Chapter 14 Mixtures & Solutions
Mixture Types Heterogeneous Mixtures: (don’t blend smoothly, parts are visibly distinct) – Suspensions: mixtures with particles that settle out when left to stand. Can be filtered. – Thixotropic mixtures: substances that separate into solid and liquid parts. When mixed, they flow but can be still like a solid. (toothpaste)
Mixture Types – Colloids:
These particles do not settle out and cannot be filtered but the particles can be made to clump together and settle. • Smaller size than suspension particles.
– Brownian
Motion: random movements of particles in a liquid colloid. – Tyndall effect: a beam of light will spread out through a colloid. • Ex: Sunbeam in dust.
(Homogeneous Mixtures) Solution Formation
Solutions depend on: – What the solute and the solvent are – Whether a substance will dissolve or not. – How much will dissolve. Any 2 states of matter can make a mixture. Some mixtures will have multiple solutes. Water is the most common solvent.
Liquids Solutions are not always a solid dissolving in a liquid. Miscible means the that two liquids can dissolve in each other. Immiscible means they can’t.
Concentration This is a measure of the amount of solute dissolved in a certain amount of solvent. A concentrated solution has a large amount of solute. A dilute solution has a small amount of solute.
Concentration Ex: 1 g NaCl per 100 g H2O might be described as dilute. 30 g of NaCl per 100 g H2O would be described a concentrated compared to the one above. - Liquids, though, are not usually measured in grams. - The measurement is usually liters. - Also, the amount of solute is usually in moles.
Molarity The number of moles of solute in 1 Liter of the solution. M = moles/Liter What is the molarity of a solution with 2.0 moles of NaCl in 4.0 Liters of solution? (0.50 M) What is the molarity of a solution with 3.0 moles dissolved in 250 mL of solution? (12 M)
Making solutions 10.3 g of KI are dissolved in a small amount of water then diluted to 250 mL. What is the concentration? 10.3 g KI
1 mol KI = 0.0620 mol KI 166.00 g KI
0.0620 mol KI 0.25 L
= 0.25 M
Making Solutions How many grams of KI are needed to make 1.5 L of a 0.50 M KI solution? 1.5 L
0.50 mol KI 1L
= .75 mol KI
.75 mol KI 166.00 g KI = 120 g KI 1 mol KI
Making Dilutions Standard solutions of certain concentrations are found in the lab, but you may not need such a strong concentration. To get the concentration you need, you have to know how much solute to add. (solute doesn’t change)
Making Dilutions You take the molarity of the solution you have and the volume and molarity of the solution you need to determine how much you will dilute. Ex: How many milliliters of a stock solution of 2.00 M MgSO4 would you need to prepare 100.0 mL of 0.400 M MgSO4?
Making Dilutions To do this you will use this formula , M 1 x V1 = M 2 x V2 Knowns: Unknowns: M1 = 2.00M MgSO4 V1 = ? M2 = 0.400M MgSO4 V2 = 100.0 mL
Making Dilutions Substitute into the formula: V1 = M 2 x V2 M1 V1 = 0.400M x 100.0mL 2.00M V1 = 20.0mL
Practice 2.0 L of a 0.88 M solution are diluted to 3.8 L. What is the new molarity? You have 150 mL of 6.0 M HCl. What volume of 1.3 M HCl can you make?
Practice 2.0 L of a 0.88 M solution are diluted to 3.8 L. What is the new molarity? (2.0 L)(0.88 M) = (3.8 L)(M2) M2 = 0.46 M You have 150 mL of 6.0 M HCl. What volume of 1.3 M HCl can you make? (6.0 M)(150 mL) = (1.3 M)(V2) V2 = 692 mL
Percent solutions Percent means per 100 so‌ Percent by volume = Volume of solute x 100% Volume of solution indicated %(v/v) What is the percent solution if 25 mL of CH3OH is diluted to 150 mL with % volume = 25 mL x 100 = 16.67%(v/v) water? 150 mL
Percent solutions Percent by mass = Mass of solute(g) x 100% mass of solution(g) Indicated %(m/v) **for water 1 mL = 1 g 4.8 g of NaCl are dissolved in 82 mL of water. What is the percent of the solution? % mass = 4.8 g x 100 = 5.53%(m/v) 82 g + 4.8 g
Mole Fractions Another way to express solution concentration is with mole fractions. To get the mole fraction of a solute or solvent, you have to know the moles of each that are present. To find the mole fraction of solute, take the moles of solute over the total moles of solute and solvent.
Mole Fractions Example: Compute the mole fraction of each component in a solution of 1.25 mol ethylene glycol (EG) and 4.00 mol of water. Mole fraction EG = 1.25 mol = 0.238 1.25 mol + 4.00 mol Mole fraction H2O =
4.00 mol
= 0.762
1.25 mol + 4.00 mol
Solution Formation/ Solvation In order to dissolve the solvent, molecules must come in contact with the solute. -Stirring moves fresh solvent next to the solute. -The solvent touches more of the surface area of the solute if it is ground up.
Dissolving ionic/molecular solutes in water
Heat of Solution When the attractive forces in a solute are overcome, energy is required to break those attractions. When the solute and solvent meet, energy is released. – If more energy is absorbed to break the attractions in the solute, the reaction will be endothermic. (feels cold) – If more energy is released when the solute and solvent mix, the reaction is exothermic. (feels hot)
Temperature and Solutions A higher temperature makes the molecules of the solvent move around faster and contact the solute harder and more often. - It speeds up dissolving, and usually increases the amount that will dissolve.
