out of solution. Let's assume you have a supersaturated solution that contains 35.00 g of litium sulfate dissolved in 100 g of water at a temperature of 30°C as shown by point A (×) on the graph below:
Reading off the light blue solubility curve for Li2SO4 at a temperature of 30°C, we see that a saturated solution will contain 34.20 g of Li2SO4 dissolved in 100 g of water. We predict that the "excess" mass of Li2SO4 in the supersaturated solution will precipitate out of solution (crystallise) until the solution becomes saturated:
We predict that 0.80 g of Li2SO(s) will precipitate, or crystallise, out of this solution.
Do you understand this? Join AUS-e-TUTE! Take the test now! Chris the Chemist has dissolved 19.50 g of copper(II) sulfate in 50.00 g of water in a 250 mL conical (erlenmeyer) flask at a temperature of 60°C. What is the question asking you to do?
What information have you been given in the question?
What is the relationship between what you have been given and what you need to find out?
Perform the calculations
Is your answer plausible? Mass of crystals formed must be less than 19.50 g because we can't crystallise out more CuSO4 than was present in the original solution. 3.5 g < 19.50 g so we are confident our answer is plausible. The mass of CuSO4(s) must be a positive value, that is, mass CuSO4 dissolved in "hot" solution must be greater than the mass of CuSO4 dissolved at lower temperature in order for some solid to precipitate (crystallise) out of solution. Our value for mass of CuSO4(s) is positive so we are confident our answer is plausible. State your solution to the problem. 3.50 g of CuSO4(s) precipitates (crystallises) out of solution at 20°C.
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Which two compounds show a decrease in solubility from 0°c to 100°c
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