PRINCIPLES OF CHEMISTRY II CHEM 1212 CHAPTER 12

PRINCIPLES OF CHEMISTRY II CHEM 1212 CHAPTER 12
DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university

CHAPTER 12 SOLUTIONS

SOLUTION - A homogeneous mixture of two or more substances Solvent
- The substance present in the greatest quantity Solute - The other substance(s) dissolved in the solvent

SOLUTION - Solutions can exist in any of the physical states
Solid Solution - dental fillings, metal alloys (steel), polymers Liquid Solution - sugar in water, salt in water, wine, vinegar Gas Solution - air (O2, Ar, etc. in N2), - NOx, SO2, CO2 in the atmosphere

CONCENTRATION OF SOLUTIONS
- The amount of solute dissolved in a given quantity of solvent or solution Molarity (M: molar) - The number of moles of solute per liter of solution - A solution of 1.00 M (read as 1.00 molar) contains 1.00 mole of solute per liter of solution

Calculate the molarity of a solution made by dissolving 2.56 g of NaCl in enough water to make 2.00 L of solution - Calculate moles of NaCl using grams and molar mass Convert volume of solution to liters - Calculate molarity using moles and liters

After dissolving 1.56 g of NaOH in a certain volume of water, the resulting solution had a concentration of 1.60 M. Calculate the volume of the resulting NaOH solution - Convert grams NaOH to moles using molar mass - Calculate volume (L) using moles and molarity

- Fraction of moles of a component of solution
CONCENTRATION OF SOLUTIONS Mole Fraction (χ) - Fraction of moles of a component of solution The sum of mole fractions of all components = 1

Given that the total moles of an aqueous solution of NaCl and other solutes is 1.75 mol. Calculate the mole fraction of NaCl if the solution contains 4.56 g NaCl.

Percent Concentration
CONCENTRATION OF SOLUTIONS Percent Concentration - Percent by mass [mass-mass percent, %(m/m)] mass of solution = mass of solute + mass of solvent

A sugar solution is made by dissolving 5.8 g of sugar in 82.5 g of water. Calculate the percent by mass concentration of sugar.

CONCENTRATION OF SOLUTIONS Percent Concentration - Percent by volume [volume-volume percent, %(v/v)] volume of solution ≠ volume of solvent + volume of solute - Due differences in bond lengths and angles

Calculate the volume percent of solute if 345 mL of ethyl alcohol is dissolved in enough water to produce 1257 mL of solution

CONCENTRATION OF SOLUTIONS Percent Concentration - Mass-volume percent [%(m/v)] - Units are specified because they do not cancel

The concentration of a solution of NaCl is 0.92 %(m/v) used to dissolve drugs for intravenous use. What is the amount, in grams, of NaCl needed to prepare mL of the solution? g solute = [%(m/v)] x [volume of solution (mL)]/[100 %] = [(0.92 % g/mL) x (41.50 mL)]/(100 %) = 0.38 g

PARTS PER MILLION (PPM)
Percent can be defined as parts per hundred 1 ppm ≈ 1 µg/mL or 1 mg/L

PARTS PER MILLION (PPM)
If L of aqueous solution with a density of 1.00 g/mL contains 13.7 μg of pesticide, express the concentration of pesticide in ppm ppm = µg/mL 0.250 L = 250 mL Density = 1.00 g/mL Implies mass solution = 250 g

PARTS PER BILLION (PPB)
1 ppb ≈ 1 ng/mL or 1 µg/L

PARTS PER MILLION (PPB)
If L of aqueous solution with a density of 1.00 g/mL contains 13.7 μg of pesticide, express the concentration of pesticide in ppb ppm = µg/L Volume of solution = L Density = 1.00 g/mL Implies mass solution = 250 g

Moles of solute per kg of solvent
MOLALITY (m) Moles of solute per kg of solvent Unit: m or molal

MOLALITY (m) What is the molality of a solution that contains
2.50 g NaCl in g water? - Calculate moles NaCl - Convert g water to kg water - Divide to get molality

CONVERTING CONCENTRATION UNITS
Calculate the molality of a 6.75 %(m/m) solution of ethanol (C2H5OH) in water Mass water = 100 g solution – 6.75 g ethanol = g water

Calculate the mole fraction of a 6.75 %(m/m) solution of ethanol (C2H5OH) in water Mass water = 100 g solution – 6.75 g ethanol = g water

Practice Question Given that the mole fraction of ammonia (NH3) in water is 0.088 Calculate the molality of the ammonia solution

