Prepared by PhD Halina FalfushynskaPrepared by PhD Halina Falfushynska

Lecture 4. Solutions and their Lecture 4. Solutions and their coligative propertiescoligative properties

A space-filling model of the water molecule.

If one looks at the boiling points of the hydrides of many elements, water, ammonia, and hydrogen fluoride have uniquelyHigh boiling points. For example water is projected to have a Boiling point of only about -80oC in stead of 100oC!

Polar water molecules interact with the positive and negative ions of a salt,

assisting with the dissolving process.

The ethanol molecule contains a polar O-H bond similar to those in the water molecule. (b) The polar water molecule

interacts strongly with the polar O-H bond in ethanol.

The Role of Water as a Solvent: The solubility of Ionic Compounds

Electrical conductivity - The flow of electrical current in a solution is a measure of the solubility of ionic compounds or a measurement of the presence of ions in solution.

Electrolyte - A substance that conducts a current when dissolved in water. Soluble ionic compound dissociate completely and may conduct a large current, and are called Strong Electrolytes.

NaCl(s) + H2O(l) Na+(aq) + Cl -

(aq)

When Sodium Chloride dissolves into water the ions become solvated,and are surrounded by water molecules. These ions are called “aqueous”and are free to move through out the solution, and are conducting electricity, or helping electrons to move through out the solution

Electrical Conductivity of Ionic Solutions

Electrical Conductivity

HCL (aq) is completely ionized.

An aqueous solution of sodium hydroxide.

Acetic acid (HC2H3O2) exists in water mostly as undissociated molecules.

The reaction of NH3 in water.

Comparison of a Concentrated and Dilute Solution

Molarity (Concentration of Solutions)= M

M = = Moles of Solute MolesLiters of Solution L

solute = material dissolved into the solvent

In air , Nitrogen is the solvent and oxygen, carbon dioxide, etc. are the solutes.In sea water , Water is the solvent, and salt, magnesium chloride, etc. are the solutes.In brass , Copper is the solvent (90%), and Zinc is the solute(10%)

LIKE EXAMPLE

Calculate the Molarity of a solution prepared by bubbling 3.68g of Gaseous ammonia into 75.7 ml of solution.

Solution:

Calculate the number of moles of ammonia:

3.68g NH3 X = 0.216 mol NH3

1 mol NH3

17.03g

Change the volume of the solution into liters:

75.7 ml X = 0.0757 L 1 L1000 mL

Finally, we divide the number of moles of solute by the volume of the solution:

Molarity = = ____________ M NH3

0.216 mol NH3

0.0757 L

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• Quantitative concentration term

– Molarity is the ratio of moles solute per liter of solution

– Symbols: M or [ ]

– Different forms of molarity equation

L

molM

M

molL LMmol

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Calculate the molarity of a solution

prepared by dissolving 45.00 grams of

KI into a total volume of 500.0 mL.

Copyright McGraw-Hill 2009 21

Calculate the molarity of a solution

prepared by dissolving 45.00 grams of

KI into a total volume of 500.0 mL.

M 5422.0L 1mL 1000

KI g 166.0KI mol 1

mL 500.0KI g 00.45

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How many milliliters of 3.50 M NaOH can

be prepared from 75.00 grams of the

solid?

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How many milliliters of 3.50 M NaOH can

be prepared from 75.00 grams of the

solid?

mL 536L 1mL 1000

NaOH mol 3.50L 1

NaOH g 40.00NaOH mol 1

NaOH g 00.75

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• Dilution– Process of preparing a less concentrated

solution from a more concentrated one.

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For the next experiment the class will

need 250. mL of 0.10 M CuCl2. There is

a bottle of 2.0 M CuCl2. Describe how toprepare this solution. How much of the2.0 M solution do we need?

Preparing a Solution - I

• Prepare a solution of Sodium Phosphate by dissolving 3.95g of Sodium Phosphate into water and diluting it to 300.0 ml or 0.300 l !

• What is the Molarity of the salt and each of the ions?

