Colligative Properties & Solubility Rules Ariel Autrey, Katie Elicker, and Alyssa Truong

Colligative Properties Properties that depend on the number of particles dissolved in a given mass of solvent are called colligative properties They do not depend on the chemical nature of the solute or the solvent

The Four Colligative Properties 1.Vapor Pressure Reduction 2.Osmotic Pressure 3.Freezing Point Depression 4.Boiling Point Elevation

Vapor Pressure Reduction Nonvolatile solute: a solute that can not escape to the vapor phase. The solvent will go to the vapor phase, and the solution's volume will increase because the solvent will go into the vapor phase more easily than the solution will. The solution's volume increases in an attempt to decrease the vapor pressure.

Vapor Pressure Reduction Raoult's Law: A nonvolatile solute lowers the vapor pressure of a solvent because solute particles become surrounded by solvent particles when dissolved. This causes the particles of solvent to evaporate less. Therefore, the vapor pressure of a solution is less than that of the pure solvent it was created with.

Osmotic Pressure Osmotic pressure is the pressure required to prevent solvent from passing through a semipermeable membrane from the higher concentration to the lower concentration of solute. The result is a net flow of solvent molecules from the less concentrated solution to the more concentrated solution.

Osmotic Pressure As the solvent moves back and forth through the membrane, the levels of solution become uneven, and the process continues until the difference of pressures is large. Two solutions with identical osmotic pressures are called isotonic. Two solutions with a different osmotic pressures are said to be hypotonic.

Freezing Point Depression Ability of a dissolved solute to lower the freezing point of its solution Freezing point: the temperature at which the vapor pressures of the solid and liquid phases are the same solute + solvent solute molecules disrupt the orderly pattern of the solid More kinetic energy must be withdrawn from the solution than from the pure solvent for it to solidify

Freezing Point Depression Problem 1.60 g of naphthalene (C 10 H 8 ) is dissolved in 20.0 g of benzene. The freezing point of pure benzene is 5.5 o C, and the freezing point of the mixture is 2.8 o C. What is the molal freezing point depression constant, K f of benzene?

Step 1 Calculate the freezing point depression of benzene. T f = (Freezing point of pure solvent) - (Freezing point of solution) (5.5 o C) - (2.8 o C) = 2.7 o C

Step 2 Calculate the molal concentration of the solution -molality = moles of solute / kg of solvent -moles of naphthalene = (1.60 g) (1 mol / 128 g) = mol naphthalene -molality of solution = ( mol) / ( kg) = 0.625m

Answer Calculate K f of the solution. T f = (K f ) (m) (2.7 o C) = (K f ) (0.625 m) K f = 4.3 o C/m

Practice Problems 1.What is the freezing point depression when 85.3 g of oxygen gas is dissolved in 1500 g of water? K f (ºC/molal) is Ethylene glycol (C2H6O2) is the principal ingredient in antifreeze. How many grams of ethylene glycol will be needed to lower the freezing point of 2100 g of water by 20ºC? K f (ºC/molal) is

Answers to Practice Problems ºC kg

Boiling Point Elevation The difference in temperature between the boiling points of a solution and a pure solvent Boiling point of a substance is the temperature at which the vapor pressure of a liquid is equal to the external pressure on its surface. (one example of external pressure is atmospheric pressure)

Boiling Point Elevation T b = imK b T b = change in boiling point i =Van't Hoff factor m = molality (moles solute/ kg solvent) K b = boiling point elevation constant (water K b = o C/m) normal boiling point of water = 100. o C

Boiling Point Elevation Adding a nonvolatile solute reduces the vapor pressure of the solution A higher temperature is now necessary to get the vapor pressure of the solution up to atmospheric pressure so that the solution boils The amount by which the boiling temp. is raised is the boiling point elevation

Problem (boiling point elevation) g of sodium chloride is added to mL of water at 34 °C. How will this affect the boiling of the water? Assume the sodium chloride completely dissociates in the water. Given: density of water at 35 °C = g/mL K f water = 1.86 °C kg/mol

Step 1 žCalculate the molality of the NaCl molality (m) of NaCl = moles of NaCl/kg water m NaCl = moles of NaCl/kg water m NaCl = mol/0.219 kg m NaCl = mol/kg

Step 2 Determine the Van't Hoff Factor The Van't Hoff Factor, i, is a constant associated with the amount of dissociation of the solute in the solvent. For substances that do not dissociate in water, such as sugar, i = 1. For solutes that completely dissociate into two ions, i = 2. NaCl completely dissociates into the two ions, Na + and Cl -. Therefore, i = 2.

