Chapter 14: Solutions

Effect of Particle Size

(Rate of Dissolving Solids) A solid can only dissolve at a surface that is in contact with the solvent.Since smaller crystals have a higher surface to volume area, smaller crystals dissolve faster than larger ones.

Effect of Temperature

(Rate of Dissolving Solids) Increasing the temperature normally increases the rate of dissolution of most compounds.Solvent molecules strike the solid surface more often, causing the solid to dissolve more rapidly. The solute molecules are more easily separate from the solid due to a higher kinetic energy.

Effect of Solute Concentration

(Rate of Dissolving Solids) Rate is highest at higher concentration and decreases at lower concentration.As the solution approaches the saturation point, the rate of solute dissolving decreases.

Effect of Agitation/Stirring

(Rate of Dissolving Solids) Stirring a solution briskly breaks up a solid into smaller pieces, increasing surface area, thereby increasing the rate of dissolution.

Common Colligative Properties

1. Vapor Pressure Lowering Solutions have lower vapor pressures than pure solvent. 2. Boiling Point Elevation Solutions have higher boiling points than pure solvent. 3. Freezing Point Depression Solutions have lower freezing points than pure solvent. 4. Osmosis and Osmotic Pressure

Properties of a True Solution

1.A homogeneous mixture of two or more components whose ratio can be varied. 2. The dissolved solute is molecule or ionic in size (< 1 nm). 3.Can be colored or colorless, though solutions are usually transparent. 4. The solute remains dissolved and does not settle (precipitate) out of solution over time. 5. The solute can be separated from solvent by physical means (usually evaporation).

Molarity

A common unit for solution concentration due to convenience. ________ = mol solute/ L solution

b. water

A solution is prepared by adding 25 mL of ethyl alcohol to 75 mL of water. Which substance is the solvent? a. ethyl alcohol b. water

5.00% propanol

A solution is prepared by mixing 20.0 mL of propanol with enough water to produce 400.0 mL of solution. What is the volume percent of propanol? a. 20.0 % b. 2.00 % c. 5.00 % d. 10.0 % 20.0 mL propanol (solute volume) 400.0 mL solution (solution volume) Solve for volume percent Formula: volume % = (volume solute/ volume solution) x 100 (Number% propanol)

Colligative Property

A solution property that depends only on the number of solute particles not the nature of the particles

Dilution

Adding a solvent to a concentrated solution to make the solution less concentrated (i.e. dilute). When a solution is diluted, only the volume changes. The number of moles of solute remains constant. moles before dilution = moles after dilution Molarity 1 x Volume 1 = Molarity 2 x Volume 2 M1× V1= M2× V2

Vapor Pressure Lowering

Dissolving a solute in a solvent lowers the vapor pressure of the solvent. As a result, the solvent's boiling point is increased, while the freezing point of the solvent is lowered

Solutions: A Reaction Medium

The purpose of dissolving reactants in a solution is often to allow them to come in close contact to react. Example: Solid-solid reactions are generally very slow at ambient temperature KCl(s) + AgNO₃(s) → No Reaction By dissolving both compounds in water, the ions can collide with one another and react to form an insoluble compound. KCl(aq) + AgNO₃(aq) → AgCl (s) + KNO₃ (aq) K⁺(aq) + Cl⁻(aq) + Ag⁺(aq) + NO₃⁻(aq) → AgCl(s) +K⁺(aq) + NO₃⁻(aq)

Saturated and Unsaturated Solutions

There are limits to the solubility of a compound at a given temperature.

Demonstrating Osmotic Pressure

Water flows through the membrane into the more concentrated sugar solution, causing the solution to rise in the tube.

100.1 c

What is the boiling point of a solution prepared by dissolving 0.10 mol of sugar in 0.50 kg of water? (Boiling point of water is 100.0 ºC and Kb = 0.512 ºC/m) 0.10 mol/0.50 kg H₂O = 20 m ΔTb = 0.20 m x 0.512 ºC/m = 0.10 ºC Boiling point is always elevated by added solute, so the change in temperature is added to the boiling point of pure water. Tb = 100.0 ºC + 0.10 ºC =

b. 102.05 ºC

What is the boiling point of an aqueous solution that is 4.00 m in solute? Tb (pure water) = 100.0 ºC and Kb = 0.512 ºC/m ΔTb = 4.00 m x 0.512 ºC/m = 2.05 ºC Boiling point is always elevated by added solute, so the change in temperature is added to the boiling point of pure water. Tb = 100.0 ºC + 2.05 ºC = 102.05 ºC a. 100.00 ºC b. 102.05 ºC c. 97.95 ºC d. 2.05 ºC

a. -15.0 ºC

What is the freezing point of 100. g of C₂H₆O₂ dissolved in 200 g of H₂O? Tf(pure water) = 0.0 ºC and Kf = 1.86 ºC/m) ΔTf = 8.05 m x 1.86 ºC/m = 15.0 ºC Freezing point is always depressed by added solute, so the change in temperature is subtracted from the freezing point of pure water. a. -15.0 ºC b. 15.0 ºC c. -0.015 ºC d. -7.35 ºC

