Solutions and Units of Concentration May 25, 2015
What is Solubility? Solubility: the maximum amount of solute that will dissolve in a certain amount of solvent at a given temperature Example: grams of salt (NaCl) in 100 g of water at C.
Factors Affecting Solubility 1. Nature of Solute / Solvent: Like dissolves like 2. Temperature: a. Solids/Liquids- Solubility increases with Temperature - Increase K.E. increases motion and collision between solute / solvent b. Gases - Solubility decreases with Temperature - Increase K.E. result in gas escaping to atmosphere
Factors Affecting Solubility 3. Pressure Factor: a. Solids/Liquids - Very little effect (esp. solids in liquids) - Solids and Liquids are already close together, extra pressure will not increase solubility b. Gases - Solubility increases with Pressure - Increase pressure squeezes gas solute into solvent. - HENRY S LAW: S = Solubility P = Press
Temperature & the Solubility of Gases The solubility of gases DECREASES at higher temperatures
SOLUBILITY FORMULAS amount of solute amount of solvent Given solubility = amount of solute amount of solvent unknown
SOLUBILITY PROBLEMS 1. The solubility of a solid is 15g / 100g of water. How many grams of the solid must be dissolved in 1 kg of water to make a saturated solution? (1 kg = 1000 g)
SOLUBILITY PROBLEMS amount of solute amount of solvent = amount of solute amount of solvent 15 g X ------- = ------- X=150g 100g 1000g
SOLUBILITY EXAMPLES 2. If you have 50g in 500g of water, using solubility from problem #1 (15 g / 100 g), is the solution saturated? 15g ------ 100g = X ------ 500g X = 75g is saturated No, 50 g is not saturated!
To read the graph: 1. Find the line for the substance. 2. The amount that dissolves at a given temp. is on the y- axis.
How much KNO 3 dissolves in 100g H 2 O at 50 o C? 1. Find the line (red) 2. Find the temperature and follow up to the line. (green) 3. Read across to the y- axis and this is the answer. (blue) 4. Since it is above the ½-way between 80 and 90, 87g KNO 3 will dissolve.
Types of Solutions: Saturated solution: point on the line Contains maximum amount of solute at given temp Contains what it should hold Supersaturated: above the line Contains more solute than a saturated solution Contains more than it should hold Unsaturated: below the line is Contains less solute than saturated solution Contains less solute than it could hold
To do Calculations: To calculate how much extra has been dissolved: Extra = Dissolved amt - saturated in soln (given value) line value @ that temp. To calculate how much more can be dissolved:? Much more = saturated - given line value @ temp value
Example 1: How much less KCl is dissolved at 20 o C than at 60 o C in 100g H 2 O? Read the line value: 32g at 20 o C Subtract it from the given value: 45g 32g = 13 g
Example 2: How much more KCl is required to saturate the solution if 25g are dissolved at 40 o C? Read the line value: 40g at 40 o C Subtract the given value: 40g -25 g = 15 g
Your turn! Use your graph 1) What is the solubility of potassium nitrate at 30 0 C? 46 g/100 g 2) How many grams of ammonia can I dissolve in 100 grams of water at a temperature of 45 0 C? 32 g 3) At what temperature is the solubility of sodium chloride the same as the solubility of potassium chloride? 35 deg C 4) How many grams of ammonium chloride would I need to make 100 grams of a saturated solution at 70 0 C? 60 g 5) What do all of the compounds that decreased in solubility over the temperature range in the graph have in common? All gases 6) What compound is least soluble at 40 0 C? SO 2
Colligative Properties When dissolving a solute in a solvent, the properties of the solvent are modified. Vapor pressure decreases Melting point Boiling point decreases increases Osmosis is possible (osmotic pressure) Colligative Properties are properties of a liquid that change when a solute is added The magnitude of the change depends on the number of solute particles in the solution, NOT on the identity of the solute particles.
Vapor Pressure Lowering for a Solution The diagram below shows how a phase diagram is affected by dissolving a solute in a solvent.
Adding a solute increases the boiling point.
Freezing Point Depression and Boiling Point Elevation Boiling Point Elevation T b =mk b (for water k b =0.51 o C/m) Freezing Point Depression T f =mk f (for water k f =1.86 o C/m) Note: m is the molality of the particles, so if the solute is ionic, multiply by the #of particles it dissociates to.
Example 1: Find the new freezing point of 3m NaCl in water. Freezing Point Depression T f =mk f (for water k f =1.86 o C/m) T f =(3m)(1.86 o C/m) 5.58 o C
Example 2: How many moles of ethylene glycol (C 2 H 6 O 2 ) should be added to 25 kg of water to lower the freezing point by 45 C? T f =mk f (for water k f =1.86 o C/m) 45=(m)(1.86 o C/m) = 24.2 m m = moles 24.2 = moles kg solvent 25 = 604 moles