- Molarity is temperature dependent (changes with change in temperature) - Volume increases with increase in temperature hence molarity decreases On the other hand - Molality - Mass percent - Mole fraction are temperature independent

SOLUBILITY - A measure of how much of a solute can be dissolved in a solvent - Grams of solute per 100 mL of solvent - Units: grams/100 mL Three factors that affect solubility - Temperature - Pressure - Polarity

Supersaturated Solution
SOLUBILITY Unsaturated Solution - More solute can still be dissolved at a given temperature - Concentration of the solute is less than the solubility Saturated Solution - No more solute can be dissolved at a given temperature - Concentration of the solute is equal to the solubility - Dynamic equilibrium is reached Supersaturated Solution - Too much solute has temporarily been dissolved - Concentration of solute is temporarily greater than the solubility - Unstable condition

DISSOLUTION The process of dissolving (known as dissolution) is
contributed by factors such as - Enthalpy change due to solute-solvent interactions and - Change in disorder

SOLUTE-SOLVENT INTERACTIONS
- Change in enthalpy arises mainly from changes in intermolecular attractions Three types of intermolecular attractions are involved - Solute-solute - Solvent-solvent - Solute-solvent

- The relative strengths of these interactions determine the formation of a solution by two substances - Substances with similar properties (strong solute-solvent interactions) tend to form solutions - Like dissolves like

- Solvent molecules move apart to accommodate solute molecules - Energy is required to separate solvent molecules attracting each other (ΔH1) - Energy is also required to separate solute molecules (ΔH2) - Energy is released when solvent and solute molecules come together due to attractive forces between them (ΔH3)

Enthalpy of Solution - The overall enthalpy change that accompanies the dissolution of one mole of a solution ΔHsoln = ΔH1 + ΔH2 + ΔH3

Endothermic Heat of Solution - Energy released by solute-solvent interactions is less than the energy absorbed by separating the solvent and solute molecules ΔHsoln is positive ΔH3 < (ΔH1 + ΔH2) Example Ammonium nitrate in water

Endothermic Heat of Solution
ENTHALPY OF SOLUTION Endothermic Heat of Solution Separated solute Separated solvent + ∆H2 Enthalpy Solute + Separated solvent ∆H3 ∆H1 Solution ∆Hsoln Solute + Solvent

Exothermic Heat of Solution - Energy released by solute-solvent interactions is greater than the energy absorbed by separating the solvent and solute molecules ΔHsoln is negative ΔH3 > (ΔH1 + ΔH2) Example Sulfuric acid in water NaOH in water

Exothermic Heat of Solution
ENTHALPY OF SOLUTION Exothermic Heat of Solution Separated solute Separated solvent + ∆H2 Enthalpy Solute + Separated solvent ∆H3 ∆H1 Solute + Solvent ∆Hsoln Solution

Generally - Substances with similar intermolecular forces and hence similar properties have strong solute-solvent interactions - Such substances tend to form solutions - Like dissolves like Example CH3OH readily dissolves in H2O (hydrogen bonding in both) CCl4 readily dissolves in C7H16 (London forces in both)

- Increase in disorder on mixing is another contributing factor in the dissolution process - Increase in disorder tends to occur spontaneously in processes - The main driving force in the formation of solutions Consider NH4NO3 in H2O (used in cold packs) - Enthalpy change on mixing is positive (+28 kJ/mol) - NH4NO3 dissolves to form solution due to increase in disorder

Spontaneous Process - Takes place with no apparent cause Nonspontaneous Process - Requires something to be applied in order for it to occur (usually in the form of energy)

SOLUBILITY OF IONIC COMPOUNDS
- Strong electrostatic attractions between oppositely charged ions hold ionic solids together - For soluble ionic compounds the enthalpy of attraction between solvent molecules and ions must be comparable to the crystal lattice enthalpy in the solid Example - NaCl solution contains Na+ and Cl- ions - Each ion is surrounded by water molecules - Good conductor of electricity

Solvation Process (Hydration)
SOLUBILITY OF IONIC COMPOUNDS Solvation Process (Hydration) - Ions in aqueous solution are surrounded by the H2O molecules The O atom in each H2O molecule has partial negative charge and attract cations - The H atoms have partial positive charge and attract anions - Cations and anions are prevented from recombining - About 4 to 10 water molecules surround each cation

SOLUBILITY OF IONIC COMPOUNDS
- There is an increase in disorder of the solute as it separates into ions - There is an increase or decrease in disorder of the solvent depending on the solute - Solubilities are difficult to predict due to these several contributing factors

