Chapter 16 Solutions



Chapter 16 Solutions

Properties of Solutions

A solution is a homogeneous mixture of substances in the same physical state.

Contain atoms, ions or molecules of one substance spread uniformly throughout the second substance.

May be solid, liquid or gas

The composition of the solvent and solute will determine whether a substance will dissolve.

What determines how fast a substance will dissolve?

Rate of dissolving

Agitation – determines the rate at which solid dissolves in solute not how much will dissolve. If it’s insoluble – it’s insoluble

Temperature – influences the rate – higher temperature – kinetic energy of water molecules is greater = more collisions between solute and solvent.

Surface area of the dissolving particles –smaller particles have more surface area – dissolving is a surface phenomenon. More surface – faster rate.

Solubility

According to the kinetic theory – water molecules are in constant motion.

As many solid particles are going into solution as coming out – equilibrium

Temperature will affect equilibrium

Saturated solution – max amount of solute for a given quantity of solvent at a constant temp and pressure.

Solubility

Solubility of a substance is the amount of solute that dissolves in a given quantity of solvent at a specified temperature and pressure to produce a saturated solution.

Expressed as Grams of solute per 100 grams of solute.

Solubility of a gas = g/L (grams per liter of solution)

Solution that contains less solid than a saturated solution is = unsaturated.

Miscible liquids

Water and ethanol

Ethylene glycol and water

Any amount will dissolve in a given volume

Dissolve in all proportions

Liquid in larger amount is the solvent

Immiscible = liquids that are insoluble in each other.

Factors affecting solubility

Solids, liquids and gases are affected by temperature

Gaseous solutes are affected by both temperature and pressure.

Temperature – as temperature of solvent increases, solubility of solute increases of most solid substances.

Exception – ytterbium sulfate ( Yb2(SO4)3 )

Factors affecting solubility

Supersaturated solutions

Contain more solute than it can hold at a given temperature.

Crystallization – seed crystal or scratching the inside of the container.

Rock candy

Solubility on gases in liquid solvents

Solubility of gases greater in cold liquids

Thermal pollution

Pressure

Changes in pressure – little affect on liquids and solids

Changes in pressure greatly affects solubility of gases.

Gas solubility increases as the partial pressure of the gas above the solution increases.

Carbonated beverages bottles under high pressure – forces large amounts of gas into solution.

Henry’s law

At a given temperature, the solubility (S) of a gas in a liquid, is directly proportional to the pressure (P) of the gas above the liquid.

As the pressure of the gas above the liquid increases, the solubility of the gas increases.

S1 = S2

P1 P2

Practice

The solubility of a gas in water is 0.16g/L at 104kPa. What is the solubility when the pressure of the gas is increased to 288kPa? Assume temperature remains constant.

Molarity

Expressing the concentration of a solution

The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent.

A dilute solution is one that contains a small amount of solute.

A concentrated solution contains a large amount of solute

Molarity

Molarity (M) is the number of moles of solute dissolved in one liter of solution.

Calculating molarity:

Divide moles of solute by the volume of solution

Molarity (M) = moles of solute

liters of solution

Volume is the total volume of solution not the volume of the solvent alone.

Molarity

3 M NaCl = 3 molar solution of salt

Practice:

One saline solution contains 0.90 g of NaCl in exactly 100 ml of solution.

What is the molarity of the solution?

Solution concentration = 0.90 g NaCl/100 ml

Molar mass of NaCl = 58.5 g/mol

? Solution concentration

Solve : convert the concentration from g/100ml to mol/L

g/100ml ( mol/100ml ( mol/L

Calculating Molarity

Solution concentration =

0.90gNaCl X 1mol NaCl X 1000ml

100ml 58.5gNaCl 1L

solution concentration = 0.15 mol/L

= 0.15M

(0.9g/100ml = 9.0 g / 1000 ml = 9.0 g /L)

Molarity Practice

A solution has a volume of 2.0L abd contains 36.0 g of glucose (C6H12O6)

If the molar mass of glucose is 180g/mol, what is the molarity of the solution?

Molarity Practice

A solution has a volume of 250 mL and contains 0.70 mol NaCl. What is its molarity?

More Practice

Household laundry bleach is a dilute aqueous solution of sodium hypochlorite

(NaClO). How many moles of solute are present in 1.5L of 0.70 M NaClO?

Volume of solution ( moles of solute

Practice

How many moles of ammonium nitrate are in 335 ml of 0.425 M NH4NO3?

Making Dilutions

Diluting a solution reduces the number of moles of solute per unit volume, but the total number of moles of solute in solution does not change.

Moles of solute before dilution = moles of solute after dilution.

Moles os solute = M1 x V1 = M2 x V2

initial final

Practice Making dilutions

How many milliliters of a solution of 4.00M KI are needed to prepare 250.0 mL of 0.760 M KI?

Practice making dilutions

How many milliliters of aqueous 2.00 M MgSO4 solution must be diluted with water to prepare 100.0 mL of aqueous 0.400 M MgSO4?

Percent Solutions

Describe solutions by the percent of a solute in the solvent.

As the ratio of the volume of the solute to the volume of the solution

As the ratio of the mass of the solute to the mass of the solution.

Percent by volume (%V/V) = volume of solute x 100%

vol of solution

Practice percent solutions

What is the percent by volume of ethanol in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water?

