Unit 6: Solution Chemistry

Content Outline: Colligative Properties: Boiling Point Elevation (6.9)

  1. Colligative Properties
  1. These are the properties of solutions that are dependent upon the concentration of solute, but not on the solution identity (name).
  1. Concentration for colligative properties is molality(m). Think mass.
  1. Molality is the number of moles of solute/kg (mass) of solvent.
  1. Boiling Point Elevation
  1. The difference between the Boiling points of the pure solvent and a solution containing a non-volatile, non-electrolyte solute in the same solvent.
  2. Boiling Pointis where the vapor pressure of the liquid is equal to the atmospheric pressure.
  1. If you change the atmospheric pressure, you change the boiling point of a liquid.

For example, water boils at a lower temperature in Denver, Colorado (the “Mile High City”… 1 mile above sea level) than at Mobile, Alabama. Remember, as the altitude increases, the pressure decreases. If the pressure decreases, the boiling point drops.

  1. Boiling Point Elevation isdirectly proportional to the molality concentration of the solution.
  1. If Molality increases the amount of Elevation increases (the temperature rises).

This elevation is referring to temperature; not altitude.

  1. Mathematically expressed as:

Δtb= Kbm

Kb = Boiling Point Constant for a pure solvent (OC/molality)

M = Molality of the solution (moles of solute/ kg solvent)

Tb = Numeric Amount of Temperature change (OC)

  1. Molal Boiling Point Constant (Kb)
  1. The Boiling Point Elevation of the solvent when a 1-molal solution is created by dissolving 1 mole of non-volatile, non-electrolyte solute in 1 kg of a solvent at 1 atm.

For example: The normal Boiling Point of Water is 100.0OC. The Boiling Point Constant for water is

0.51OC. So if a 1 molal solution is dissolved in water, the new Boiling Point is 0.51OC higher. If a 2-molal solution is dissolved in water, the new boiling Point becomes 1.02OC (2 x 0.51) and so forth. The Boiling Point is being elevated (rising).

1. The elevation is the result of more solvent molecules being “actively” involved in making

solvent bubbles. They cannotbreak the inter-molecular attractions with the solute to “free

from” the liquid and “escape” as a gas.

  1. The solute being dissolved in the solvent also causes the vapor pressure to drop. If the pressure drops, it will require more energy (heat) to get the solution to boil. You should be able to see the boiling point line on a phase diagram drops, and thus increases the boiling point on the graph.

a. Solute in solvent lowers vapor pressure  requires more energy input  raises Boiling Pt.

  1. Each individual solvent has its own unique molal boiling point constant (Kb).
  1. Calculating Freezing Point Depression
  1. The equation is: Δtb = Kbm
  2. If looking for Kb = Δtb/m

If looking for m= Δtb/Kb

Let’s use the same example as for freezing point as it is the same thought process in calculating.

For example, Antifreeze (Ethylene Glycol) is C2H6O2. It is a non-volatile, non-electrolyte solute that is added to pure water to elevate (increase) the boiling point of water so that it does not overheat in your car/truck engine and crack the engine block (metal) or spit the rubber hosing. If a 250.0 gram amount of antifreeze is added to 750.0 mL of H2O. What is the new Boiling Point?

Remember, for water 1 mL = 1 gram

Step 1: Determine the molar Mass for each substance.

For Antifreeze: C = 12.0 AMUs x 2 = 24.0 AMUs

H = 1.0 AMU x 6 = 6.0 AMUs

O = 16.0 AMUs x 2 = 32.0 AMUs

Total = 62.0 AMUs

For water: H = 1.0 AMU x 2 = 2.0 AMUs

O = 16.0 AMUs x 1 = 16.0 AMUs

Total = 18.0 AMUs

Remember: Unit Given X Unit wanted = Unit wanted

Unit Given

Step 2: Calculate the molarity for Antifreeze.

250.0 (g) C3H8O3 X 1 mol C2H6O2 = 4.03 mol of C2H6O2

62.0 g C2H6O2

Step 3: Calculate the molality for the amount of solvent provided.

4.03moles/.750 kg = 5.37 molal

Step 4: Take the molality and plug it into the equation. As the solvent is water, use the Kf of water: (-1.86).

Δtf= Kbm = (0.51) x(5.37) = 2.74OC

Step 5: Add the elevation value to the pure boiling point value for the solvent.

100.0OC (water) + 2.74OC = 102.74OC (New Boiling Point of water)

  1. Calculating Boiling Point Elevation for Electrolytes(Ionic solutions)
  1. Follow Steps 1 – 5 from above but in step 4 use the molal solution of the ions in place of the molal concentration for the non-electrolyte.

For example, a 0.100 molal solution of K2SO4  0.2 for K+ (as the are double) and 0.1 for SO4-

So… (0.51) (0.3) = 0.15OC

100.0OC + 0.15OC = 100.15 (New Boiling Point of ionized water solution)