Module 5Solution Chemistry

Name:Mark: /20

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Solution ChemistryMost lab experiments in high school are completed in solutions. In fact most

reactions involve solutions where ever possible because they are convenient and more easily manipulated (diluted, transferred etc.). Some problems with other phases include:

1) Gas Phase: Need to make sure the reaction is air tight so no air gets in from the environment and no gas from the experiment escapes.

2) Solid State: The reaction is too slow or inert so the reaction can’t occur at all.

3) Pure Liquids: Are too concentrated and react uncontrollably.

Before we start, lets make sure we are clear on the vocabulary…

A solution is a homogeneous mixture while a suspension or mechanical mixture are heterogeneous mixtures. This makes it different to a pure substance because a solution is made up of at least 2 types of molecules mixed uniformly while a pure substance is composed only of one type of molecule. A solvent is the component in a solution which exists in greater quantity. Water is the universal solvent. While a solute is the component in a solution which exists in the smaller quantity. A solute is soluble in a solvent if the solute and solvent mix to form a uniform appearance (ie. a homogeneous mixture) – that is the solute has dissolved in the solvent. An example of this is sugar in water. When the solute does not dissolve in the solvent it is said to be insoluble. An example of this is oil in water. When iced tea is made from crystals, the crystals are the solute and water is the solvent. You add one spoon of iced tea crystals to a cup of water. The solution you have made is not sweet enough so you add more crystals. You can continue to sweeten this solution because it is unsaturated, it can dissolve more of the solute being added. Is there a limit to how much crystal powder can be added? What happens when you add too much? Excess settles at the bottom. A solvent is said to be saturated with a particular solute if the solvent has dissolved as much of the solute as possible. We can observe this because the extra solute settles at the bottom. We could probably increase the amount of solute if we stirred or heated the solution. These observations can be used to describe our solvent (in this example water). The solubility of a solute is the maximum amount of the solute which can dissolve in a given amount of solvent at a given temperature, or the amount of substance needed to saturate a solution. For example “the solubility of a sweet glass of iced tea is 4 teaspoons of crystals in 1 cup of water at room temperature”.

The solubility of Ba(NO2)2∙H2O is 63g/100mL of H2O at 20oC109.6g/100mL of H2O a 80oC1.6g/100mL of alcohol at 20oC

In summary of the paragraph above, what 5 things must be specified when defining the solubility of a solution?

1. solute2. amount of solute3. solvent 4. amount of solvent5. temperature of the solution

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Solubility Start up Questions

1. Which substance is the most soluble at room temperature (21oC)?

2. Which substance is the least soluble at room temperature? Explain.

3. Which substance has the greatest change in solubility as temperature increases?

4. Which substance has the least change in solubility as temperature increases?

5. Which substance has a decrease in solubility as temperature increases?

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6. Types of compounds:Ionic Compounds: metals-nonmetal (NaCl) or metals-radicals (NaNO3)Molecular Compounds: non metal-non metal; compounds of covalent bonds only

(C6H12O6)When these compounds are put in water a physical change occurs. Molecular compounds are often hydrated with water – for our purposes remain unchanged except for the state.

Eg. C4H10(s) C4H10(aq) Remember aqueous mean dissolved in waterWhen ionic compounds are put in water, the compounds dissociate to their ionic state and float around in the solvent as ions. We can show this in an equation. This type of equation is called a Dissociation Equation (DE):

Eg1. NaI(s) Na+(aq) + I - (aq)Eg2. BaF2 Ba2+(aq) + 2F - (aq)

Notice now we must * indicate the states of all parts of the equation* indicate the charge of the ions* balance the equation

Try some questions yourself:What is the dissociation equation for Al2(CO2)3(s) in water?

