Question

Potassium sulfate has a solubility of 15g/100g water at 40 Celsius. A solution is prepared by adding 39.0g of potassium sulfate to 225g water, carefully heating the solution, and cooling it to 40 Celsius. A homogeneous solution is obtained. Is this solution saturated, unsaturated, or supersaturated? The beaker is shaken and precipitation occurs. How many grams of potassium sulfate would you except to crystallize out?

Answer 1

Based from the given information, the maximum amount of solute a 100 g of water can dissolve at 40 degrees Celsius is 15 g of potassium sulfate, K2SO4. The ratio of which 15 g K2SO4/100 g H2O is equal to 0.15. A solution consisting of 39 g K2SO4 per 225 g H2O results to a solubility of 0.173. This is greater than the maximum amount of solute that can be dissolved at the given temperature of 40 degrees Celsius. This means that the resulting solution is supersaturated. It should also be noted that supersaturated solutions are usually prepared by heating the solution in order to dissolve the remaining solute then consequently cooling the solution to the desired temperature.

To determine the amount of K2SO4 precipitated out, use the formula:

(39 - x) g K2SO4/ 225 g H2O = 15 g K2SO4/ 100 g H2O

where, x is the grams of K2SO4 that would crystallize out. After applying the formula, we obtain 5.25 g of potassium sulfate crystals.