Solution Chemistry

Characteristics of solutions:

solute - the material that is being dissolved into something else.

Example : when we dissolve sugar into a cup of coffee, the sugar is the solute.

solvent - the material into which a solute is being dissolved. In the above example, the coffee is a solvent. Solvents can be almost any material or in any state but for the most part, we will be using water as a solvent.

solution - a combination of a solute and a solvent. We can not distinguish the identify a solute and a solvent- it is a HOMOGENEOUS MIXTURE!

A solution can be in any combination of states:

- a solid/liquid combination (dissolving a solid in water such as salt dissolved in water).

- a liquid/liquid combination (table vinegar is a mixture of water and acetic acid).- solid/solid combination (amalgam or any type of alloy).

- Air is an example of gas solution: nitrogen 78.08%oxygen 20.955argon 0.93%Carbon dioxide 0.03%

-solutions having water as the solvent is referred to as AQUEOUS solution (blood, saliva, etc).

There comes a point where no more solute can dissolve into the solution. At this point in time the solution is said to be saturated.

- When solute can still dissolve in a solution, it is said to be unsaturated. - To increase the solubility of the chemical, we can usually heat the solution up. - The result will be a solution that contains more solute than what would normally be dissolved in the solvent. This is called a supersaturated solution.

Solubility of a solute is the maximum amount of solute that can dissolve in a given amount of solution at a given temperature.

- Solubility is temperature dependent. Usually an increase in temperature increases the solubility and increases how fast a solid dissolves in water.

- Solubility is measured in terms of g/L, g/mL, or mol/L (at a given temp.)

- Solubility of solids in liquids vary widely:

e.g. Solubility of NaCl in water at 25oC: ~6 mol/LSolubility of NaCl in ethanol at 25oC: ~ 0.0009 mol/L

e.g. Solubility of AgNO3 in water at 25oC: ~1 mol/LSolubility of AgCl in water at 25oC: ~0.00001mol/L

- There is an upper limit to the solubility of a solid. It important to note that even the least soluble solid will have a few particles dissolve per litre of solution.

- So, we will say that a substance is soluble if 0.1 mol of the substance dissolves per litre.

When there is no apparent limit to the solubility of one substance in another, the components are said to be MISCIBLE.

Electrical Conductivity:

Metals (solid and liquid phase): conduct electricity

Non-metals (solid and liquid phase): non-conductor of electricity

Ionic compound in solid phase: non-conductore.g. NaCl(s),

Ionic compound in liquid and aqueous phase: conductor

Covalent compounds in solid and liquid phase: non-conductor (usually!)

Acid and Base: conductor

Examples: Which of the following would you expect to form conducting (ionic) solutions?

a) KI b) ICl c) HBr d) CH4 e) LiOH

f) CH3OCH3 g) N2O h) HCl

Examples: Which of the following would you expect to conduct electricity?

a) NaCl(s) b) CH3OH(l) c) Cu(s) d) Fe(l)

e) K2CrO4(aq) f) C14H10(s) g) CO2(s) h) KBr(aq)

i) KBr(s) j) Ag(s) k) Hg(l) l) Na(s)

Polarity of Molecules:

A. Van der Waal’s Forces (intermolecular interaction):

1. Dipole-dipole interactions:

molecules that are dipolar are attracted electrostatically with one another.

Polar molecules have a partial positive and a partial negative portion of the molecule.

They are made of atoms having differences in ELECTRONEGATIVITY.

Polar molecule must also be ASYMETRICAL.

2. London Forces:

are weak attractive forces which arise as the result of temporary dipolar attraction between neighbouring atoms.

London Forces are always present, even in species that have dipolar attraction.

This is a weaker attraction than dipole-dipole forces.

B. Hydrogen Bonding (intermolecular interaction):

Hydrogen bonding occurs when hydrogen is bonded covalently to N, O, or F (high electronegativity). Highly polar.

Hydrogen bonding is stronger than the Van der Waal’s Forces.

The presence of H-O- , H-N-, H-F indicate a highly polar section of the molecule.

Asymetrical molecules are POLAR molecules.