SOLUBILITY OF MOLECULAR COMPOUNDS
- Most molecular compounds do not form ions in solution - The molecules disperse throughout the solution Example - Sucrose in water solution contains neutral sucrose molecules - Each molecule is surrounded by water molecules - Poor conductor of electricity - A few molecular compounds form ions in aqueous solution - HCl dissociates into H+(aq) and Cl-(aq) - HNO3 dissociates into H+(aq) and NO3-(aq)

Consider mixing hydrocarbons such as C6H14 and C7H16 - London dispersion forces dominate within the molecules - Attraction between C6H14 and C7H16 molecules are also due to London dispersion forces - These two substances mix because the attractions are close in energy - Increase in disorder is the controlling factor

Consider mixing water and a hydrocarbon - Strong hydrogen bonding dominates intermolecular attractions between water molecules - London dispersion forces dominate intermolecular attractions between hydrocarbon molecules - Attraction between water and hydrocarbon molecules are due to weak London dispersion forces - Increase in disorder is not sufficient to overcome the unfavorable enthalpy change hence very low solubility results

EFFECT OF PRESSURE ON SOLUBILITY
- Solubilities of gases in liquids are sensitive to pressure changes - Increase in pressure increases solubility of gases - An increase in pressure of a saturated solution results in dissolving more gas molecules - Solubilities of liquids and solids change very little with pressure due to very little change in volume

EFFECT OF PRESSURE ON SOLUBILITY
Henry’s Law - The solubility of a gas is directly proportional to its partial pressure at any given temperature C = kP C = concentration of the gaseous solute k = proportionality constant (units depend on units of C) P = partial pressure of gaseous solute

EFFECT OF TEMPERATURE ON SOLUBILITY
- Effect of temperature depends on the sign of the enthalpy change - Solubility increases with increasing temperature when the enthalpy change is positive (+∆H, endothermic process) - Solubility decreases with increasing temperature when the enthalpy change is negative (-∆H, exothermic process) Generally - The more positive the ∆H the greater the change in solubility with temperature

EFFECT OF TEMPERATURE ON SOLUBILITY
- Solubility of most ionic solids increase with increase in temperature - Solubility of most gases decrease with increase in - Enthalpy of solution of most gases in water is negative - There is little or no attraction between gas molecules but there are attractions between solvent and solute molecules - Hence the negative enthalpy change

COLLIGATIVE PROPERTIES
- The physical properties of a solution differ from those of the pure solvent - The physical properties of a solvent change when a solute is added to a solvent - Four physical properties change based on the amount of solute added but not the solute’s chemical identity

- These are known as the Colligative Properties - Vapor-pressure lowering - Boiling-point elevation - Freezing-point depression - Osmotic pressure

Vapor-Pressure Depression - When a nonvolatile solute (low tendency to vaporize) is added to a solvent, the vapor pressure of the resulting solution is lowered below that of the pure solvent at the same temperature

Vapor-Pressure Depression
COLLIGATIVE PROPERTIES Vapor-Pressure Depression Raoult’s Law - The partial pressure of a substance in equilibrium with a solution is equal to the product of its mole fraction in the solution and the vapor pressure of the pure substance Psolv = partial pressure exerted by solvent’s vapor above a solution Xsolv = mole fraction of the solvent in the solution Posolv = vapor pressure of the pure solvent

Vapor-Pressure Depression

Vapor-Pressure Depression - The vapor pressure depression (∆P) is proportional to the mole fraction (concentration) of solute - Raoult’s law only applies to dilute solutions

Vapor-Pressure Depression At 25 oC, the vapor pressure of pure benzene is 93.9 torr. What is the solute concentration in a benzene solution that has a vapor pressure of 92.1 torr? ∆P = torr – 92.1 torr = 1.8 torr

Boiling-Point Elevation
COLLIGATIVE PROPERTIES Boiling-Point Elevation - When a nonvolatile solute (low tendency to vaporize) is added to a solvent, the boiling point of the resulting solution is raised above that of the pure solvent - Since vapor pressure is lowered, a higher temperature is needed to raise the depressed vapor pressure to atmospheric pressure ( a condition required for boiling)

Boiling-Point Elevation
COLLIGATIVE PROPERTIES Boiling-Point Elevation ΔTb = mkb ΔTb = increase in boiling point kb = molal boiling-point elevation constant (depends only on solvent) m = molality (molal concentration) - Applies to dilute solutions only