-0.37 ºC

What is the freezing point of a solution prepared by dissolving 0.10 mol of sugar in 0.50 kg of water? (Freezing point of water is 0.0 ºC and Kf = 1.86 ºC/m) Molarity = 0.10 mol / 0.50 kg H2O= 0.20 m ΔTf = 0.20 m x 1.86 ºC/m = 0.37 ºC Freezing point is always depressed by added solute, so the change in temperature is subtracted from the freezing point of pure water. Tb = 0.0 ºC - 0.37 ºC = -0.37 ºC

43.1 g Na2CO3

What is the mass of Na₂CO₃ needed to make 350.0 g of a 12.3% aqueous solution? Knowns: 12.3% solution (mass %) 350.0 g solution (solute + solvent mass) Solve for mass of solute (Na₂CO₃) Formula: Mass % = mass solute/ mass solution x 100 Mass solute = (mass % x mass soln)/ 100 = (Number g Na2CO3)

c. 3.37 m

What is the molality (m) of a solution prepared by dissolving 2.70 g of methanol (CH₃OH) in 25.0 g of water? a. 0.812 m b. 6.74 m c. 3.37 m d. 1.69 m

Example

What is the molality of a solution prepared by dissolving 0.10 mol of starch in 0.50 kg of water?

4.2 mL

What volume of 12 M HCl is needed to make 500.0 mL of a 0.10 M HCl? Knowns: 12 M HCl M₁ 10.10 M HCl M₂ 500.0 mL V₂ Solve for: volume of 12 M HCl V1 Formula: M1 × V1= M2 × V2 (Number mL)

b. 330. mL soln

What volume of a 3.0% H₂O₂ solution will contain 10.0 g of H₂O₂? 10.0 g H₂O₂ (desired solute mass) 3.0 m/v% Solve for volume of solution Formula: m/v% = (g solute / mL solution) x 100 mL solution = (g solute/ m/v %) x 100 a. 33.0 mL soln b. 330. mL soln c. 3.00 L soln d. 165 mL soln

3300 mL soda

What volume of soda that is 6.0 % by volume alcohol contains 200.0 mL of ethanol (CH₃CH₂OH)? 20.0 mL ethanol (solute volume) 6.0 volume % Solve for volume of solution Formula: volume % = (volume solute/ volume solution) x 100 volume soln = (volume solute / volume %) x 100 (Number mL soda)

Miscible

when two liquids dissolve in each other.

Immiscible

when two liquids do not dissolve one another. A mixture of oil and water is __________.

25% NaCl

Calculate the mass % of NaCl in a solution prepared by dissolving 50.0 g of NaCl in 150.0 g of H₂O. Knowns: 50.0 g NaCl (solute mass) 150.0 g H2O (solvent mass) 200.0 g solution (solute + solvent mass) Formula: Mass % = mass solute/ mass solution x 100 = (Number% NaCl)

0.080 M NaOH

Calculate the molarity of a NaOH solution prepared by mixing 100. mL of 0.20 M NaOH with 150 mL of H₂O. a. 2.0 M NaOH b. 0.050 M NaOH c. 0.080 M NaOH d. 12.5 M NaOH Knowns: 0.20 M NaOH M1 100 mL sln V1 100 + 150 = 250 mL V2 Solving for: molarity NaOH M2 Formula: M1× V1= M2× V2 (Number M NaOH)

0.502 M KCl

Calculate the molarity of a solution prepared by dissolving 9.35 g of KCl in enough water to prepare a 250.0 mL solution. 9.35 g KCl (solute mass) 250.0 mL solution (solution volume) Solve for molarity Formula: First... grams → moles. Then... Molarity = mol solute/ L solution (Number M KCl)

Colligative Properties

Calculating the change in boiling/freezing point of a solution ΔT = m x K

Molarity

Example: To prepare a 1.0 M KCl solution, 1.0 mol of KCl is dissolved in enough water to make 1.0 L of solution.

How is this possible?

Heating a solution can allow more to dissolve.Upon cooling to ambient temperature, the solution is supersaturated. These solutions are unstable -- disturbing the solutions can cause precipitation of solute. Some hotpacks release heat by crystallization of a supersaturated solution of sodium acetate.

a. 172 g AgCl

How many grams of AgCl will form by adding enough AgNO₃ to react fully with 1500. mL of 0.400 M BaCl₂ solution? 2 AgNO₃ (aq) + BaCl₂(aq) → 2 AgCl (s) + Ba(NO₃)₂ (aq) Plan: Volume BaCl₂ → mol BaCl₂ → mol AgCl → g AgCl a. 172 g AgCl b. 86.0 g AgCl c. 8.37 x 10⁻³ g AgCl d. 36.0 g AgCl

15.1 g KOH

How many grams of KOH are required to prepare 600.0 mL of a 0.450 M KOH solution? 600 mL (solution volume) 0.450 M (solution molarity) First... Solve for moles. Then...solve for grams using molarity and molar mass as conversion factors Formula: Molarity = mol solute/ L solution Mol solute = molarity x L soln (Number g KOH)