Al2(CO2)3(s) 2Al3+(aq) + 3CO22-(aq)

A) Classify the following compounds as ionic (I) or molecular (M):1. SCl2 _______2. CaS _______3. SiO2 _______4. LiClO ______5. CH3OH _____

6. MnO2 ______7. PbO _______8. KCl _______9. SO3 _______10. C2H6 _______

B) Write the dissociation equation by completing the following: (All solids)1. C6H12O6 2. Ba(NO3)2

3. Na2S4. Li3PO4

5. CuCl26. SiO2

7. GeCl4How much do you remember?C) 7.42g of sodium nitrate is dissolved in water to form 80.0mL of solution1. Calculate the molarity of the solution

2. Write the dissociation equation

D) A beaker contains 42.0mL of 0.950M calcium chlorate1. Write the dissociation equation (DE).

2. Calculate the mass of solid present if the water is evaporated.

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Double Replacement Reactions – A second look

The Solubility Chart can give us more detail for our double replacement reactions. It allows you to predict the form of your products assuming a reaction occurs.

For example, what is the chemical reaction when NaCl(aq) and Pb(NO3)2 (aq) react?

2NaCl + Pb(NO3)2 2NaNO3 + PbCl2

We now have the tools to indicate whether a precipitate (solid) is formed or not.

Checking the solubility chart if the compound formed is said to be “soluble” then keep it in aqueous form ie. (aq). If the compound formed is said to have “low solubility”, a precipitate (ppt) is formed in other words, the compound is solid (s).

Lets add this information to the above equation.

2NaCl(aq) + Pb(NO3)2(aq) 2NaNO3 (aq) + PbCl2 (s)

Look at the solubility chart: NaCl is an aqueus solution because from the solubility chart, we can see Cl- forms a soluble

compound with ALL metals (except Ag, Pb, Cu) Pb(NO3)2 is an aqueous solution because ANY compound with nitrate is soluble NaNO3 is aqueous by the same token as Pb(NO3)2 PbCl2 is as solid because is has low solubility when Cl- reacts with Pb2+

This equation which shows the double replacement reaction in terms of aqueous and solid is called the Complete Reaction Equation (CRE)

Complete Ionic Equation (CIE) takes the complete reaction equation and writes all aqueous solution in their ion form. Solids are included as the unaffected compound.

2Na+(aq) + 2Cl-(aq) + Pb2+(aq) + 2 NO3-(aq) 2Na+(aq) + 2NO3-(aq) + PbCl2(s)

Finally we have the Net Ionic Equation (NIE). This takes the complete ionic equation and eliminates the spectator ions (those that do not react), leaving the actual reaction occurring.

Pb2+(aq) + 2Cl- (aq) PbCl2(s)

Consider Na2SO3(aq) reacting with AgNO3(aq)

1. Write the complete reaction equation. Name the ppt formed. Balance the equation.

2. Write the complete ionic equation.

3. Write the net ionic equation

Consider sodium iodide reacting with silver nitrate1. Write the complete reaction equation. Name the ppt formed. Balance the equation.

NaI(aq) + AgNO3(aq) NaNO3(aq) + AgI(s)2. Write the complete ionic equation.

Na+(aq) + I-(aq) Ag+(aq) + NO3-(aq) Na+(aq) + NO3-(aq) + AgI(s)3. Write the net ionic equation

Ag+(aq) + I-(s) AgI(s)

Consult the Solubility Tables and name the ppt formed, if any, on mixing the following solutions. Write the balanced equation if a reaction occurs.

1. barium hydroxide and cadmium chloride

2. ammonium phosphate and aluminum nitrate

3. sodium sulfite and silver nitrate

4. sodium acetate and calcium chloride

5. copper(II) nitrate and sodium chloride

6. strontium nitrate and sulphuric acid

7. sodium hydroxide and manganese(II) acetate

8. sodium iodide and silver nitrate

9. chromic nitrate and potassium sufide

10. A solution of lithium chloride is added to a solution of lead(II) nitratea. Write the complete balanced equation and circle the ppt.

