B. Boiling Point Elevation, Chang p. 520, Eq 12.6 T = Kb m or T = i Kb m Calculating Change in Boiling Point The change in boiling point is T in the formula above. The key to doing these problems is the calculation of molality and choosing the right constant (Kb) from the table - [Table 12.2 p. 521] (Always choose the constant based on the solvent and be sure to choose the boiling point constant for boiling point problems and freezing point constant for freezing point problems. Example 6: What is the change in boiling point if 3.69 g of K2SO4 was added to 100.0 mL of water? Solution: T = Kb m Kb = boiling point constant for water = 0.52 °C/m m= moles K2SO4 kg water (solvent only) moles K2SO4= 3.69 g x 1 mole = 0.0212 mole 174.20 g kg water = 100.0 mL x 1.00 g x 1 kg = 0.100 kg 1 mL 1000 g m K2SO4 = 0.0212 mol = 0.212 m 0.100 kg Remember that you need the molality of particles and when K2SO4 dissolves in water it dissociates into two K+ ions and one SO4 2– ion. Therefore, i = 3 and molality of particles = i m = 3 (0.212 m) = 0.636 m T = i m( 0.52 °C /m); T = 0.636 m (0.52 °C /m) T = 0.33 °C The answer has two significant figures because the Kb only has two. Calculating the Actual Boiling Point Problems involving actual boiling point are different because they ask for the boiling point of the solution rather than for the change in boiling point of the solvent. The process is initially the same. You must still determine the change in boiling point, but there is an extra step. The extra step is to add the change in boiling point, T, to the boiling point of the pure solvent. It will help you to remember to add if you remember the name of this colligative property is boiling point elevation. Example 7: What is the boiling point of a solution made by mixing 13.63 g ethanol and 540.00 g water? Solution: The sneaky part of this question has to do with both ethanol and water being listed as solvents. Remember that the solvent is the substance in greater quantity. In this case, it is obviously water. solute = 13.63 g ethanol (C2H5OH) solvent = 540.00 g water (H2O) T = Kb m The value of Kb is the value of the solvent (for water) = 0.52 °C/m The value of m is the value of particles of solute. Since ethanol is a nonelectrolyte, i = 1. The molality of ethanol is the same as the molality of particles. m = mole solute = mole ethanol kg solvent kg water mol ethanol = 13.63 g C2H5OH x 1 mole = 0.2959 mol 46.07 g kg water = 540.00 g x 1 kg = 0.54000 kg 1000 g m = 0.2959 mol = 0.5479 m 0.54000 kg T = Kb m [ Kb = 0.52 °C/ m]; T = 0.5479 m (0.52 °C/ m ) = 0.28 °C Don't forget (and it is the most common mistake) that this does not answer the question! boiling point (bp) solution = bp solvent + T = 100 °C + 0.28 °C = 100.28 °C Recall that the 100°C was defined as the boiling point of water; therefore it is an exact number with infinite significant figures. The significant figure rule for addition is to use the value with the fewest decimal places.