• Na3PO4 (s) + H2O(solvent) = 3 Na+(aq) + PO4

-3(aq)

Preparing a Solution - II

• Mol wt of Na3PO4 = 163.94 g / mol

• 3.95 g / 163.94 g/mol = 0.0241 mol Na3PO4

• dissolve and dilute to 300.0 ml

• M = 0.0241 mol Na3PO4 / 0.300 l = 0.0803 M

Na3PO4

• for PO4-3 ions = ______________ M

• for Na+ ions = 3 x 0.0803 M = ___________ M

Like ExampleAn isotonic solution, one with the same ionic content as blood is about0.14 M NaCl. Calculate the volume of blood that would contain 2.5 mgOf NaCl?

Find the moles in 1.0 mg NaCl:

2.5 mg NaCl x x = 4.28 x 10-5 mol NaCl

1 g NaCl 1000 mg NaCl

1 mol NaCl58.45g NaCl

What volume of 0.14 M NaCl that would contain the amount of NaCl(4.28 x 10-5 mol NaCl):

V x = 4.28 x 10-5 mol NaCl0.14 M NaCl L solution

Solving for Volume gives:

V = = ______________________ L 4.28 x 10-5 mol NaCl 0.14 mol NaCl L solution

Or _________ ml of Blood!

Steps involved in the preparation of a standard solution.

Like Example 4.4 (P 98)A Chemist must prepare a 1.00 L of a 0.375 M solution of Ammonium Carbonate, what mass of (NH4)2CO3 must be weighed out to prepare this solution?

First, determine the moles of Ammonium Carbonate required:

1.00 L x = 0.375 M (NH4)2CO3 0.375 M (NH4)2CO3

L solution

This amount can be converted to grams by using the molar mass:

0.375 M (NH4)2CO3 x = 35.276 g (NH4)2CO3

94.07 g (NH4)2CO3

mol (NH4)2CO3

Or, to make 1.00L of solution, one must weigh out 35.3 g of (NH4)2CO3, put this into a 1.00 L volumetric flask, and add waterto the mark on the flask.

Make a Solution of Potassium Permanganate

Potassium Permanganate is KMnO4 and has a molecular mass of 158.04 g / mole

Problem: Prepare a solution by dissolving 1.58 grams of KMnO4

into sufficient water to make 250.00 ml of solution.

1.58 g KMnO4 x = 0.0100 moles KMnO4 1 mole KMnO4

158.04 g KMnO4

Molarity = = ______________ M0.0100 moles KMnO4

0.250 liters

Molarity of K+ ion = [K+] ion = [MnO4-] ion = _____________ M

(a) A measuring pipette (b) A volumetric pipette.

(a) A measuring pipette (b) Water is added to the flask. (c) The resulting

solution is 1 M acetic acid.

Dilution of Solutions

• Take 25.00 ml of the 0.0400 M KMnO4

• Dilute the 25.00 ml to 1.000 l - What is the resulting Molarity of the diluted solution?

• # moles = Vol x M

• 0.0250 l x 0.0400 M = 0.00100 Moles

• 0.00100 Mol / 1.00 l = _______________ M

Reactant solutions: (a) Ba(NO3)3(aq)

Reactant solutions: (b) K2CrO4(aq).

When yellow aqueous potassium chromate is added to a colorless barium nitrate solution, yellow barium chromate precipitates.

Reaction of K2CrO4 (aq) and Ba(NO3)2 (aq).

Precipitation of silver chloride by mixing solutions of silver nitrate and potassium chloride.

Photos and molecular-level representations illustrating the reaction of KCL(aq) with

AgNO3(aq) to form AgCl(s).

Precipitation ofsilver chromate byadding potassiumchromate to a solution of silver nitrate.

K2CrO4 (aq) + 2 AgNO3 (aq)

Ag2CrO4 (s) + 2 KNO3 (aq)

Simple Rules for Solubility of Salts in Water

1. Most nitrate (NO3-) salts are soluble.

2. Most salts of Na+, K+, and NH4+ are soluble.

3. Most chloride salts are soluble. Notable exceptions are AgCl, PbCl2, and Hg2Cl2.4. Most sulfate salts are soluble. Notable exceptions are BaSO4, PbSO4, and CaSO4.5. Most hydroxide salts are only slightly soluble. The important soluble hydroxides are NaOH, KOH, and Ca(OH)2 (marginally soluble).6. Most sulfide (S2-), carbonate (CO3

2-), and phosphate (PO43-)

salts are only slightly soluble.