b. Write the net ionic equation11. A solution of magnesium nitrate is added to a solution of ammonium sulphite

a. Write the complete ionic equation for the reaction

b. Write the net ionic equation12. A solution of hydrochloric acid is added to a solution of sodium hydroxide in a neutralization

reactiona. Write the CRE

b. Write the net ionic equation13. In a neutralization reaction, the salt formed is potassium sulphate

a. Write the neutralization equation for this reaction

b. Write the NIE14. 10.0mL of unknown strength HCl is neutralized by 18.4mL of 0.192M NaOH.

a. Write the neutralization equation

b. Calculate the mol of NaOH reacted

c. Calculate the mol of HCl neutralized

d. Calculate the molarity of the HCl solution

15. A 10.0 mL solution of sulphuric acid is neutralized by 24.4mL of 0.280M NaOH. Calculate the molarity of the acid solution

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Stoichiometry in Solution Chemistry

Recall that stoichiometry involves calculating the amounts of reactants and produces in chemical reactions. If you know the atoms or ions in a formula or a reaction, you can use stoichiometry to determine the amounts of these atoms or ions that react. Solving stoichiometry problems in solution chemistry involves the same strategies you learned in Unit 2. Calculations involving solutions sometimes require a few additional steps, however. For example, if a precipitate forms, the net ionic equation may be easier to use than the chemical equation. Also, some problems may require you to calculate the amount oaf a reactant, given the volume and concentration of the solution. Take your time working through the next three Sample Problems. Make sure that you understanding how to arrive at the solutions. 1. The Concentration of Ions Calculate the concentration (in mol/L) of chloride ions in each solution.(a) 19.8g of potassium chloride dissolved in 100 mL of solution

(b) 26.5g of calcium chloride dissolved in 150 mL of solution

(c) a mixture of the two solutions in parts (a) and (b), assuming that the volumes are additive.

2. Finding the Minimum Volume to Precipitate Aqueous solutions that contain silver ions are usually treated with chloride ions to recover silver chloride. What is the minimum volume of 0.25 mol/L magnesium chloride, MgCl2(aq) , needed to precipitate all the silver ions in 60mL of 0.30mol/L silver nitrate, AgNO3(aq) ?Assume that silver chloride is completely insoluble in water.

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3. Finding the Mass of a Precipitated Compound Mercury salts have a number of important uses in industry and in chemical analysis. Because mercury compounds are poisonous, however, the mercury ions must be removed from the waste water. Suppose that 25.00 m L of 0.085 mol/L aqueous sodium sulfide is added to 56.5 mL of 0.10 mol/L mercury (II) nitrate. What mass of mercury (II) sulfide, HgS(s), precipitates ?

4. Finding the Mass of Another Precipitated Compound Silver chromate, Ag2CrO4, is insoluble. It forms a brick-red precipitate. Calculate the mass of silver chromate that forms when 50.0 mL of 0.100 mol/L silver nitrate reacts with 25.0 mL of 0,150 mol/L sodium chromate.

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Homework: Please Hand in Next Day Name:___________________1. Equal volumes of 0.120 mol/L potassium nitrate and 0.160 mol/L iron(III) nitrate are mixed

together. What is the concentration of nitrate ions in the mixture?

2. An excess of aluminum foil is added to a certain volume of 0.675 mol/L aqueous copper(II) sulfate solution. The mass of solid copper that precipitates is measured and found to be 4.88g. What was the volume of the copper(II) sulfate solution?

3. Your stomach secretes hydrochloric acid to help you digest the food you have eaten. If too much HCl is secreted, however, you may need to take an antacid to neutralize the excess. One antacid product contains the compound magnesium hydroxide, Mg(OH)2 .

(a) Predict the reaction that takes place when magnesium hydroxide reacts with hydrochloric acid. (Hint: This is a double-displacement reaction.)

(b) Imagine that you are a chemical analyst testing the effectiveness of antacids. If 0.10 mol/L HCl serves as your model for stomach acid, how many litres will react with an antacid that contains 0.10g of magnesium hydroxide ?