Polar and Non-Polar Solvents:

Polar Molecules: - Dipole is present- Molecule is ASYMETRICAL (not symmetrical).

eg. HCl Chloromethane HBr Ammonia

eg. formaldehyde acetylene propanol dimethyl ether

Polar Solvent: Water Methanol ethanol

Non-polar Solvent: benzene chloroform

“Like dissolves like” :

polar solvents dissolve polar solutes and ionic compounds Non-polar solvents dissolve non-polar solvents

A. Polar and ionic molecules are soluble in polar solvents.

For example: Salt (ionic) dissolves in water (polar solvent).For example: I2(s) is more soluble in water than gasoline.

non-polar molecules are soluble in non-polar solvents.

B. Polar and ionic compounds are generally held together quite tightly.

Polar solvents are strongly attracted to the ions or the polar compounds causing separation of the solute out of its crystal structure and into solution.

Non-polar solvents do not have this attraction, therefore, polar solutes are unable to separate into a non-polar solution.

C. Non-polar solutes are held together by London Forces.

Solvents with enough London Forces to attract the solutes and break off the solid crystal structure are capable of dissolving the solid. Generally speaking, non-polar solvents have high London Forces, and dissolve other non-polar compounds.

Hydrogen bonding holds water molecules together. For example, water molecules undergo hydrogen bonding. When water freezes, the six sided structure of water crystals is due to hydrogen bonding.

The nature of Solutions of ions:

The formation of solution depends on the ability of the solute to dissolve in the solvent.

Ionic solid are crystal structure made up of ions.

Covalent solids are crystal structure made up of covalent molecules.

Crystal lattice is an orderly arrangement of particles that exists within a crystal.

Water: “Universal Solvent”, polar, V-shaped molecule. Its positive end attracts the negative ions and the negative end attracts the positive ions.

Dissociation: reaction that involves the separation of ions in an ionic compound.

KBr(s) ------- > K+(aq) + Br-

(aq)

Ionization: reaction that involves the breaking up of a neutral molecule into ions.

CH3COOH(l) ----------- > CH3COO - (aq) + H +

(aq)

Assignment: read p. 133-148, Zumdahl. Exercises:p. 180-181 # 11, 13, 15, 17, 25

Calculating The concentration of Ions in Solution:

concentration - this is the amount of solute that is in a certain amount of solution. Because 'amount of solute' can be interpreted in a number of different ways (i.e. grams, moles, molecules etc.) we will define a value for concentration : moles solute (moles

concentration = ------------------------------- volume solution (Litre)

This is also defined as MOLARITY.

moles of soluteMolarity = ------------------------------------

volume (in litres) solution

The units for molarity are moles/litre, mol/L or M for short.

Examples : 1) What is the concentration of a solution containing 0.055 moles of HCl dissolved in water to make 250.0 mL of solution ?

Solution: moles 0.055 mol

M = -------- -- = --------- --- = 0.22 M HCl volume 0.2500 L

2) What is the concentration of a solution containing 0.055 grams of HCl dissolved in water to make 250.0 mL of solution ?

Solution : 0.055 g x 1 mol

moles HCl = ---------- = 0.0015 mol HCl 36.5 g

0.0015 molmolarity = ----------- = 0.0060 M HCl

0.250 L

3) How many molecules of NaOH are dissolved in 125.3 mL of a 3.00 M NaOH solution ?

Solution :

moles NaOH = Molarity X volume = (3.00 M)(.1253 L) = 0.376 mol NaOH

molecules NaOH = moles X Avogadro's number= (0.376 mol)(6.02 x 1023)= 2.26 x 1023 molecules NaOH

How to make a solution of a certain concentration:

In order to make a solution of certain concentration, we do NOT dissolve the amount of solute needed in the volume of solvent. If we did that, the volume would increase somewhat and we would not have the precise concentration we require. In order to accomplish this, we simply determine the amount of solute needed, dissolve this in about 1/2 the amount of solvent and then fill up to the proper amount of solution. It is best to use a volumetric flask for this purpose.

Example : Describe how you would make 125 mL of a 0.250 M solution of sodium hydroxide.