The Solubility of Ionic Compounds in Water

The solubility of Ionic Compounds in water depends upon the relative strengths of the electrostatic forces between ions in the ionic compoundand the attractive forces between the ions and water molecules in the solvent. There is a tremendous range in the solubility of ionic compounds in water! The solubility of so called “insoluble” compoundsmay be several orders of magnitude less than ones that are called“soluble” in water, for example:

Solubility of NaCl in water at 20oC = 365 g/LSolubility of MgCl2 in water at 20oC = 542.5 g/LSolubility of AlCl3 in water at 20oC = 699 g/LSolubility of PbCl2 in water at 20oC = 9.9 g/LSolubility of AgCl in water at 20oC = 0.009 g/LSolubility of CuCl in water at 20oC = 0.0062 g/L

The Solubility of Covalent Compounds in Water

The covalent compounds that are very soluble in water are the oneswith -OH group in them and are called “Polar” and can have strong polar (electrostatic)interactions with water. Examples are compound such as table sugar, sucrose (C12H22O11); beverage alcohol, ethanol (C2H5-OH); and ethylene glycol (C2H6O2) in antifreeze.

CH

H

H

O H Methanol = Methyl Alcohol

Other covalent compounds that do not contain a polar center, or the-OH group are considered “Non-Polar” , and have little or no interactions with water molecules. Examples are the hydrocarbons in Gasoline and Oil. This leads to the obvious problems in Oil spills, wherethe oil will not mix with the water and forms a layer on the surface!

Octane = C8H18 and / or Benzene = C6H6

When a solution of Na2SO4 (aq) is added to a solution of Pb(NO3)2, the white solid

PbSO4(s) forms.

Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - I

Problem: How many moles of each ion are in each of the following:

a) 4.0 moles of sodium carbonate dissolved in water b) 46.5 g of rubidium fluoride dissolved in water c) 5.14 x 1021 formula units of iron (III) chloride dissolved in water d) 75.0 ml of 0.56M scandium bromide dissolved in water e) 7.8 moles of ammonium sulfate dissolved in water

a) Na2CO3 (s) 2 Na+(aq) + CO3

-2(aq)

moles of Na+ = 4.0 moles Na2CO3 x

= 8.0 moles Na+ and 4.0 moles of CO3-2 are present

H2O

2 mol Na+

1 mol Na2CO3

Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - II

b) RbF(s) Rb+(aq) + F -

(aq) H2O

moles of RbF = 46.5 g RbF x = 0.445 moles RbF 1 mol RbF 104.47 g RbF

thus, 0.445 mol Rb+ and 0.445 mol F - are present

c) FeCl3 (s) Fe+3(aq) + 3 Cl -

(aq)

H2O

moles of FeCl3 = 9.32 x 1021 formula units 1 mol FeCl3

6.022 x 1023 formula units FeCl3

x

= 0.0155 mol FeCl3

moles of Cl - = 0.0155 mol FeCl3 x = _________ mol Cl - 3 mol Cl -

1 mol FeCl3

and ____________ mol Fe+3 are also present.

Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - III

d) ScBr3 (s) Sc+3(aq) + 3 Br -

(aq)

H2O

Converting from volume to moles:

Moles of ScBr3 = 75.0 ml x x = 0.042 mol ScBr3 1 L103 ml

0.56 mol ScBr3

1 L

Moles of Br - = 0.042 mol ScBr3 x = 0.126 mol Br - 3 mol Br -

1 mol ScBr3

0.042 mol Sc+3 are also present

e) (NH4)2SO4 (s) 2 NH4+

(aq) + SO4- 2

(aq)

H2O

Moles of NH4+ = 7.8 moles (NH4)2SO4 x = ____ mol NH4

+ 2 mol NH4+

1 mol(NH4)2SO4

and ______ mol SO4- 2 are also present.

Precipitation Reactions: Will a Precipitate form?