4. Even though lead is toxic, many lead compounds are still used as paint pigments

(colourings). What volume of 1.50 mol/L lead (II) acetate contains 0.400 mol Pb2+ ions.

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Molarity and concentration of ions – a Pseudo – labName:

Purpose: To find the concentration of a solution and the ions present

Observations:

Volume of aluminium sulphate solution used 10.0mL

Mass of empty evaporating dish and watch glass 65.16g

Mass of evaporating dish, watch glass and solid 65.56g

Calculations:

1. Calculate the mass of solid collected

2. Calculate the mole of solid collected

3. Determine the molartiy of the aluminium sulphate solution.

4. Write the dissociation equation for aluminium sulphate

5. Calculate the concentrations of the two ions in this solution

[Al3+] = ______________

[SO42-] = _____________

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Questions:

1. 3.22g of magnesium nitrate is dissolved in water to form 120mL of solutiona. Write the dissociation equation

b. Calculate the molarity of this solution

c. Calculate the concentration of the ions in this solution

2. A solution of aluminium fluoride is known to have a fluoride ion concentration of 0.450Ma. Write the dissociating equation for aluminium fluoride

b. If there are 200mL of solution available, calculate the mass of solid needed to prepare this solution

3. In an experiment 0.62g of magnesium hydrogen carbonate was formed on evaporating a 30.0mL solutiona. Write the dissociation equation for the solid dissolving in water

b. Calculate the molarity of this solution

c. Calculate the concentrations of the ions present

4. A 50.0mL of a solution of barium chloride has a [Cl-] = 0.400M. Calculate the mass of solute dissolved in this solution

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Dilution Problems – an alternate lookWe’ve done this before – let’s see what you remember…

Calculate the concentration of the final solution resulting from adding 350mL of water to 50.0mL of 1.50M silver nitrate.

1. First calculate the number of moles: n =MV = (0.050L)(1.50M) = 0.075 mole AgNO32. Now calculate the new concentration given the new volume

M = n = 0.075 mol AgNO3 = 0.188MV 0.400L

[AgNO3] = 0.188M

Since the number of moles does not change, only the volume, we can state this mathematically… (remember M = n/V)

ninitial = VinitialMinitial nfinal = VfinalMfinalnumber of moles does not change in other words

ninitial = nfinal or

VinitialMinitial = VfinalMfinal

* This only works for dilution problems, but it is a great alternate method*

1. Calculate the volume of water that must be added to 10.0 mL of sulphuric acid to form a solution that is 2.50M sulphuric acid.

Vinitial = 0.0100L Vfinal ?Minitial = 5.60M Mfinal 2.50M

MiVi = MfVf(0.0100L)(5.60M) = (2.50M)Vf

Vf = 0.0224L = 22.4mL

Don’t forget 22.4mL is to total final solution – we need to account for the fact that wealready had 10.0mL to begin with so…

22.4mL – 10.0mL = 12.4mL

You need to add 12.4mL H2O

2. A stock solution of hydrochloric acid is 9.25M. Calculate the volume of this stock solution in mL needed to prepare 1.25L of a 0.220M hydrochloric acid solution.

MiVi = MfVf(9.25M)Vi = (1.25L)(0.220M)

Vi = 0.0297L = 29.7mL

29.7mL of stock solution is needed

Calculating the Concentrations of ions in Solution

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Now it’s time to put it all together. I’ve been refreshing your memory on calculating moles and concentration…and I’ve introduced you to the idea of dissociation equations and ions in solution. This gives you the tools to calculate the concentration of ions in solution.