Solution :

mol NaOH = Molarity X volume = (0.250 M)(0.125 L) = 0.03125 mol NaOH

mass NaOH = moles X molar mass= (0.03125 mol)(40.0 g/mol)= 1.25 g NaOH

Dissolve 1.25 g of NaOH into about 60 mL distilledwater. Add more distilled water to raise volume upto 125 mL.

- Assignment :- Solution Worksheet #1 - Molarity

Ionic concentrations

When a solid dissolves in a solvent it can act one of three ways - it can either dissociate into its ions (only if it is an ionic compound);it can remain as a molecule dissolved in water (if it is not an ionic compound);or it can react with the water (we will not consider this at the present time). For the sake of Chemistry 11, we will discuss only what happens when an ionic substance dissolves in a solvent.

First of all, why do some ionic compounds dissolve readily in water while some others appear to not be as readily soluble?

The reasons for this are due to the relative shapes and structures of both the solute and solvent molecules. Sometimes the molecules have a slight charge on one side of the molecule compared to the other. This type of molecule is called POLAR. Water is a good example of this. A NON-POLAR molecule, such as carbon tetrachloride, can also be used as a solvent. The polarity of the molecules is the basis of dissolution, but we will not get into that aspect of chemistry in this course.

When an ionic compound dissolves in water, it breaks into its ions. Example :

NaCl (s) ----> Na+(aq) + Cl-(aq)

CaF2 (s) ----> Ca2+(aq) + 2 F-(aq)

We are going to assume that all of the compound breaks into its ion state. This is not always true, but we will leave that for Chem 12.

If all of the compound ionizes, then we actually have none of the solute molecules left in the solution. By looking at the ratios between the ions and compound (from the dissociation equation) we can determine the concentrations of each individual ion in the solution.(The abbreviation for concentration of an ion is to place the symbol for the ion inside square brackets. For instance, shorthand for 'the concentration of the hydrogen ion' is [H+].)

Examples :Find the concentration of each ion in the following solutions :

a) 12.0 M HCL b) 18.0 M H2SO4 c) 1.15 M Cr2(SO4)3Solutions :a) HCl ---> H+ + Cl- [H+] = 12.0 M [Cl-] = 12.0 M

b) H2SO4 ---> 2H+ + SO42- [H+] = 36.0 M [SO42-] = 18.0 M

c) Cr2(SO4)3 ---> 2 Cr3+ + 3 SO42- [Cr3+] = 2.30 M [SO42-] = 4.45 M

Dilutions

When a solution of known concentration has its volume changed, it stands to reason that its concentration changes also. If the liquid we are adding does not contain any of the original solute OR if a reaction does NOT take place, the amount of dissolved solute does not change - only the volume changes. In fact the concentration will change by a factor equal to the change in volume. In short, the new concentration can be found by using the following formula :

C1V1=C2V2

old volumenew concentration = old concentration X --------------------

new volume

Examples :

1) 125 mL of 3.00 M HCl has 35.8 mL of water added to it. What is the new concentration ?Solution :

old volumenew M = old M X --------------------

new volume

25 mL = 3.00 M X -------------- = 2.33 M

160.8 mL

2) 15.0 mL of a solution of 1.50 M HCl has some water added to it changing the concentration of the acid to 1.15 M. What volume of water was added ?

Solution : old volume

new M = old M X ---------------------- therefore rearranging new volume we get the following

old M new volume = old volume X ------

new M 1.50 M

= 15.0 mL X ------- 1.15 M

= 19.6 mL

If the new volume was 19.6 mL and the old volume was 15.0 mL, then I guess we added 4.6 mL of water to the acid.

3) How much 6.00 M NaOH would we have to use to produce 275 mL of 0.500 M NaOH ?

Solution : old volumenew M = old M X ----------------- therefore rearranging

new volume we get the following

new Mold volume = new volume X -------------

old M 0.500 M = 275 mL X --------------- 6.00 M

= 22.9 mL of 6.00 M NaOH needed. (Iguess we would also need about252.1 mL of water but that was notasked.)

- Assignment : Begin Solution worksheet #2 (to be completed next class)

- Quiz on Molarity worksheet #1

Mixing together solutions of different concentrations

As we said last class, if we mix together two solutions of different concentrations together, the formula we used can NOT be used. In order to find the new concentration, we must find the amount of solute in each solution.