If we add a solution containing Potassium Chloride to a solution containing Ammonium Nitrate, will we get a precipitate?

KCl(aq) + NH4NO3 (aq) = K+(aq) + Cl-

(aq) + NH4+

(aq) + NO3-(aq)

By exchanging cations and anions we see that we could have PotassiumChloride and Ammonium Nitrate, or Potassium Nitrate and Ammonium Chloride. In looking at the solubility table it shows all possible products as soluble, so there is no net reaction!

KCl(aq) + NH4NO3 (aq) = No. Reaction!

If we mix a solution of Sodium sulfate with a solution of Barium Nitrate,will we get a precipitate? From the solubility table it shows that Barium Sulfate is insoluble, therefore we will get a precipitate!

Na2SO4 (aq) + Ba(NO3)2 (aq) BaSO4 (s) + 2 NaNO3 (aq)

Precipitation Reactions: A solid product is formed

When ever two aqueous solutions are mixed, there is the possibility of forming an insoluble compound. Let us look at some examples to see how we can predict the result of adding two different solutions together.

Pb(NO3)2 (aq) + NaI(aq) Pb+2(aq) + 2 NO3

-(aq) + Na+

(aq) + I-(aq)

When we add These two solutions together, the ions can combine in the way they came into the solution, or they can exchange partners. In this case we could have Lead Nitrate and Sodium Iodide, or Lead Iodide andSodium Nitrate formed, to determine which will happen we must look atthe solubility table(P 141) to determine what could form. The table indicates that Lead Iodide will be insoluble, so a precipitate will form!

Pb(NO3)2 (aq) + 2 NaI(aq) PbI2 (s) + 2 NaNO3 (aq)

Predicting Whether a Precipitation Reaction Occurs; Writing Equations:

a) Calcium Nitrate and Sodium Sulfate solutions are added together.

Ca(NO3)2 (aq) + Na2SO4 (aq) CaSO4 (s) +2 NaNO3 (aq)

Ca2+(aq)+2 NO3

-(aq) + 2 Na+

(aq)+ SO4-2

(aq) CaSO4 (s) + 2 Na+(aq+) 2 NO3

-(aq)

Molecular Equation

Total Ionic Equation

Net Ionic Equation

Ca2+(aq) + SO4

-2(aq) CaSO4 (s)

Spectator Ions are Na+ and NO3-

b) Ammonium Sulfate and Magnesium Chloride are added together.

In exchanging ions, no precipitates will be formed, so there will be no Chemical reactions occurring! All ions are spectator ions!

Acids - A group of Covalent molecules which lose Hydrogen ions to water molecules in solution

HI(g) + H2O(L) H+(aq) + I -

(aq)

When gaseous hydrogen Iodide dissolves in water, the attraction of theoxygen atom of the water molecule for the hydrogen atom in HI is greater that the attraction of the of the Iodide ion for the hydrogen atom, and it is lost to the water molecule to form an Hydronium ion and an Iodide ion in solution. We can write the Hydrogen atom in solution as either H+

(aq) or as H3O+(aq) they mean the same thing in solution. The

presence of a Hydrogen atom that is easily lost in solution is an “Acid”and is called an “acidic” solution. The water (H2O) could also be written above the arrow indicating that the solvent was water in which the HIwas dissolved.

HI(g) + H2O(L) H3O+(aq) + I -

(aq)

HI(g) H+(aq) + I -

(aq)H2O

Strong Acids and the Molarity of H+ Ions in Aqueous Solutions of AcidsProblem: In aqueous solutions, each molecule of sulfuric acid will loose two protons to yield two Hydronium ions, and one sulfate ion. What is the molarity of the sulfate and Hydronium ions in a solution prepared by dissolving 155g of concentrate sulfuric acid into sufficient water to produce 2.30 Liters of acid solution?Plan: Determine the number of moles of sulfuric acid, divide the moles by the volume to get the molarity of the acid and the sulfate ion. The hydronium ions concentration will be twice the acid molarity.Solution: Two moles of H+ are released for every mole of acid:

H2SO4 (l) + 2 H2O(l) 2 H3O+(aq) + SO4

- 2(aq)

Molarity of H+ = 2 x 0.687 mol H+ = 1.37 Molar in H+

Moles H2SO4 = = 1.58 moles H2SO4155 g H2SO41 mole H2SO4

98.09 g H2SO4

x

1.58 mol SO4-2

2.30 l solutionMolarity of SO4

- 2 = = 0.687 Molar in SO4- 2

Acid - Base Reactions : Neutralization Rxns.

