UNIT 4.1: SOLUTIONSMolarity, Molality, Colligative Properties, Solubility

Solutions• Solution – a homogeneous mixture

• Remember that a homogeneous mixture is one in which all of the parts are evenly distributed and not easily distinguished

• Solutions are made of 2 parts: the solute and the solvent

• Solute – dissolved particles in a solution• What is being dissolved

• Solvent – the dissolving medium in a solution• what the solute is being dissolved in

Concentration•Concentration – a measurement of the amount of solute

that is dissolved in a given quantity of solvent

•Concentration can be measured in two major ways•Molarity•Molality

•Molarity is the more popular way to express concentration

MOLARITY

Molarity

•Molarity – symbolized by M; the concentration of solute in a solution as expressed as the number of moles of solute dissolved in 1 L of solution

•The units are VERY important!

𝑴 =𝒏

𝑳=

𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆

𝑳𝒊𝒕𝒆𝒓𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏

Measurement Variable Unit

Molarity MM

(mol/L)

Number of Moles

n mol

Liters of Solution

LL

(Liters)

Practice Problems

1. What is the molarity of a solution that contains 0.70 mol of NaCl dissolved in 0.250 L of solution?

Practice Problems

2. What is the molarity of a solution that contains 0.90 g of NaCl in exactly 2.5 L of solution?

Practice Problems

3. How many moles of NH4NO3 are in 335 mL of 0.425 M solution?

Practice Problems4. How many liters of solution are prepared if the

solution is 0.75 M and made from 12.3 moles of KCl?

MOLALITY

Molality•Molality – symbolized by m; the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 kg (1000 g) of solvent

•The units are VERY important!

•𝒎 =𝒏

𝒌𝒈=

𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆

𝒌𝒊𝒍𝒐𝒈𝒓𝒂𝒎𝒔 𝒐𝒇 𝒔𝒐𝒍𝒗𝒆𝒏𝒕

Measurement Variable Unit

Molality m m (mol/kg)

Number of Moles

n mol

Kilograms of Solvent

kgkg

(kilograms)

Practice Problems1. Calculate the molality of a solution prepared by

dissolving 0.575 moles of NaCl in 0.675 kg of water.

Practice Problems2. Calculate the molality of a solution prepared by

dissolving 10.0 g of NaCl in 600. g of water.

Practice Problems3. How many kg of water must be used to dissolve

2.35 moles of MgCl2 to produce a 0.750 molalsolution?

Practice Problems4. How many grams of NaF are needed to prepare a

0.400 m solution that contains 750. grams of water?

COLLIGATIVE PROPERTIES

Colligative Properties• When a solute is added to a solvent, the solution will have

different properties

• Colligative properties: a property of a solution that depends only on the NUMBER of solute particles and not upon their identities

• Examples of colligative properties are:• Vapor-pressure lowering

• Freezing-point depression

• Boiling-point elevation

Vapor-Pressure Lowering• Vapor pressure: the pressure

exerted by a vapor that is in dynamic equilibrium with its liquid in a closed system• a solution that contains a solute

that is nonvolatile (does not easily vaporize) always has a lower vapor pressure than the pure solvent

• The decrease in a solution’s vapor pressure is proportional to the number of particles the solute makes in solution

Vapor-Pressure Lowering• When a solute is added to a solvent, the solute particles

become surrounded by the solvent particles, not allowing them as much energy to be able to escape to the surface

Freezing-Point Depression• Freezing-point depression: the

difference in temperature between the freezing point of a solution and the freezing point of the pure solvent

• The magnitude of the freezing-point depression is proportional to the number of solute particles dissolved in the solvent and DOES NOT DEPEND ON THEIR IDENTITY

Freezing-Point Depression• When a substance freezes, the particles of the solid take on a

orderly pattern• When a solute is added, the solute particles disrupt this

pattern, which means it will take even more energy to cause the solution to solidify

Boiling-Point Elevation•Boiling-point elevation: the difference in temperature

between the boiling point of a solution and the boiling point of the pure solvent

•The magnitude of the boiling-point elevation is proportional to the number of solute particles dissolved in solution

•Because of the decrease in vapor pressure, it will take more energy to raise the vapor pressure and initiate boiling• Think of the cats

Number of Particles in Solution•The number of particles that dissolve in a solution depends on the type of solute – whether it is ionic or covalent

•Ionic vs. Covalent•Ionic – metal and a non-mental, charges are important ,

DO DISSOCIATE IN SOLUTION•Covalent – two non-metals, charges are not important, DO NOT DISSOCIATE IN SOLUTION

Dissociation Factor• The number of particles produced by a solute will be referred to as the

dissociation factor (df)

• How many particles will the be in solution from the following compounds?

•KCl•CaBr2

•C3H4O3

•NaNO3

•CH4

Calculating the Freezing-Point Depression and Boiling-Point Elevation•The magnitude of the freezing-point depression and the boiling-point elevation of a solution are DIRECTLY proportional to the molal concentration

•Freezing-Point Depression•ΔTf = Kf x m x df

•Boiling-Point Elevation•ΔTb = Kb x m x df

Equations, Variables, and Units, OH MY!Freezing-Point Depression

ΔTf = Kf x m x df

Variable Measurement Unit

ΔTf

Change in freezing point

°C

Kf

Freezing-Point depression

constant°C/m

m molalitym

(mol/kg)

df

Dissociationfactor

# ofparticles

Boiling-Point ElevationΔTb = Kb x m x df

Variable Measurement Unit

ΔTb

Change in boiling point

°C

Kb

Boiling-Point elevation constant

°C/m

m molalitym

(mol/kg)

df

Dissociation factor

# of particles

Practice

•What is the boiling point elevation and the boiling point of a 1.50 m NaCl solution? Kb = 0.512 °C.

Practice• Antifreeze protects a car from freezing. It also protects it from

over heating. Calculate the freezing point depression and the new freezing point of a solution containing 1.61 moles of antifreeze (ethylene glycol = C2H6O2) in 0.500 kg of water. Kf = 1.86 °C.

SOLUBILITY

Solution Formation•The composition (what it’s made of) of the solvent and solute determine whether the solute will dissolve in the solvent

•Other factors that determine how fast the solute dissolves•Agitation (stirring) •Temperature•Surface area

Solubility• The solubility of a substance is the amount of solute that

dissolves in a given quantity of a solvent at a specific temperature and pressure to produce a saturated solution

• There are three main categories of solutions• Saturated: contains the maximum amount of solute for a given

quantity of solvent

• Unsaturated: contains less solute than a saturated solution• Can dissolve more solute

• Supersaturated: contains more solute than it can theoretically hold

Temperature & Solubility

•Temperature affects the solubility of solid, liquid, and gaseous solutes in a solvent

•The solubility of MOST solidsubstances increases as the temperature increases•This is not true for all solids,

there are some exceptions

Solubility Curves•Solubility is usually measured in grams of solute per 100 grams of solvent (typically water)

•The graphs give us a visual representation of how the solubility changes are temperature changes

Reading Solubility Curves• Which substance is the most soluble at 0 °C?

• Which substance is the least soluble at 0 °C?

• What is the solubility of KNO3 at 50 °C?

• What is the solubility of KClO3 at 20 °C?

• At what temperature is the solubility of KNO3 110 g in 100 g of water?

• At what temperature is the solubility of NH4Cl 60 g in 100 g of water?

• If 20 g of KClO3 is dissolved in 100 g of water at 30 °C, is the solution saturated, unsaturated, or supersaturated?

• If 110 g of NaNO3 is dissolved in 100 g of water at 50 °C, is the solution saturated, unsaturated, or supersaturated?