Example : Determine the new concentration of HCl if225 mL of 1.50 M HCl is mixed with 115 mL of2.50 M HCl.

Solution:moles HCl from first solution

mol = c X V = (1.50 M)(0.225 L) = 0.3375 mol HCl

moles HCl from second solution = (2.50 M)(0.115 L) = 0.2875 mol HCl

total moles HCl = 0.3375 mol + 0.2875 mol = 0.625 mol HCl

total volume = 225 mL + 115 mL = 340 mL

total moles 0.625 molfinal concentration = ----------------- = - -----------------

total volume 0.340 L

= 1.84 M

- complete worksheet #3

Reactions in SolutionsNet Ionic Equations

When two solutions are mixed together, a reaction might occur. (Demo : 0.1 M Pb(NO3)2 + 0.1 M KI). The reactions that we are going to talk about here are all double replacement reactions so if a reaction does take place, we can determine the products.

Pb(NO3)2 + 2 KI ---> PbI2 + 2 KNO3

The above equation is called a chemical equation. We have done these in the past. However, we have recently seen that we don't actually have any Pb(NO3)2 or KI in the solution, just their ions. Also, on the product side of the equation we have the yellow solid being PbI2 (we'll figure out how to determine which is the precipitate later) and a solution of KNO3. Rewriting the equation to show this we get

Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq) ---> PbI2 (s) + 2K+(aq) + 2NO3-(aq)

This is called an overall ionic equation and is not used that often because of a number of reasons - not the least of which is that it takes up too much space !!

If we look carefully at the overall ionic equation, we see that there are 2NO3-(aq) ions and 2K+(aq) ions on either side of the equation. If we cancel them out, we get

Pb2+(aq) + 2I-(aq) ---> PbI2 (s)

This is called a NET IONIC EQUATION for the reaction shown above. It shows exactly what is happening in the solution. The NO3-(aq) ions and the K+(aq) ions are actually doing nothing in the reaction. They can be thought of as just watching the lead ions and the iodide ions get together. For this reason, they are referred to as SPECTATOR IONS. Spectator ions are never included in net ionic equations.

Another example :

CuSO4 + 2 NaOH ---> Cu(OH)2 + Na2SO4 assume that theprecipitate is Cu(OH)2

overall ionic equation

Cu2+(aq) + SO42-(aq) + 2Na+(aq) + 2OH-(aq) --->Cu(OH)2 (s) + 2Na+(aq) + SO42-(aq)

net ionic equation

Cu2+(aq) + 2OH-(aq) ---> Cu(OH)2 (s)

If you look carefully at the net ionic equation of any precipitation reaction, you will find that the equation is simply how we would go about making the solid. We do not always have to

go about all three equations in order to obtain the net ionic equation. All we have to do is to identify the precipitate (don't worry, we'll get to it) and simply write how we would produce it using ions.

Example : if we knew we got a precipitate of Ca3(PO4)2 in areaction, the net ionic equation would be

3 Ca2+(aq) + 2 PO43-(aq) ---> Ca3(PO4)2 (s)

This would be the net ionic equation for the formation of the precipitate Ca3(PO4)2 no matter what solutions were used in the first place. The other ions would just be spectator ions.

There is one type of net ionic equation that occurs when an acid and a base are mixed together. A precipitate is NOT formed but a reaction does take place.

Acids we know - HCl, H2SO4 , HNO3 etc. If we look at them carefully, all acids contain an H atom. More specifically, when it ionizes, an H+ ion is released.

Bases can be identified by the presence of an OH- ion.When an acid is added to a base, water and a salt are formed (review from types of

reactions). Let's see what happens in the reaction between NaOH and HCl.

NaOH + HCl ---> H2O + NaCl

Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) ---> H2O (l) + Na+(aq) + Cl-(aq)

H+(aq) + OH-(aq) ---> H2O (l)

This is the net ionic equation for the reaction between any acid and any base.

- Assignment :

Prepare 2 data tables for 16-DThe tables should be 7x7 matrices - do NOT have the students write in the chemicals

before they come into the lab. They might find the chemicals are different. The data tables are their 'ticket' into the class - no

data tables - no entry !!!!