What is the molar concentration of the chloride ions in 0.25 M AlCl3(aq)?AlCl3(aq) Al3+(aq) + 3Cl-(aq)

0.25M = 0.25 mol/LSince the number of moles is three times more, so too is the molarity

0.250 mol AlCl3 x 3 mol Cl - = 0.75 M Cl-L 1 mol AlCl3

[Cl-] = 0.75M

What is the concentration of each type of ion in a solution made by mixing 50.0mL of 0.240M AlBr3 and 25.0mL of 0.300M CaBr2?***assume no reaction occurs ***AlBr3(aq) Al3+(aq) + 3 Br-(aq) CaBr2(aq) Ca2+(aq) + 2Br-(aq)

***Note: you could do the complete ionic equation, but since there is not reaction, it is simpler to deal with them separately***MiVi = MfVf MiVi = MfVf(2.40M)(0.050L) = Mf(0.075L) (0.0250L)(0.300M) = Mf(0.075L)[AlBr3] = 0.160M [CaBr2] = 0.100MTherefore [Al3+] = 0.160M therefore [Ca2+] = 0.100MAnd [Br-] = 0.480M [Br-] = 0.200M

Now add the two sources of Br- together0.480M + 0.200M = 0.680M

The concentrations in the final solution are as follows:[Al3+] = 0.160M[Ca2+] = 0.100M[Br-] = 0.680M

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Now it’s time to try some on your own:

1. 300. mL of water added to 500. mL of 0.800M of ferrous chloride. Calculate the [ions] in the final solution

2. To 25.0mL of 4.75M HCl is added water until the final volume is 800. mL. Calculate the [ions] in the final solution

3. 100. mL of 0.500M sodium chloride is added to 400mL of 0.400M calcium chloride. Calculate the [ions] in the final solution.

4. A solution was prepared by adding water to 150.0mL of 3.25M magnesium nitrate. Calculate the amount of water added if the concentration of the nitrate ion in the final solution is 2.00M

5. 50.0mL of water was added to 100mL of an unknown strength aluminium chloride solution producing a chloride ion concentration of 0.66M. Calculate the molarity of the 100mL of unknown aluminium chloride solution.

6. To 25.0mL of sodium nitrate is added 50.0mL of 0.210M barium chloride. The [NO3-] is 0.340M in the final solution. Calculate the initial concentration of the sodium nitrate solution.

7. When 350.0mL of 0.250M MgCl2 is boiled down to a final volume of 275.0mL, what is the [Cl -] in the resulting solution?

8. A chemistry student dissolves 3.25g of K2CrO4 and 1.75g of K2Cr2O7 in water and dilutes the mixture to a total of 100.0mL. What is the concentration of all the ions in the solution?

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Concentration of ions when mixing solutions

So with our new equation M1V1 = M2V2 dealing with dilutions by adding water should be more approachable for you and you can now take it one step further to calculate the [ions] present…What if we mix two solutions together (ie. not just add water) such as adding salt water to sugar water can we still calculate the concentrations of the ions? Sure we can. You follow exactly the same steps, just add the volume of the second solution to the first to get the final volume – the fact that there are ions floating around in it doesn’t affect it. The only thing you have to watch out for is there is more than one source for the ions. So often you will add at the end.

Having said all that – if you understand the intuition here you can still calculate this will the knowledge of molarity and mole we had before we started this module.

1. 200mL of 0.400M NaCl is added to 300mL of 0.600M CaCl2a. Calculate the molarity of the NaCl in the final solution

M1V1 = M2V2 (0.400M)(0.200L) = M2(0.500L)[NaCl] = 0.160M

b. Calculate the [Na+] and the [Cl-] in the final solutionNaCl Na+ + Cl-Since [NaCl] = 0.160M and there is a 1:1 ratio, the ion concentrations are the same[Na+] = 0.160M[Cl-] = 0.160M

c. Calculate the molarity of the CaCl2 solutionM1V1 = M2V2(0.600M)(0.300L) = M2(0.500L)[CaCl2] = 0.360M

d. Calculate the [Ca2+] and the [Cl-] in the final solutionCaCl2 Ca2+ + 2Cl-[Ca2+] = 0.360 M[Cl-] = 0.720M

e. What is the total [Cl-] in the final solution?[Cl-]total = 0.160M + 0.720M = 0.880M2. 100mL of 0.300M barium nitrate solution is added to 200ml of 0.660M ferric nitrate

solution.a. Calculate the [ions] as a result of the dilution of the barium nitrate solution

