Chemistry 1 Chapter 18 - Solution Formation

Factors affecting rate of solution: Agitation brings fresh solvent in contact w/ solute does not affect amount that dissolves

Factors affecting rate of solution: Temperature heat increases rate of solution by increasing kinetic energy and motion of particles

Factors affecting rate of solution: Particle size smaller particles dissolve faster because of greater surface area

Solubility - a measure of the amount of solute that will dissolve in a given amount of solvent. we usually use g solute/100 g solvent *See table of solubility on page 504 0ºC 20ºC 100ºC Ba(OH)2 1.67 31.89 ------ NaCl 35.7 36.0 39.2 Sucrose 179 230.9 487

Factors affecting solubility: Nature of the solute and the solvent like dissolves like

Factors affecting solubility: Temperature Most solids become more soluble at a higher temperature. A few become less soluble.

Factors affecting solubility: solubility of gases decreases as temperature increases

Factors affecting solubility: Pressure (for solutions of gases in liquids) increasing pressure increases solubility of gas

Saturated solution -contains the maximum amount of solute that will dissolve in a given amount of solvent at a given temperature. Unsaturated solution- more solute could be dissolved -indicated by no solid present Supersaturated solution- more solute is dissolved than is theoretically possible at a given temperature.

EXAMPLE: How could you determine if a solution is saturated, unsaturated or supersaturated? If solid is present – it is Saturated If no solid is present – it is unsaturated or supersaturated To determine which, add a crystal Supersaturated will crystallize Unsaturated will dissolve the crystal DEMO

Cloud seeding is a practical example of supersaturation.

Miscible - liquids that are soluble in each other Ex. ethanol and water Immiscible- liquids that are not soluble in each other Ex. oil and water

Henry’s Law- At a given temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. S1 = S2 P1 P2

EXAMPLE: At 25oC, the solubility of a certain gas in H2O is 0.75 g/L of H2O at a pressure of 1.0 atm. What is the solubility of the same gas at 10.0 atm pressure? S1 = 0.75 g/L P1 = 1.0 atm S2 = ? P2 = 10.0 atm S1 S2 P1 P2 = 0.75 x 1.0 10.0 x = 7.5 g/L =

Concentration -a measure of the amount of solute per given amount of solvent or solution. Dilute solution- small amount of solute Concentrated solution- large amount of solute

Molarity(M) = moles of solute liters of solution A 1 M solution contains 1 mole of solute per 1 L of solution. A 0.5 M NaCl solution has 0.5 mol NaCl in 1 L total solution. n V =

EXAMPLES: What is the concentration in molarity of a solution made with 1.25 mol NaOH in 4.0 L of solution? n V # mol # L M = = M = ? n = 1.25 mol V = 4.0 L M = 1.25 mol 4.0 L = 0.3125 M = 0.31 M

How would you make 500. mL of 6.0 M NaOH? How many grams of NaOH would be necessary? n V V = 500. mL = 0.500 L M = M = 6.0 M n = M ∙ V = 6.0 M ∙ 0.500 L n = ? = 3.0 mol Na 23.0 O 16.0 H 1.0 3 mol 40.0 g = 120 g NaOH 1 mol 40.0 g/mol

Dilutions Diluting a solution won’t change the number of moles of solute. It only changes the concentration. M1V1 = M2V2 This formula works only for dilution of a solution!!!! Don’t use it for a reaction!

EXAMPLE: For a lab we need 500. mL of 4.25 M HCl. Concentrated HCl is 12.0 M. How do we make up the solution? M1V1 = M2V2 M1 = 4.25 M V1 = 500. mL M2 = 12.0 M V2 = ? (4.25M)(500. mL) = (12.0 M)(V2) V2 = 177 mL

Solution Stoichiometry use Molarity as a conversion factor# # mol 1 L 1 L # mol

EXAMPLES: What volume of 0.246 M nitric acid is required to react completely with 0.386 L of 0.0515 M Ba(OH)2? 2HNO3 + Ba(OH)2 → Ba(NO3)2 + 2H2O 0.386 L Ba(OH)2 0.0515 mol Ba(OH)2 2 mol HNO3 1 L HNO3 1 L Ba(OH)2 1 mol Ba(OH)2 0.246 mol HNO3 molarity mol-mol ratio molarity = 0.162 L HNO3

Calculate the number of grams of carbon dioxide that can react with 0 Calculate the number of grams of carbon dioxide that can react with 0.135 L of a 0.357 M solution of potassium hydroxide, according to the following reaction: 2KOH + CO2  K2CO3 + H2O 0.135 L KOH 0.357 mol KOH 1 mol CO2 44.0 g CO2 1 L KOH 2 mol KOH 1 mol CO2 = 1.06 g CO2 molarity mol-mol ratio GFM

% Solutions Percent by volume % v/v = volume solute (mL) x 100 volume solution (mL) Percent mass/volume % m/v = mass solute(g) x 100

EXAMPLES: Rubbing alcohol is 70.0% isopropyl alcohol by volume. How many mL of isopropyl alcohol are in a 250 mL bottle of this solution? % v/v = volume solute (mL) x 100 volume solution (mL) % v/v = 70.0 % v solution = 250 mL v solute = X 70.0 % = __X__ x 100 250 mL X = 175 mL

If the % mass/volume (m/v) for a solute is 6 If the % mass/volume (m/v) for a solute is 6.80 %, and the volume of the solution is 42.6 mL, what mass of the salt is present? % m/v = mass solute (g) x 100 volume solution (mL) % v/v = 6.80 % v solution = 42.6 mL g solute = X 6.80 % = __X__ x 100 42.6 mL X = 2.8968 g = 2.90 g

Colligative Properties- properties that depend on the number of dissolved particles freezing point depression boiling point elevation vapor pressure lowering osmotic pressure

Most automobile engines use a water cooling system to dissipate engine heat. However, water freezes at 0 °C. With pure water as the coolant, automobiles could operate only in regions with temperatures above 0 °C . Instead, ethylene glycol or propylene glycol is mixed with the water as an "antifreeze." 