An Acid is a substance that produces H+ (H3O+) ions when dissolved in water, and is a proton donorA Base is a substance that produces OH - ions when dissolved in water. the OH- ions react with the H+ ions to produce water, H2O, and are therefore proton acceptors.

Acids and Bases are electrolytes, and their strength is categorized in terms of their degree of dissociation in water to make hydronium orhydroxide ions. Strong acids and bases dissociate completely, and are strong electrolytes. Weak acids and bases dissociate weakly and are weak electrolytes.

The generalized reaction between an Acid and a Base is:

HX(aq) + MOH(aq) MX(aq) + H2O(L)

Acid + Base = Salt + Water

Selected Acids and Bases

Acids Bases

Strong Strong Hydrochloric, HCl Sodium hydroxide, NaOH Hydrobromic, HBr Potassium hydroxide, KOH Hydroiodoic, HI Calcium hydroxide, Ca(OH)2

Nitric acid, HNO3 Strontium hydroxide, Sr(OH)2

Sulfuric acid, H2SO4 Barium hydroxide, Ba(OH)2

Perchloric acid, HClO4

Weak Weak Hydrofluoric, HF Ammonia, NH3

Phosphoric acid, H3PO4

Acetic acid, CH3COOH (or HC2H3O2)

Writing Balanced Equations for Neutralization Reactions - I

Problem: Write balanced chemical reactions (molecular, total ionic, and net ionic) for the following Chemical reactions: a) Calcium Hydroxide(aq) and Hydroiodoic acid(aq) b) Lithium Hydroxide(aq) and Nitric acid(aq) c) Barium Hydroxide(aq) and Sulfuric acid(aq)Plan: These are all strong acids and bases, therefore they will make water and the corresponding salts.Solution:

a) Ca(OH)2 (aq) + 2HI(aq) CaI2 (aq) + 2H2O(l)

Ca2+(aq) + 2 OH -

(aq) + 2 H+(aq) + 2 I -

(aq)

Ca2+

(aq) + 2 I -(aq) + 2 H2O(l)

2 OH -(aq) + 2 H+

(aq) 2 H2O(l)

Writing Balanced Equations for Neutralization Reactions - II

b) LiOH(aq) + HNO3 (aq) LiNO3 (aq) + H2O(l)

Li+(aq) + OH -

(aq) + H+(aq) + NO3

-(aq)

Li+

(aq) + NO3-(aq) + H2O(l)

OH -(aq) + H+

(aq) H2O(l)

c) Ba(OH)2 (aq) + H2SO4 (aq) BaSO4 (s) + 2 H2O(l)

Ba2+(aq) + 2 OH -

(aq) + 2 H+(aq) + SO4

2-(aq) BaSO4 (s) + 2 H2O(l)

Ba2+(aq) + 2 OH -

(aq) + 2 H+(aq) + SO4

2-(aq) BaSO4 (s) + 2 H2O(l)

GGEENNEERRAAL PL PRROOPPEERRTTIIEESS O OFF S SOOLLUUTTIIOONNSS1. A solution is a homogeneous mixture of two or

more components.

2. It has variable composition.

3. The dissolved solute is molecular or ionic in size.

4. A solution may be either colored or colorless nut is generally transparent.

5. The solute remains uniformly distributed throughout the solution and will not settle out through time.

6. The solute can be separated from the solvent by physical methods.

Colligative Properties of Colligative Properties of SolutionsSolutions Depends on the concentration of the

solute particles and not on the identity of the solute.

Dissolved particles alter and interfere with the dynamic process of a solution.