Ba(NO3)2 Ba2+(aq) + 2NO3-(aq)M1V1 = M2V2(0.300M)(0.100L)=M2(0.300L)[Ba(NO3)2] = 0.100MTherefore [Ba2+]=0.100M and [NO3

-]=0.200Mb. Calculate the [ions] as a result of the dilution of the ferric nitrate solution

Fe(NO3)3 Fe(aq)3+ + 3NO3-(aq)M1V1 = M2V2(0.660M)(0.200L) = M2(0.300L)[Fe(NO3)3] = 0.440MTherefore [Fe3+]=0.440M and [NO3

-] = 1.32Mc. What is the [NO3 -] in the final solution?

[NO3-] = 0.200M + 1.32M = 1.52M

3. 30.0mL of 0.500M MgCl2 is mixed with 0.700M Na2S solution. The [Cl-] in the final solution is 0.300Ma. Calculate the volume of the Na2S solution added

MgCl2 Mg2+ + 2Cl-Since we know the final [Cl-]=0.300M, the ratios of the above equation tell us [MgCl2] is half that ie. [MgCl2] = 0.150M – this is what we use for our calculationsM1V1 = M2V2

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(0.500M)(0.0300L) = (0.150M)V2V2 = 0.100L = 100mL is the total volume so…Na2S added = 100mL = 30.0mL = 70.0mL Na2S added

b. Calculate the [Na+] and [S2-] in the final solutionNa2S 2Na+ + S2-

M1V1 = M2V2(0.700M)(0.0700L) = M2(0.100L)[Na2S] = 0.490MTherefore [Na+] = 0.980M [S2-] = 0.490M

Mass of Precipitate

Now that we predict if a precipitate is formed when two solutions mix, lets take it one step further to determine the mass of the solid formed.

Example: Calculate the mass (g) of the precipitate formed when 40.0mL of 0.420M H2SO4 is added to 60.0mL of 0.400M Ba(NO3)2.

Step 1:Determine the Net Ionic EquationBa2+(aq) + SO42-(aq) BaSO4 (s)

Step 2:Determine the concentration of ions that were involved in the NIEIn this example this means [SO42-] from the [H2SO4] after mixing is…H2SO4 2H+ + SO4-

M1V1 = M2V2(0.420M)(0.0400L = M2(0.100L)[H2SO4] = [SO42-] = 0.168M

And [Ba2+] from the [Ba(NO3)2] after mixing is…Ba(NO3)2 Ba2+ + 2NO3-

M1V1 = M2V2(0.400M)(0.0600L) = M2(0.100L)[Ba(NO3)2] = [Ba2+] = 0.240M

Step 3:Set up ICE table to determine limiting factor

Ba2+(aq) + SO4

2-(aq) BaSO4

Initial molarity 0.240 0.168Consumed 0.168 0.168 0.168End molarity (subtract) 0.072 0 0.168

SO42- is the limiting factorStep 4:Calculate mass of precipitate using total volume.

n = MV = (0.168M)(0.100L) = 0.0168mol

0.0168mol x 233.3g = 3.22g BaSO41 mol

Why is it ok for us to compare molarity here when before I was adamant that it had to be moles?- because the volumes are the same- we could multiply by the volume to give moles, but we would be multiplying by the same

number.