Step 3 Find ΔT ΔT = imK f ΔT = 2 x 0.51 °C kg/mol x mol/kg ΔT = 2.53 °C

Answer Adding g of NaCl to mL of water will lower the boiling point 2.53°C This means the new boiling point is 36.53°C o Because the original boiling point was 34°C, you have to add 2.53°C to the original boiling point to get the new boiling point. ( Elevation-Example-Problem.htm

Practice Problems 1. What is the boiling point elevation when 11.4 g of ammonia (NH 3 ) is dissolved in 200. g of water? K b for water is 0.52 °C/m. 2. Determine the molecular mass of a non- ionizing, non-volatile solute if 0.546g of it are dissolved in 15.0g of benzene ( K f =5.12 ºC/ F.P. = 5.5ºC) and the freezing point depression was 0.240ºC.

Answer to Practice Problem 1 1. Δt = 1.74 °C A) Determine molality of 11.4 g of ammonia in 200. g of water: 11.4 g / g/mol = mol mol / kg = m B) Determine bp elevation: Δt = i K b m Δt = (1) (0.52 °C/m) ( m) Δt = 1.74 °C

Answer to Practice Problem g/mol m= T/kf m=0.240C/5.12C/m m= mol/kg mol/kgx0.0150kg=7.03x10^-4mol 0.546g/ mol=776g/mol

Solubility Rules The solubility of a substance varies from one substance to the next. The solubility of a substance can be determined on the solubility rules table. If a substance is mainly soluble in water, it is aqueous. If a substance is mainly insoluble in water, it is a solid.

Solubility Rules An example of a soluble substance is KCl because according to the solubility rules table, all chlorates are soluble. An example of an insoluble substance is MgSO 3 because according to the solubility rules table, all sulfites are insoluble except those of the IA elements and NH 4 + o Because Mg +2 is not part of the IA elements, the substance is insoluble.

Solubility Problems Use the solubility rules table to find out if the substance is soluble or insoluble. 1.HCl 2.Ba(OH) 2 3.(NH 4 ) 2 CrO 4 4.CaCrO 4 5.AgI

Complete Ionic Equations Complete Ionic Equations (CIE) are used to describe the complete reaction and show all of the reactants and products as ions. Only split the aqueous substances on both sides of the reaction.

How to Write a CIE 1.Start with a balanced equation. 2.Break up all of the aqueous compounds into ions on both sides of the equation. 3.Write all of the ions together to form a CIE. 4.Reminders: a. Make sure you count for the coefficients in front of the compounds. b. Leave all solid, liquid, and gaseous compounds alone. c. Be sure to keep all of the (s), (l), (g), (aq) labels on the ions and compounds!

Example: Writing a CIE Mixing together the aqueous solutions Sodium Phosphate and Calcium Chloride makes an insoluble white solid, Calcium Phosphate. The balanced equation looks like this: 2 Na 3 PO 4 (aq) + 3 CaCl 2 (aq) 6 NaCl (aq) + Ca 3 (PO 4 ) 2 (s)

Example: Writing a CIE Break up the aqueous compounds on the reactants side of the equation: 2 Na 3 PO 4 (aq) has 6 sodium ions and 2 phosphate anions. This looks like 6 Na +1 (aq) + 2 PO 4 -3 (aq) 3 CaCl 2 (aq) has 3 calcium ions and 6 chloride anions. This looks like 3 Ca +2 (aq) + 6 Cl -1 (aq)

Example: Writing a CIE Break up the aqueous compounds on the products side of the equation: 6 NaCl (aq) has 6 sodium ions and 6 chloride anions. This looks like 6 Na +1 (aq) + 6 Cl -1 (aq) Because Ca 3 (PO 4 ) 2 (s) is a solid, the compound does not change; it stays together.

