Chapter 15: solutions

Solutions

Types of solutions

Factors Affecting Solubility

Factors Affecting the Rate of Dissolution

Saturation

Ways of Expressing Concentration

Objective: 1Understand the process of dissolving

A homogeneous mixture of two or more substances

Can be described by two terms Solute the component that is present in lesser quantity

Solvent -- the component that is present in greater quantity -- It determines the state of matter in which the solution

exist

Example

Salt solution (salt in water)

Sugar solution (sugar in water)

Solubility

Maximum amount of solute expressed in grams that can be dissolved in 100g of water at specific temperature

Soluble

substance that dissolves in solvent

Insoluble

Substance that does not dissolve in solvent

Example

Soluble ( miscible)

Water and alcohol

salt and water

Insoluble (Immiscible)

water and alcohol

salt and gasoline

Objective:2 Differentiate the Types of Solutions

Types of solutionsSolute Solvent Example

Gaseous

Solution

Gas Gas Air

Liquid

Solution

Gas

Liquid

Solid

Liquid

Liquid

Liquid

Carbonated drinks

Alcohol in water

Sugar solution

Solid Solution Liquid

Solid

Solid

Solid

Dental amalgam (mercury and silver)

Brass (zinc and copper)

Steel (Carbon & iron)

Objective:3Explain the Factors Affecting Solubility

Nature of Solute and Solvent

Temperature

Pressure

1.Nature of Solute & Solvent

Like dissolves like Polar solute dissolves polar solvent

( alcohol dissolves in water) Both polars Nonpolar solute dissolves nonpolar solvent

( Gasoline dissolves in Carbon tetrachloride)

Solubility based on polarity

Solute Solvent Soluble or insoluble

Sugar ( polar) Water (polar)

Sugar (polar) Kerosene (nonpolar)

Salt (ionic) Water (polar)

Salt (ionic) Kerosene (nonpolar)

Oil (nonpolar) Water (polar)

Oil( nonpolar) Kerosene (nonpolar)

2.Temperature

For gases dissolving in liquids

An increase in temperature decreases solubility

For solids

Increase in temperature increases solubility

Pressure

increases the solubility of gases in liquids but not solids in liquids or liquid in liquid

Solubility decreases as pressure decreases Ex: when a bottle of soft drink is opened; the

pressure of the gas in the bottle is reduced and the solubility of carbon dioxide is decreased. Thus the gas bubbles out the solution

Objective:4Explain the Factors Affecting the Rate of DissolutionParticle size

Stirring or agitation

Application of heat

1. Particle size

When the total surface area of the solute particles is increased, the solute dissolves more rapidly

Finely divided solute has a large surface area, which allows more contact with the solvent molecules

Fine table salt dissolves faster than rock salt

2. Stirring or Agitation

increases the rate of dissolution Hastens the contact between the surface of

the solute and the solvent particles Does not affect the amount of solute that

dissolves :

an insoluble substance remains undissolved no matter how much you stir the solution

3. Application of heat

Where do coffee dissolves faster in hot or cold water?

Why? KE of hot molecules is higher than cold

molecules. Hot molecules move around faster and come in contact more frequently

with the solute particles

Objective 5: Differentiate the Types of Saturation

Saturated

Unsaturated

Supersaturated

Saturated

The amount of solute particles dissolving equals the amount of solute particles precipitating

How do you know that the solution is saturated or saturation point is reached?

When undissolved solid appears

Unsaturated

The amount of dissolved solute is less than the maximum the solvent can dissolve

Supersaturated

A solution that contains more solute or concentration of solute is greater than a saturated solution

Ex: Gout or painful inflammation of joints-> occurs when the uric acid concentration in the

plasma exceeds its maximum solubility. Crystals of uric acid deposit in the joints causing inflammation and severe pains

Sources of high level uric acid: beans, some meats, mushrooms, and small fishes.