Now you try a few on your own1. Consider the NIE: Ca2+ + SO32- CaSO4

[Ca2+] = 0.450M and [SO42-] = 0.225M in 200mLa. Determine the limiting factor

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b. Determine the concentration of the ion in excess

c. Calculate the mass of the precipitate formed

2. Consider the NIE: Al3+ + PO43- AlPO4[Al3+] = 0.250M and [PO43-] = 0.310M in 300mLa. Determine the limiting factor ion

b. Determine the concentration of the ion in excess

c. Calculate the mass of the precipitate formed

3. 200mL of 0.400M Pb(NO3)2 solution is added to 300mL of 0.200M CaS solutiona. Determine the concentrations of the ions in the lead(II) nitrate solution after mixing

but before reaction

b. Determine the concentration of the ions in the CaS solution after mixing but before reaction

c. Determine the limiting factor ion

d. Determine the concentration of the ion in excess

e. Calculate the mass of the precipitate formed.

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Conductivity of Aqueous Solutions – a Quick Lab

Using a light Bulb and Battery determine whether the following solutions conduct electricity or not.

Solution Conducts? Dissociation equation.

Distilled H2O

Metnanol(CH3OH)

NaCl

Sugar (C2H12O11)

HCl

NaOH

H2SO4

Conclusion: Based on what you know about ions in solution explain your results relating conductivity and solution.

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Solution Chemistry Review

1. Classify the following compounds as ionic (I) or molecular (M)a. SCl2 _____ b. CaS ____ c. SiO2 ____ d. LiClO ____ e. CH3OH ____f. MnO2 ____ g. PbO ____ h. KCl ____ i. SO3 ____ j.C2H6 ____ k.NH3 ___

2. Write the dissociation equation by completing the following: (all solids)a. C6H12O6( )

b. Ba(NO3)2 ( )

c. IBr ( )

d. Li3PO4 ( )

e. C12H26 ( )

3. 7.42 g of sodium nitrate is dissolved in water to form 80.0mL of solutiona. Calculate the molarity of this solution

b. Calculate the concentration of the ions in this solution.

4. A beaker contains 42.0mL of 0.950M calcium chloratea. Write the dissociation equation

b. Calculate the mass of solid present if the water is evaporated.

5. To 40.0mL of 1.25M calcium iodide is added 60.0mL of water. Calculate the [ions] in the final solution

6. To 50.0mL of 0.400M NaCl is added 150mL of CaCl2 solution. The [Cl -] = 0.550M in the final solution. Calculate the initial concentration of the CaCl2

7. A solution of barium nitrate is added to a solution of sodium carbonatea. Write the complete reaction equation and circle the precipitate formedb. Write the net ionic equation

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8. Calculate the concentrations of all ions in the following. 100.mL of 2.20M potassium iodide is added to 400mL of 0.800M aluminum iodide.

9. A solution of magnesium chloride is added to a solution of rubidium sulfitea. Write the CRE

b. Write the CIE

c. Write the NIE

10. To 30.0mL of 0.400M strontium nitrate is added 70.0mL of 0.300M sulphuric acid. What is the mass of the precipitate formed? What is the concentration of the ion in excess?

11. Using your solubility table, devise a procedure for separating a mixture of Ba2+ and Pb2+ by precipitating them individually.

12. A solution contains a mixture of Cl- and OH-. Devise a procedure for precipitating each of these separately.

13. Classify the bond in each as ionic, polar covalent or non polar covalent by calculating their electronegativities. Draw the molecule show it’s dipole where applicable.a. HBr

b. Cl2

c. PCl3

d. HI

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14. Each of the following molecules have a calculated dipole. Speculate the shape of the molecule (lewis dot structure) to determine if it is asymmetrical and therefore still polar. In the non polar examples show the line of symmetry.

a. CH3Cl c. OF2

b. H2O2 d. HCN

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16. Which of the following solutions would you expect to conduct electricity?a.HBr b.CH4 c.CH3OCH3 d.NO3 e. FeCl2 f.LiBr g. Br2 h. NaBr

Detergents used for washing clothes have a similar structure to the phospholipids in the cell membrane. They contain a polar head and non polar tail and can form a radial type structure called a micelle. Considering grease and fat are non-polar speculate how detergents work to remove dirt from your clothing.

Polar head

Non polar tail

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