- Lab 16-D

Formation of precipitates

We saw in the lab yesterday that some compounds are soluble in water and some are not. We will use the definition here that soluble means that we can make at least a 0.1 M solution of the compound at 25oC. How can we predict if a precipitate will form ? The answer is not easy. In fact we don't have too many rules that always apply. Chemists have done thousands of experiments similar to the one you did yesterday and have collated the results into pages and pages of data tables. We will use a portion of those tables and rules - the data table on page 459 of the text book. (The students will not be required to memorize this table - they will be provided with one for the exams.)

To use this table, we determine what the product will be in a reaction (remember, the reactions here are double replacement reactions.) From here we must check out both products to see if there is a precipitate formed.

Examples : 1) Identify the precipitate (if any) when the following solutions are mixed:a) lead (II) nitrate and sodium chlorideb) ammonium hydroxide and copper (I) sulfatec) potassium phosphate and cesium sulfide

Answers : the following are the products formed. The one that is underlined is the precipitate.a) lead (II) chloride and sodium nitrateb) ammonium sulfate and copper (I) hydroxidec) potassium sulfide and cesium phosphate

2) Write the net ionic equation for the formation of any precipitate when solutions of the following are mixed. If no precipitate is formed, write NO REACTION.

a) barium sulfide and ammonium carbonate

b) silver nitrate and sodium acetate

c) iron (III) chloride and copper (II) chloride

d) lead (II) acetate and lithium phosphate

Answers :

a) Ba2+(aq) + CO32-(aq) ---> BaCO3 (s)b) Ag+(aq) + CH3COO-(aq) ---> AgCH3COO (s)c) NO REACTIONd) 3 Pb2+(aq) + 2 PO43-(aq) ---> Pb3(PO4)2 (s)

Precipitation Formation and Ionic Concentrations

As we can see from the above and from the lab, sometimes we get a precipitate formed in a reaction. What does this do to the concentrations of the ions in solution ? It lowers those involved. We will, for the sake of argument here, assume that one of the ions is completely used up in the reaction while some of the other one may still remain. If this sounds familiar, it should. This is really a stoichiometric INXS calculation. It is best set up in the following example :

Example : 25.0 mL of 1.00 M NaCl is mixed with 35.0 mL of2.20 M AgNO3. A precipitate is formed and falls out of solution. Calculate the concentrations of all ions remaining in solution after the precipitation stops.

Answer : It is best to first write the net ionic equation for the reaction that occurs.Ag+(aq) + Cl-(aq) ---> AgCl (s)

Set up a table similar to the one below :

Ion Moles Before Moles After Volume Concentration

Na+ 0.0250 mol 0.0250 mol 0.0600 L 0.417 M

Cl- 0.0250 mol 0 mol 0.0600 L 0.000 M

Ag+ 0.0770 mol 0.0520 mol 0.0600 L 0.867 M

NO3- 0.0770 mol 0.0770 mol 0.0600 L 1.28 M

Example : 20.0 mL of 0.500 M AgNO3 is mixed with 15.2 mL of0.750 M Na2S. Calculate the concentrations of allions remaining in solution.

Answer : 2 Ag+(aq) + S2-(aq) ---> Ag2S (s)

Ion Moles Before Moles After Volume Concentration

Ag+ 0.0100 mol 0 mol 0.0352 L 0.000 M

NO3- 0.0100 mol 0.0100 mol 0.0352 L 0.284 M

Na+ 0.0228 mol 0.0228 mol 0.0352 L 0.648 M

S2- 0.0114 mol 0.0064 mol 0.0352 L 0.182 M

- Assignment : Read p. 157, exercise p. 182 # 29-32, 35,37, 39, 43

Solution worksheet #4

Lab: Introduction of Acids and Bases, (lab 20-A)

read p. 158-164

Acids and BasesFrom the lab, we should be able to identify an acid and a base from its properties and its

formula. In short, the chemical properties of solutions of acids and bases are as follows :

Acids :

1) taste sour2) conduct an electrical current3) cause certain dyes to change colour (e.g. litmus goes red)4) liberates hydrogen when it reacts with certain metals5) loses the above properties when mixed with a base though the resulting solution conducts electricity.