Example: Writing a CIE Now, put all of the information together. 6 Na +1 (aq) + 6 Cl -1 (aq) + 3 Ca +2 (aq) + 2 PO 4 -3 (aq) 6 Na +1 (aq) + 6 Cl -1 (aq) + Ca 3 (PO 4 ) 2 (s) This is your complete ionic equation.

Net Ionic Equations A net ionic equation (NIE) is a little different than a CIE. In a NIE, the ions that are not used in the equation to make a solid, called spectator ions, are not written. The spectator ions are present during the reaction, but were not actively participating in the reaction.

How to write a NIE 1.Start with a balanced equation. 2.Write a complete ionic equation. 3.Remove all the spectator ions from both sides of the reaction. 4.All of the leftover ions make the NIE. a. Reminder: Don't forget the (aq), (s), (l), (g) labels! b. Hint: the spectator ions always look exactly the same on both sides of the reaction. They will be in the same state and have the same formula, charge, and number of ions.

Example: Writing a NIE Using the same equation we used to write a CIE, write a NIE. Since we have already written the CIE within the last example, let's remind ourselves what it looks like: 6 Na +1 (aq) + 6 Cl -1 (aq) + 3 Ca +2 (aq) + 2 PO 4 -3 (aq) 6 Na +1 (aq) + 6 Cl -1 (aq) + Ca 3 PO 4 (s)

Example: Writing a NIE Identify the spectator ions, and cross them out. 6 Na +1 (aq) + 6 Cl -1 (aq) + 3 Ca +2 (aq) + 2 PO 4 -3 (aq) 6 Na +1 (aq) + 6 Cl -1 (aq) + Ca 3 (PO 4 ) 2 (s) When the spectators are crossed out: 6 Na +1 (aq) + 6 Cl -1 (aq) + 3 Ca +2 (aq) + 2 PO 4 -3 (aq) 6 Na +1 (aq) + 6 Cl -1 (aq) + Ca 3 (PO 4 ) 2 (s)

Example: Writing a NIE This will leave you with the net ionic equation: 3 Ca +2 (aq) + 2 PO 4 -3 (aq) Ca 3 (PO 4 ) 2 (s)

Practice: Writing CEIs and NEIs Balance the equations, if needed, and write a complete ionic equation and a net ionic equation. NaOH (aq) + H 2 SO 4 (aq) Na 2 SO 4 (aq) + H 2 O (l) (NH 4 ) 2 CO 3 (aq) + Al(NO 3 ) 3 (aq) NH 4 NO 3 (aq) + Al 2 (CO 3 ) 3 (s)

Complete Ionic Equation: 2 Na + (aq) + 2 OH - (aq) + 2 H + (aq) + SO 4 2- (aq) 2 Na + (aq) + SO 4 2- (aq) + 2 H 2 O (l) Net Ionic Equation: 2 OH - (aq) + 2 H + (aq) 2 H 2 O (l) Which Simplifies To: OH - (aq) + H + (aq) H 2 O (l) Answers to #1

Wrong Answer? Make sure these spectator ions are the ones you crossed out: 2 Na + (aq) + 2 OH - (aq) + 2 H + (aq) + SO 4 2- (aq) 2 Na + (aq) + SO 4 2- (aq) + 2 H 2 O (l) That should leave you with this net ionic equation: 2 OH - (aq) + 2 H + (aq) 2 H 2 O (l) Because of the same coefficients in front, the NIE becomes: OH - (aq) + H + (aq) H 2 O (l)

Answers to #2 Complete Ionic Equation: 6 NH 4 + (aq) + 3 CO 3 2- (aq) + 2 Al 3+ (aq) + 6 NO 3 - (aq) 6 NH 4 + (aq) + 6 NO 3 - (aq) + Al 2 (CO 3 ) 3 (s) Net Ionic Equation: 2 Al 3+ (aq) + 3 CO 3 2- (aq) Al 2 (CO 3 ) 3 (s) m

Wrong Answer? Make sure these spectator ions are the ones you crossed out: 6 NH 4 + (aq) + 3 CO 3 2- (aq) + 2 Al 3+ (aq) + 6 NO 3 - (aq) 6 NH 4 + (aq) + 6 NO 3 - (aq) + Al 2 (CO 3 ) 3 (s) That should leave you with this net ionic equation: 2 Al 3+ (aq) + 3 CO 3 2- (aq) Al 2 (CO 3 ) 3 (s)