Boiling-Point Elevation What is the boiling point of a 0.21-molal aqueous solution of sodium chloride at 1 atm (kb of water = oC/m)? ΔTb = mkb = (0.21)(0.512 oC/m) = 0.11 oC Boiling point of pure water = 100 oC Tb = 100 oC oC = oC

Freezing-Point Depression
COLLIGATIVE PROPERTIES Freezing-Point Depression - When a nonvolatile solute (low tendency to vaporize) is added to a solvent, the freezing point of the resulting solution is lowered below that of the pure solvent - The triple-point temperature of a solution decreases with increasing concentration of solute (due to decrease in vapor pressure) - The solid-liquid equilibrium line moves to lower temperatures

COLLIGATIVE PROPERTIES Freezing-Point Depression ΔTf = mkf ΔTf = decrease in freezing point kf = molal freezing-point-depression constant (depends only on solvent) m = molality (molal concentration) - Applies to dilute solutions only

COLLIGATIVE PROPERTIES Freezing-Point Depression What is the freezing point of a 0.21-molal aqueous solution of sodium chloride at 1 atm (kf of water = 1.86 oC/m)? ΔTf = mkf = (0.21)(1.86 oC/m) = 0.39 oC Freezing point of pure water = 0 oC Tf = 0 oC – 0.39 oC = – 0.39 oC

Osmosis - The passage of a solvent through a semipermeable membrane that separates a solution of lower solute concentration from a solution of higher solute concentration - Flow of solvent is in both directions so it is actually a net flow Semipermeable Membrane - Allows certain types of molecules to pass through but prohibits other types of molecules (usually based on size)

Osmotic Pressure - Pressure required to prevent osmosis by pure solvent - Pressure difference needed for no net transfer of solvent - Very sensitive and useful for measuring molar mass of large molecules - Aqueous solutions with higher osmotic pressure take up more water than aqueous solutions with lower osmotic pressure

Osmotic Pressure πV = nRT (similar to the ideal gas law) π = osmotic pressure V = volume of solution n = number of moles of solute R = ideal-gas constant T = absolute temperature M = molarity of solution

Electrolyte Solutions Electrolytes dissociate into ions in solution NaCl(aq) → Na+(aq) + Cl-(aq) 1 mole of NaCl in solution produces 2 moles of ions AlCl3(aq) → Al3+(aq) + 3Cl-(aq) 1 mole of AlCl3 in solution produces 4 moles of ions For example - The osmotic pressure of NaCl is twice that of glucose - Glucose does not dissociate in solution

Electrolyte Solutions van‘t Hoff Factor (i) i = number of particles produced from the dissociation of one formula unit of solute (for dilute solutions) - The number of particles present determines the measured colligative property

Electrolyte Solutions Taking van‘t Hoff Factor (i) into account ΔTb = imkb ΔTf = imkf

Nonideal Solutions - The value of i tends to be smaller than expected at greater solution concentrations (> 0.01 m) - Some ions cluster in solution and behave as a single unit as a result of strong electrostatic attractions

MIXTURES OF VOLATILE SUBSTANCES
Consider a solution of two volatile substances A and B - The vapor phase in equilibrium with the solution contains all the volatile components - According to Raoult’s law and

- Total vapor pressure is the sum of the partial pressures of the components PT = PA + PB - The vapor in equilibrium with the mixture is richer in the more volatile component - This is applied in fractional distillation to separate volatile substances

Ideal Solution - Obeys Raoult’s law throughout the entire range of composition Considering the mixture of substances A and B - Solution is ideal when A-B attractions are closer to A-A and B-B attractions - Nearly true for similar liquids Benzene and toleune Hexane and heptane - Strictly for very dilute solutions

MIXTURES OF VOLATILE SUBSTANCES Positive Deviation from Raoult’s Law
- Occurs when the A-B attractions are weaker than the average A-A and B-B attractions - The observed vapor pressure is greater than expected Pressure 0.5 1 Mole fraction - Straight lines (dotted) in ideal situation become curves

MIXTURES OF VOLATILE SUBSTANCES Negative Deviation from Raoult’s Law
- Occurs when the A-B attractions are stronger than the average A-A and B-B attractions - The observed vapor pressure is less than expected Pressure 0.5 1 Mole fraction - Straight lines (dotted) in ideal situation become curves

PRINCIPLES OF CHEMISTRY II CHEM 1212 CHAPTER 12

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PRINCIPLES OF CHEMISTRY II CHEM 1212 CHAPTER 12

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