How Much Solute Dissolves? Solubility- The maximum amount of substance that will dissolve at that temperature (usually g/L). Saturated solution- Contains the maximum amount of solid dissolved. Unsaturated solution- Can dissolve more solute. Supersaturated- A solution that is temporarily holding more solute than it can at that temperature; a seed crystal will make it come out.
Changing Saturation The saturation level of solution can be changed in several ways. A saturated solution can be made unsaturated by adding solvent or by increasing the temperature of the solution. An unsaturated solution can be made saturated by adding solute.
What affects solubility? For solids in liquids as the temperature increases, the solubility increases. For gases in a liquid as the temperature increases, the solubility decreases. AND as the pressure increases, the solubility increases.
Solubility Curves -This graph shows the solubility of many solids dissolved in water. (aqueous solutions)
Solubility Curves
-This graph shows solubility curves of gases dissolve in water.
NO
Henry’s Law This law states that at a given temperature the solubility (S) of a gas in a liquid is directly proportional to the pressure (P) above the liquid. S1 = S2 P1 P2 -This formula is a relationship between solubility and pressure of a gas .
Henry’s Law As pressure increases, solubility increases, and as pressure decreases, solubility decreases. As long as you have 3 amounts from the formula, you can solve for the unknown.
Henry’s Law Ex: If the solubility of a gas in water is 0.77g/L at 3.5 atm of pressure, what is its solubility (in g/L) at 1.0 atm of pressure? (Temp. is constant) Knowns: Unknowns: P1 = 3.5 atm S2 = ? g/L S1 = 0.77 g/L P2 = 1.0 atm
Henry’s Law 0.77 g/L = S2 3.5 atm 1 atm S2 = (0.77 g/L) x 1 atm 3.5 atm S2 = 0.22 g/L
Colligative Properties These are properties that depend on how many solute particles are present. These properties include freezing point depression, boiling point elevation, and vapor pressure depression of a solution.
Vapor Pressure The bonds between the molecules keep them from escaping. In a solution, some of the solvent is busy keeping the solute dissolved, so it can’t evaporate. The solute lowers the vapor pressure. When ionic compounds are dissolved electrolytes form as the ions become free. In NaCl, there are two ions that come from the compound. (Na+ and Cl-)
Vapor Pressure The ions reduce how much kinetic energy the solvent particles have. If you have more ions, the vapor pressure is lowered even more. With CaCl2, there are 3 ions present when it breaks up. CaCl2 ďƒ Ca2+ + 2ClMore ions means an even lower vapor pressure.
Boiling Point Elevation The vapor pressure determines the boiling point. Lower vapor pressure = higher boiling point. Salt water boils above 100ºC The solute interferes with the solvent’s vaporization. It will require more energy for the solvent particles to break away.
Freezing Point Depression Solids form when molecules make an orderly pattern. The solute molecules break up the orderly pattern of a solvent. This makes the freezing point lower. Salt water freezes below 0ยบC How much the freezing point lowers depends on how many solute particles are dissolved in the solvent.
Molality Because colligative properties only depend on the concentration of the solute, we look at something called molality. This is the number of moles of a solute dissolved per kilogram of solvent. Molality = moles of solute = moles of solute kg solvent 1000g solvent NOT the same as molarity.
Molality Ex: What is the molality of a solution that has 2.0 moles of KI dissolved in 500. g of H2O? - the mass of the solvent should be in kg, so convert 500. g to kg before calculating. molality = 2.0 mol KI .500 kg
= 4.0 m KI
Molality Practice: Find the molality of a solution that has 131g of NaNO3 dissolved in 750 grams of water First convert grams of solute to moles: 131g NaNO3 1 mol NaNO3 = 1.54 mol 84.99 g NaNO3 NaNO3 Then solve for molality: 1.54 mol NaNO3 = 2.1 m NaNO3
Why use molality? The size of the change in boiling point is determined by the molality. ∆ T b = Kb x m ∆ Tb is the change in the boiling point Kb is a constant determined by the solvent (pg 500).
m is the molality of the solution. The particle molality may need to be calculated if the solute is ionic.
Why use molality? Example: What is the boiling point of a 1.50m NaCl solution? ∆ Tb = Kb x m (look up Kb for water, p. 500) NaCl Na+(aq)+ Cl-(aq) (2 moles of particles are formed when in solution)
Particle molality = 2 x 1.50m = 3.00m ∆ Tb = 0.512oC/m x 3.00m = 1.54oC (New boiling point of water is 101.54oC)
What about Freezing? The size of the change in freezing point is determined by the molality. ∆ T f = Kf x m ∆ Tf is the change in the freezing point Kf is a constant determined by the solvent (pg 502).
m is the molality of the solution. Particle molality also needs to be considered.
What about freezing point? Example: What is the freezing point of a solution that contains 2.0 mol CaCl2 in 800.0 g of water? -This problem doesn’t give you the molality, so you will have to first calculate that to get the freezing point depression.
What about freezing point? Molality = 2.00 mol CaCl2 = 2.50m CaCl2 .800 kg CaCl2 = Ca2+(aq) + 2Cl-(aq) (3 moles ions)
Παρτιχλε µολαλιτψ = 2.50m ξ 3 = 7.50m ∆ Tf = 1.86oC/m x 7.50m = 14.0oC (Since the freezing point is lowered, take this amount away from the freezing pt. of the solvent which is water)
New freezing point is -14.0oC