NOTE: T=Tf-Ti or in this case T=Tsolution-Tsolvent

Boiling point elevationBoiling point elevationFreezing point depressionFreezing point depression

OsmosisOsmosisVapor pressure loweringVapor pressure lowering

Vapor pressure lowering

• For an ideal solution, the equilibrium vapor pressure is given by Raoult's law as

• - is the vapor pressure of the pure component i (= A, B, ...) and    is the mole fraction of the component i in the solution

• For a solution with a solvent (A) and one non-volatile solute (B),    and 

• The vapor pressure lowering relative to pure solvent is  , which is proportional to the mole fraction of solute.

Boiling point elevation (ebullioscopy)

• The boiling point of a pure solvent is increased by the addition of a non-volatile solute, and the elevation can be measured by ebullioscopy.

• Here i is the van't Hoff factor as above, Kb is the ebullioscopic constant of the solvent (equal to 0.512°C kg/mol for water), and m is themolality of the solution

Freezing point depression (cryoscopy)

• The freezing point of a pure solvent is lowered by the addition of a solute which is insoluble in the solid solvent, and the measurement of this difference is called cryoscopy.

• Here Kf is the cryoscopic constant, equal to 1.86°C kg/mol for the freezing point of water. Again i is the van't Hoff factor and m the molality.

Osmotic pressure• The osmotic pressure of a solution is the

difference in pressure between the solution and the pure liquid solvent when the two are in equilibrium across a semipermeable membrane.

• If the two phases are at the same initial pressure, there is a net transfer of solvent across the membrane into the solution known as osmosis.

MOLALITYMOLALITY• Molality = moles of solute per kg of solvent

• m = nsolute / kg solvent

• If the concentration of a solution is given in terms of molality, it is referred to as a molal solutionmolal solution.

Q. Calculate the molality of a solution consisting of 25 g of KCl in Q. Calculate the molality of a solution consisting of 25 g of KCl in 250.0 mL of pure water at 20250.0 mL of pure water at 20ooC?C?

First calculate the mass in kilograms of solvent using the density of solvent:

250.0 mL of H2O (1 g/ 1 mL) = 250.0 g of H2O (1 kg / 1000 g) = 0.2500 kg of H2O

Next calculate the moles of solute using the molar mass:

25 g KCl (1 mol / 54.5 g) = 0.46 moles of solute

Lastly calculate the molality:

m = n / kg = 0.46 mol / 0.2500 kg = 1.8 1.8 mm (molal) solution (molal) solution

Freezing Point DepressionFreezing Point Depression

TTff = - k = - kff mmQ. Estimate the freezing point of a 2.00 L sample of seawater (kf = 1.86 oC kg / mol), which

has the following composition:

0.458 mol of Na+ 0.052 mol of Mg2+ 0.010 mol Ca2+

0.010 mol K+ 0.533 mol Cl- 0.002 mol HCO3

-

0.001 mol Br- 0.001 mol neutral species.

Since colligative properties are dependent on the NUMBER of particles and not the character of the particles, you must first add up all the moles of solute in the solution.

Total moles = 1.067 moles of solute

Now calculate the molality of the solution:

mm = moles of solute / kg of solvent = 1.067 mol / 2.00 kg

= 0.5335 mol/kg

Last calculate the temperature change:

Tf = - kf m = -(1.86 oC kg/mol) (0.5335 mol/kg) = 0.992 oC

The freezing point of seawater is Tsolvent - T = 0 oC - 0.992 oC

= - 0.992 oC

Boiling Point Elevation

Tb = kb mQ. The boiling point of a solution containing 40.0 g of an unknown substance dissolved in

100.0 g of water is 105.3 oC . Calculate the molar mass of the compound.

Since the solvent is water, the change in temperature (T) would be 105.3 - 100.0 oC = 5.3 oC. You can also find the kb in the table in your textbook, kb = 0.512 oC kg/mol.

From this data, you can calculate the molality:

m = Tb / kb = 5.3 oC / 0.512 oC kg/mol = 10.4 mol/kg

Molality is also defined as the moles of solute per kg of solvent:

m = n /(kg solvent), can be rearranged to be n = m (kg of solvent)

n = 10.4 mol /kg (0.1000 kg) = 1.04 mol of solute

The molar mass can be calculated by using the equation, MW = m/n

MW = 40.0 g / 1.04 mol = 38.5 g/mol