Objective :6Ways of Expressing Concentration

Mass / Volume Percent

Mole fraction

Molarity

Molality

Concentration of Solutions

Dilute

There is only a little amount of solute dissolved in a solution

Concentrated

Large amount of solute dissolved in a solution

Mass percent

Mass % = mass solute x 100%

mass solution

Ex1: A solution is prepared by dissolving 1.0g of NaCl in 48g water. Calculate the mass percent of NaCl

Solution:

Mass percent = 1g x 100%

49g

= 2%

Ex:2

Cows milk typically contains 4.5% by mass of lactose ( C12H22O11). Calculate the mass of lactose present in 175g of milk

Solution

mass% = m solute x 100%

m solution

0.045 = m solute (cross multiply)

175 g

m solute = 0.045 ( 175g)

= 7.9 g

Volume percent

Volume% = Vsolute x 100%

V solution

Ex: A solution is prepared by mixing 50mL of C2H5OH in 300 mL of distilled water, what is the volume percent concentration?

Solution:

Volume % = V solute x 100%

V solution

= 50mL x 100%

350mL

= 14.28%

Mole Fraction

the ratio of the number of moles of one component of a solution to the total number of moles of all the components

Mole fraction = mole solute

mole solute + mole solvent

Example:1What is the mole fraction of the solute in a 40% by mass ethanol (C2H6O) solution in water?Solution:

Step 1: determine the component of ethanol and water by using a 100g sample of the solution40% --- 40 g ethanol and 60g waterStep 2: Change the mass of the components to the number of

moles

Mole ethanol = 40g C2H6O x 1mole = 0.87

46g/mole

Mole water = 60gH2O x 1mole H2O = 3.33

18g H2OMole fraction = mole solute = 0.87 = 0.21 n solute + n solvent 0.87 +3.33

2.Molarity (M)

Ratio of the number of moles of solute to the volume of solution in liters

M = moles solute = n

Volume of solution (L) V What does “3M” means?

The solution has 3 moles solute for every liter of solution

Example:1

What is the molarity of the solution if 3.0 moles of solute is dissolved in 2L solution?

Solution:

M = n = 3.0mol = 1.5M

V 2.0L

Ex:2 Calculate the molarity of a solution prepared by

dissolving 11.5g of solid NaOH in enough water to make 1.50L solution

Solution M = n V Mass of solute moles of solute Molarity 11.5 g NaOH x 1molNaOH = 0.288mol NaOH 40.0gNaOH M = 0.288 mol NaOH 1.5L = 0.192M NaOH

Ex: How many grams of Ca(OH)2 are needed to prepare an 800.0 mL solution of 0.015 M concentration? Solution: M = n ,

V n = MV = ( 0.015 mol/L ) ( 0.800L) = 0.012molCalculate the mass of Ca(OH)2 mass = 0.012mol Ca(OH)2 x 74.0g = 0.89 g 1 mole

Molality (m)

The ratio of the number of moles of solute per kilogram of solvent

m = moles solute

mass solvent in kilograms

What is the molality of 30% by mass CaCl2 solution? Solution

Step 1:Determine the amount of the components of the solution by using a 100g sample

100g of 30%CaCl2 = 30g CaCl2 Therefore massH2O is 70g

Step 2: Change the mass of solute to number of moles

30gCaCl2 x 1 moleCaCl2 = 0.273mol

110g/mol

Step 3: Change the mass of solvent from g to kg

70g x 1kg = 0.0700kg

1000g

m= n solute = 0.273mol = 3.90mol/kg

mass solvent 0.700kg

Ex:2 How many grams of NaOH are needed to prepare 0.700m solution using 700.0g water?SolutionDetermine the number of moles NaOH using the molality formula m = n solute n = m x kg solute kg solute = 0.700mol x 0.700kg = 0.490mol kg Determine the mass of NaOH using the calculated n 0.490mol NaOH x 40g NaOH = 19.6 g NaOH 1mol NaOH

How to convert Molarity to Molality?

Ex: Calculate the molality of 0.763M acetic acid (C2H4O2) solution with a density of 1.004g/mL.Determine the number of moles of solute 0.763M means 0.763moles is present in 1L solutionDetermine the mass of the solvent in kilograms using D=M/V convert 1.004 g kg/L = 1.004g x 1kg x 1000mL = 1.004kg/L ml 1000g 1L convert moles solute---> mass

0.763moles C2H4O2 x 60g C2H4O2 = 45.78g x 1kg = 0.04578kg

1mole C2H4O2 1000g since mass of solute = 0.04578kg , therefore mass of solvent is = 1.004kg – 0.04578 kg = 0.95822kg

3. Determine the molality m = n solute = 0.763mol = 0.796mol/kg or 0.796m kg solvent 0.95822kg