Bases :

1) taste bitter2) conduct an electrical current3) cause certain dyes to change colour (e.g. litmus goes blue, phenolphthalein goes pink)4) feels slippery5) loses the above properties when mixed with a base though the resulting solution conducts electricity.

In their respective chemical formula, and more importantly when they dissolve in water, acids have an H+ ion while bases have an OH- ion. (This is an incomplete definition of an acid and a base but will suffice for chemistry 11.)

When an acid and a base react, the result is a salt and water. The type of reaction is called a NEUTRALIZATION reaction because the properties of the acid and the base are neutralized by each other. The net ionic reaction for this reaction is as follows :

H+(aq) + OH-(aq) ---> H2O (l) We must remember that in any neutralization reaction, the moles of H+ MUST equal the moles of OH-.

Example : What volume of 6.00 M HCl must be added to 125 mL of 1.59 M NaOH in order to neutralize it ?

Answer :NaOH + HCl ---> NaCl + H2O

moles OH- = (1.59 M)(0.125 L) = 0.199 mol OH-

moles H+ = moles OH- = 0.199 mol H+

mol HCl = mol H+ = 0.199 mol HCl

mol HCl 0.199 molvol HCl = ------- = --------- = 0.0331 L (33.1 mL)

M HCl 6.00 M

Example : What volume of 0.10 M Ca(OH)2 is needed to neutralize 115.2 mL of 0.55 M H3PO4?

Answer :mol H3PO4 = (0.55 M)(0.1152 L) = 0.063 mol H3PO4mol H+ = 3 x mol H3PO4 = (3)(0.063 mol)= 0.19 mol H+mol OH- = mol H+ = 0.19 mol OH-mol Ca(OH)2 = (mol OH-)/2 = (0.19 mol)/2 =

0.095 mol Ca(OH)2 mol Ca(OH)2 0.095 molvol Ca(OH)2 = ----------- = --------------= 0.95 L M Ca(OH)2 0.10 M (950 mL)

Titrations

A titration reaction is simply a neutralization reaction using special techniques. A given volume of an acid is placed in a flask (as well as some indicator - phenolphthalein is commonly used). The base is placed in a burette and added drop wise until the colour of the solution in the flask just changes. From here we can get the volumes needed for neutralization calculations.

- Assignment : exercise p. 183 # 45, 46, 47, 49, 51 Zumdahl!

Solution worksheet #3 - net ionic equations, - neutralization calculations- titration calculations

- Set up lab 20-C

Day 10:

- Lab 20-C - day 1

- Lab 20-C day 2

- collect lab 20-C

- go over worksheet #3

Demonstration titration labEffectiveness of an Antacid

1. Obtain several commercial antacids. (Tums, Rolaids, Maalox, Eno etc.)

2. Place one 'dose' of an antacid in an Erlenmeyer flask.If the antacid is a solid (like Tums), grind it up with a mortar and pestle first.

3. Place 25 mL of distilled water in the flask with the antacid. 4. Place 10.0 mL of 1.0 M HCl (stomach acid) in the flask.

5. Add a few drops of phenolphthalein in the flask.

6. Let the antacid react in the 'stomach' (flask) for aminute or so.

7. Titrate using 1.0 M NaOH. Record the volume of NaOHneeded to neutralize the excess stomach acid.

8. Compare the volume needed for each antacid. The smallerthe volume of NaOH needed, the more effective theantacid has been in neutralizing stomach acid. Note :if the volume of NaOH needed exceeds the 10 mL of HCladded, then the antacid actually adds acid to thestomach. (Read package of Eno carefully - it doesn'tactually say antacid - now read contents - over 55% ofthe contents are acid !!! Why ? plop plop fizz fizz.

9. A discussion might also ensue on the 'fillers' of thetablets such as Tums. Weigh two tablets. Compare themeasured mass with the mass of the 'activeingredients'. The discrepancy in mass is due to anumber of fillers.

- Assignment : Solution Worksheet #5 - Review

- go over the homework - solution review

- exam next day on solution chemistry

- Begin year end review

NOTE : the cumulative exam coming up does not include all of the material covered in the solution unit. For that reason, the cumulative exam can be given either at the end of the term or about half way through the last unit.

- solution exam