43.0 mL Hg(NO3)2

How many mL of 0.175 M Hg(NO₃)₂ are needed to precipitate 2.50 g of KI? Hg(NO₃)₂ (aq) + 2 KI (aq) → 2 KNO₃ (aq) + HgI2 (s) Plan: g KI → mol KI → mol Hg(NO₃)₂ → mL soln (Number mL Hg(NO3)2)

a. soluble

Predict the solubility of NaCl. a. soluble b. insoluble All Na⁺ salts are (a or b)

a. soluble

Predict the solubility of ammonium carbonate. a. soluble b. insoluble All NH⁴⁺ salts are (a or b).

b. insoluble

Predict the solubility of barium sulfate. a. soluble b. insoluble Most sulfates are (a or b), except Ba²⁺

b. insoluble

Predict the solubility of silver hydroxide. a. soluble b. insoluble Most hydroxides are (a or b).

a. soluble

Predict the solubility of silver nitrate. a. soluble b. insoluble All NO³⁻ salts are (a or b)

Pressure and Solubility

Pressure does not affect solubility of liquids or solids. Gas solubility in a liquid is proportional to the gas pressure over the liquid. Example: A bottle of root beer is under high pressure. As the bottle opens, the pressure decreases, and the bubbles formed indicate gas loss from the liquid.

0.45 g NaCl

Saline is a 0.9 m/v % NaCl solution. What mass of sodium chloride is needed to make 50 mL of saline? 50.0 mL solution (solution volume) 0.90 m/v% (mass/volume %) Solve for mass of solute (NaCl) Formula: m/v% = (g solute / mL solution) x 100 mass solute = (m/v % x mL soln) / 100 = (Number g NaCl)

Factors Affecting Solubility

Several ionic compounds dissolve in water, due to strong ion-dipole forces. The cation is attracted to the partially negative O atom.The anion is attracted to the partially positive H atoms.

Solution Stoichiometry

Similar to previous stoichiometry problems, but we can now use molarity as an additional conversion factor.

Molality

Since colligative properties depend on the number of particles in the solvent and not the identity, a new concentration unit is used when discussing colligative properties.

a. sugar

Soda is a mixture of sugar in water. Which substance is the solute? a. sugar b. water c. soda

Temperature and Solubility

Solubility increases with temperature for most solids. Solubility decreases with temperature for all gases. As a gas increases in temperature, the kinetic energy increases, which means it interacts less with the liquid, making it less easy to solvate.

Surface Area of Two Crystals

Surface area = 6(side)² = 6(0.1)² = 0.06 cm² 1000 cubes have a total surface area of 1000 x 0.06 cm² = 60 cm² What is the surface area of a 1 cm square crystal? Surface area = 6(side)² = 6(1)² = 6 cm²

saturated

Will a solution prepared by adding 9.0 g of KCl to 20.0 g of H₂O be saturated or unsaturated at 20 ºC? Using Table 14.3, 34.0 g of KCl will dissolve in 100.0 g of H2O at 20 ºC. 6.8 g of KCl will then dissolve in 20.0 g of water at that temperature. The KCl solution should be (saturated or unsaturated).

Air

___ Solute: gas Solvent: gas Phase: gas

Polar compounds

_____ _________ dissolve in polar solvents. Ethanol (CH₃OH) dissolves in water (HOH). "Like Dissolves Like"

Nonpolar compounds

________ _________ dissolve in nonpolar solvents. Carbon tetrachloride (CCl₄) dissolves in hexanes (CH₃(CH₂)₄CH₃). "Like Dissolves Like"

solution

a ________ does not always just refer to liquids. Example: Air is a ________ composed of N₂, O₂, Ar and CO₂N₂ is the solvent as it composes 78% of air.

Solution

a homogeneous mixture of one or more solutes and a solvent

Concentrated

a solution that contains a relatively large amount of dissolved solute. Example: A 12 M HCl solution is concentrated acid.

Dilute

a solution that contains a relatively small amount of dissolved solute. Example: A 0.1 M HCl solution is dilute acid.

Unsaturated solutions

contain less than the maximum amount of possible dissolved solute in a solvent. undissolved solute ↔ dissolved solute

Supersaturated solutions

contain more solute than needed to saturate a solution at a given temperature.

Saturated solutions

contain the maximum amount of dissolved solute in a solvent. Saturated solutions are still dynamic; dissolved solute is in equilibrium with undissolved solute. undissolved solute ↔ dissolved solute

Osmotic Pressure

difference in the amount of pressure necessary to apply to a solution to stop the flow of water due to osmosis and the atmospheric pressure.

Osmosis

diffusion of water from a dilute solution or pure water, through a semipermeable membrane into a solution of higher solute concentration.

Solvent

dissolving agent that is usually the most abundant substance in the mixture

Hypertonic

higher concentration of dissolved particles relative to cellular levels. (1.6% saline)

Hypotonic

lower concentration of dissolved particles relative to cellular levels. (0.2% saline)

Isotonic

same concentration of dissolved solute as a cell.(0.9% saline)

Solute

substance being dissolved

Solubility

the amount of a substance that will dissolve in a specific amount of solvent at a given temperature. Example 27 g KBr/100 g H₂O at 23 ºC