Ethylene glycol is non-corrosive to metal and rubber; mixes with water in all ratios, boils at 198 °C and freezes at -13 °C, has no odor, and is nonflammable and inexpensive. However, ethylene glycol is toxic, and its sweet taste is particularly attractive to pets and children. Because of its toxicity, propylene glycol is now used as an alternative to ethylene glycol. The antifreeze lowers the freezing point and raises the boiling point of the solution above that of pure water, permitting operation of the engine at a higher temperature.

An ice storm is coming. Your car’s radiator has no antifreeze in it; all the stores are closed, and you must use whatever you have around the house to save your car’s engine. It would be ideal to have antifreeze (ethylene glycol) but it’s quantity not quality that counts here.

The next morning you wonder if what you did saved you thousands of dollars on a new engine.

Rule one: What you add has to dissolve in water. Rule two: If you add a solid that dissolves in water, it doesn’t matter what it is, just the amount. Rule three: Just like antifreeze, your goal is to replace about ½ of the water with a solid or liquid that is miscible with water. Examples 1: Sugar, salt, baking soda, shampoo, laundry detergent, pancake syrup. Examples 2: Rubbing alcohol, brake fluid

How does it work? Get in the way. Create Disorder Freezing point depression How does it work? Get in the way. Create Disorder Reason 1: As the water tries to freeze, the other molecules get in the way. Reason 2: By adding these other substances, you’ve added disorder to the mixture. Nature tends to favor disorder (entropy). When water tries to freeze, it has to get organized, which will take more energy.

Vapor Pressure- Pressure of a gas above a liquid in a closed system The vapor pressure of a liquid decreases when solute particles are added because some of the H2O molecules at the surface are replaced by solute particles which don’t vaporize.

This decreases the number of H2O molecules that can escape and become gas. The more particles dissolved, the lower the vapor pressure becomes.

An ionic substance that ionized into 3 particles (Ex An ionic substance that ionized into 3 particles (Ex. CaCl2) has 3 times the effect of a molecular substance that does not ionize. CaCl2 (l) → Ca2+ (aq) + 2Cl- (aq) C12H22O11 (l) → C12H22O11 (aq)

Boiling Point - temperature at which the vapor pressure of a liquid equals atmospheric pressure The addition of a solute increases the boiling point of a solution because the vapor pressure of the solution is lower. It must reach a higher temperature before both vapor pressures are equal. The boiling point goes up 0.512ºC for every mole of particles dissolved in a kilogram of water.

Freezing Point - temperature at which the vapor pressure of the solid and liquid are equal. The addition of a solute lowers the freezing point of a solvent. Each mole of solute particles, per kg of water, decreases the freezing point by 1.86ºC. This is caused by the disruption of intermolecular forces between solvent molecules

EXAMPLE: Molality (m)- moles solute Kg solvent Calculate the molality of a solution formed by dissolving 20.0 g of NaOH in 2000.g water. m = ? mol Kg m = mol solute = 20.0 g 1 mol 0.500 mol = 40.0 g 0.5 mol 2.000 Kg m = Kg solvent = 2000. g = 2.000 Kg = 0.250 m

Calculating BP elevation & FP depression Tb = (Kb)(m)(i) Tf = (Kf)(m)(i) Tb = boiling pt. elevation Tf = freezing pt. depression m = molality Kb = molal boiling pt. elevation constant -for water = 0.512 Kf = molal freezing pt. depression constant -for water = 1.86 i = van’t Hoff factor (number of particles or ions) - for NaCl, i = 2 - for MgCl2, i = 3 - for C12H22O11, i = 1 i is always 1 for nonelectrolytes (they don’t ionize) After you find ΔT, add or subtract it to or from the normal BP or FP.

Calculate the i value for each of the following: CaCl2 CaCl2 → Ca2+ + 2Cl- Al2(SO4)3 NaCl i = 3 i = 5 i = 2

EXAMPLE: Tf = (1.86)(0.855 m)(2) Tf = 3.1806 ºC Calculate the freezing point of a solution that contains 100.g of NaCl in 2.00 kg H2O. Tf = (Kf)(m)(i) 1 mol NaCl 100. g NaCl = 1.71 mol NaCl Kf = m = i = 1.86 0.855 m 2 58.5 g NaCl Tf = (1.86)(0.855 m)(2) Tf = 3.1806 ºC Tf = 0 ºC – 3.1806 ºC = -3.1806 ºC

EXAMPLE: Tb = (0.512)(0.855 m)(2) Tb = 0.8755 ºC Calculate the boiling point. Tb = (Kb)(m)(i) Tb = (0.512)(0.855 m)(2) Kb = m = i = 0.512 0.855 m 2 Tb = 0.8755 ºC Tb = 100 ºC + 0.8755 ºC = 100.88 ºC

We can use freezing or boiling points to calculate the molecular mass of an unknown solute. Use the FP or BP to calculate molality Use molality to find moles Use g/moles to get molar mass

EXAMPLE: Find the molecular mass (GFM) of a molecular compound that has a freezing point of -0.460ºC when 5.20 g is dissolved in 500. g water. Tf = (Kf)(m)(i) 0.460 ºC = (1.86)(x)(1) mol Kg x = 0.86 m m = Tf = Kf = m = i = 0.460 ºC 1.86 x 1 mol = Kg∙m mol = (0.500 Kg)(0.86m) = 0.43 mol g mol 5.20 g 0.43 mol GFM = = = 12 g/mol