%V/V = vol of solute x 100 %

vol of solution

85/250 (100%) = 34 % ethanol (v/v)

Practice percent solutions

If 10 mL of propanone is diluted with water to a total solution volume of 200 mL, what is the percent by volume of propanone in the solution?

Concentration in percent

Mass / Mass

Mass / mass = is the number of grams of solute in 100 grams of solution

Percent by mass(%m/m) =

Mass of solute x 100%

Mass of solution

How many grams of K2SO4 would you need to prepare 1500 g of 5.0% K2SO4 solution?

Parts per million

Similar to percent composition

Compares masses

ppm = ratio between the mass of a solute and the total mass of the solution

Used with extremely dilute solutions

ppm = grams of solute x 1,000,000 ppm

grams of solution

Parts per million practice

Approximately 0.0043 g of oxygen can be dissolved in 100 mL of water at 20 °C. Express this in terms of parts per million.

Grams of solution = solvent + solute

= 100.0043 g of solution

Substitute known and solve for unknown

Preparing a solution of a known concentration

How much solute has to be added to a known volume of solvent to make a solution of a specific concentration.

Prepare a solution of a known molarity.

You must know how much solute and solvent you need to prepare the solution

Steps are very specific

Steps to prepare a solution of a know concentration

Calculate how much solute is needed to make a solution of a specific concentration.

Add the solute to a volumetric flask and some distilled water – stir – fill with water to the mark and make sure the solution is homogeneous.

How to calculate the amount of solute for a specific concentration

What mass of Na2CO3 is required to prepare 2.00 L of a 0.250 M Na2CO3 (sodium carbonate) solution?

Identify known and unknown

Known unknown

Concentration of solution mass of

Volume of solution Na2CO3 ?g

1.Determine the number of moles of solute needed by using molarity and volume

Moles = MV = molarity x liters of solution

Moles of Na2CO3 = 0.250 mol x 2.00 L

1L

= 0.500 mol

2. Convert moles to grams

Moles of Na2CO3 to grams Na2CO3

Determine the formula mass of Na2CO3

Na : 2 atoms x 23.0 amu/atom=46.0amu

C: 1 atom x 12 amu/atom = 12.0 amu

O: 3 atoms x 16.0 amu/atom = 48.0 amu

Formula mass Na2CO3 = 106.0amu

3. Change formula mass to gram formula mass

Gram formula mass = formula mass in grams = 106.0 grams /mole

4. Convert moles of Na2CO3 into grams of Na2CO3

Mass = moles x gram formula mass

Mass Na2CO3 = 0.500 mol x 106.0 g/mol

Mass Na2CO3 = 53.0 g

Colligative Properties

A property that depends only upon the number of particles in the solution and not upon their identity is called a colligative property.

3 important colligative properties of solutions

Vapor pressure lowering – vapor pressure is the pressure exerted by a vapor that is in dynamic equilibrium with its liquid in a closed system.

Boiling point elevation – the difference in temperature between the boiling point of a solution and the boiling point of the pure solvent.

Freezing point depression – the difference in temperature between the freezing point of a solution and the freezing point of the pure solvent. (antifreeze)

Colligative Properties

Freezing and boiling points of water change when nonvolatile solutes are added.

Salt + water = freezing point of water decreases.

Added salt lowers the freezing point

One mole of particles lowers the freezing point of 1000g of water by 1.86 °C.

Molecular versus ionic

Molecular – one mole of substance dissolves – one mole is in solution and the effect on freezing point is one for one.

When one mole of ionic substance is dissolved in water – ions are formed and the freezing point is depressed with relation to the number of ions formed – CaCl3 contains 3 ions so one mole of this salt will depress the freezing point 3 times as much as sugar.

Molecular versus ionic and boiling point elevation

One mole of particles will elevate the boiling point of 1000g water by 0.52°C.

One mole of sugar – 1000g water by 0.52°C.

One mole of sodium chloride – will elevate the temperature of the same amount of water by 1.04 °C.

One mole of CaCl2 – 3 x 0.52 = 1.56°C.

Colligative properties

Vapor Pressure

Molecules in a liquid are held together by weak forces – polar molecules – dipole – dipole forces.

Hydrogen and oxygen, nitrogen or fluorine – hydrogen bonds

Molecules at surface of liquid have energy to escape to gas phase.

Gas phase = vapor

Pressure vapor particles exert = vapor pressure

(table H)

Greater the vapor pressure the weaker the intermolecular forces holding it in the liquid phase.

Lower the vapor pressure the greater the intermolecular forces holding it in the liquid phase.

Colligative properties – boiling point

Temperature of liquid rises – vapor pressure increases

Boiling point of a liquid = temperature at which the vapor pressure of the liquid is 101.3kPa or standard atmospheric pressure. = 1 atm = 760 mm Hg = 760 torr.

The heat required to change 1 mol of a substance from a liquid at its boiling point to 1 mol of vapor is called the heat of vaporization.

Boiling point

Normal boiling point of water – 100 °C

Vapor pressure at boiling point is 101.3 kPa

If the pressure decreases from normal atmospheric – the boiling point will decrease also. If pressure increases the reverse is true.

What to Remember about colligative properties

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

The difference in temperature between the freezing point of a solution and the freezing point of the pure solvent is freezing point depression. Magnitude is proportional to the number of solute particles dissolved in the solvent.

The difference in temperature between the boiling point of a solution and the boiling point of the pure solvent is boiling point elevation. Magnitude is probortional to the number of solute particles dissolved in the solvent.

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