Unit III: Solubility Equilibrium

Review of SolubilityCase#1:Many barium compounds are extremely poisonousBaSO4 routinely swallowed by patients when getting X-ray pictures of intestinesWhy doesn't BaSO4 poison these patients?Case #2:AgBr kills off micro-organisms such as algae and bacteria in swimming pool which is saturated with this substance.The swimming pool water is as pure as drinking water and is harmless to people.Since silver ion is a heavy metal ion and is dangerous to our health in large doses, how is it possible that water saturated with AgBr doesn't harm us?

electrolyte: substance which dissolves to give an electrically conducting solution containing ionsex-1: (NH4)3PO4(s) → 3NH4+(aq) + PO43-(aq)ex-2: HCl(g) → H+(aq) + Cl-(aq)

demo: electrolytes and electrolyte sensorsheet: electrolyte or not?

non-electrolyte: substance which dissolves to give a non-conducting solution containing only neutral molecules.ex-1: C2H2(g) → C2H2(aq)ex-2: Br2(l) → Br2(aq)

molecular solution: contains only neutral molecules (no charge)ionic solution: contains ions (with charges)

How can you predict whether a compound will form an ionic or molecular solution?Concept Map Bkt p.1Problems: Sol Bkt p.9, p.28#1 Hebden p.74#1

General Rules for Classifying Compounds as Ionic or MolecularIonic Compounds: metal + non-metalex: FeCl3(s) → Fe3+(aq) + 3Cl-(aq)polyatomic ions: compounds made up of several species/atomsex: (NH4)2Cr2O7(s) → 2NH4+(aq) + Cr2O72-(aq)

ex-1: (NH4)2HPO4(s) → 2NH4+ + HPO42-

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NH4+:one ion present HPO42-:second ion present, P.A.I.(polyatomic ion)

ex-2: NaH3P2O7(s) → Na+ + H3P2O7-Na+ :one ion present H3P2O7-:second ion present

Molecular Compounds: non-metal + non metalex-1: CH3OH(l) → CH3OH(aq)ex-2: ClO2(l) → ClO2(aq)

Problems: Heb 12 p.74 #2 more problems Heb 11 p.74 #1, 2solution: homogenous mixturesolution: solute + solvent

ex: NaCl dissolved in water solute solvent

solvent: component which exists in greater quantitysolute: component which exists in smaller quantitysoluble: if solute and solvent mix to form a homogeneuous mixture

more than 0.1 mol of substance dissolved per litresolubility: maximum amount of a solute which can dissolve in a given amount of solvent at a given temperature.saturated soln: has dissolved the maximum amount of substance. Addition of more substance will cause this extra material to accumulate in undissolved form.

Other Definitions: Acid: begins with H except for organic acid such as acetic acid (CH3COOH)Base: ends with OHSalt: cation and anion that do not begin with H or end with OH

Problems: Sol Bkt p.7-8

Chem 12 definitions:saturated solution: dissolved substance is in equilibrium with some of the undissolved substancesolubility: equilibrium concentration of a substance in solution at a given temperature.molar solubility: concentration given in moles/litreConcept Map Bkt p.4

2 conditions to have saturated soln:1) some undissolved material is present2) equilibrium exists between dissolved and undissolved material

saturated soln written as equation showing substance in equilibrium

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with its aqueous products

Solubility Equation:AgBrO3(s) ⇌ Ag+ (aq) + BrO3-(aq)

2 possibilities:(i)dissolving rxn:AgBrO3(s) → Ag+ (aq) + BrO3-(aq)(ii)crystallization rxn:Ag+ (aq) + BrO3-(aq) → AgBrO3(s)At equilibrium: rate of dissolving rxn = rate of crystallizationProblems: Heb p.76 #3-7, Sol Bkt p.29#2-6, p.29

Dissociation and IonizationDissociation: rxn involves separating existing ions in an ionic solidSince ions are already dissolved, the solvent is merely separating the ions from one anotherNaCl(s) → Na+(aq) + Cl-(aq) Ionization: rxn involves the breaking up of a neutral molecules into ionsIons do not exist until solvent is able to react with the molecule and break it apart into ions.CH3COOH(l) → CH3COO-(aq) + H+(aq)

.Both dissociation and ionization rxns produce electricity-conducting ionic solutions

.Equations showing ionization and dissociation are identical

Example: Write an equation to show the dissociation of FeBr3(s) in water:FeBr3(s) → Fe3+(aq) + 3Br-(aq)

Example: Write an equation to show the ionization of HCN(g) in water:HCN(g) → H+(aq) + CN-(aq)

Heb (11) p. 210#28

Three Types of Chemical Equations:(i) Chemical Formula Equation-balanced and with correct states(s, l, aq, g)(ii) Complete Ionic Equation-dissociates all aqueous compounds into cations

and anions(iii) Net Ionic Equation-removal of spectator ions

spectator ions: remain unchanged during rxn (iv) Use Data bkt on p.4 to identify if a sustance is soluble or insoluble

Precipitation and Net Ionic Equations Demo: lead (II) nitrate and potassium iodide

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Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq) (Chemical Formula Eqn)

Dissociation of 3 soluble components of the rxn to form lead(II) iodide:Pb(NO3)2(s) → Pb2+(aq) + 2NO3-(aq)KI(s) → K+(aq) + I-(aq)KNO3(s) → K+(aq) + NO3-(aq)

Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq) → PbI2(s) + 2K+(aq) + 2NO3-(aq)(Complete Ionic Eqn)

spectator ions: K+(aq) & NO3-(aq):can be deleted from ionic equationPb2+(aq) + 2I-(aq) → PbI2(s) (Net Ionic Equation)Concept Map Bkt p.3Predicting the Solubility of Salts (Heb 12 p. 81, data bkt p.4)low solubility: small amount of solute dissolved.ex: PbI2 has low solubility in water, it doesn’t dissolve with water

.could form a precipitateLow solubility: if a saturated solution is less than 0.1 MProblems: Sol Bkt p.4-6, p.22-23#1-2

Example: Determine whether FeCO3(s) is soluble or has low solubilitySince Fe2+ is not listed, it must be included in the category "all others"∴ FeCO3(s): low solubilityPrecipitate: when 2 ions form a compound having low solubility, the mixing of the 2 ions will cause a precipitate to form.

Example: Will a precipitate form when 0.2 M solutions of CaS and NaSO4 are mixed?Combinations of ions might form a precipitate:

CaS → Ca2+ + S2-NaSO4→ Na+ + SO42-

Cross combinations using an ion box:Cation Anion ProductsCa+2 S2- Na2S (aq)Na+ SO4

2- CaSO4(s)Na+ and S2-:solubleCa2+ and SO42-: low solubilityprecipitate: CaSO4

Notes:1) copper: 2 ions: Cu+ and Cu2+

Cu+: low solubility and Cu2+: soluble

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2)You need to write chemical formula of compound formed from 2 ionsHeb(12) p.83 #21-23

Important Generalizations:Compounds containing alkali metals, H+, NH4+or NO3- are soluble in water1)no ppt with alkali metals, H+, NH4+or NO3-2)for soluble compound containing specific ion, include alkali metals, NH4+or NO3- to ensure the compound will be soluble3)For a particular anion needed, strongly suggested that you choose Na+ to combine with your anion, since salts containing sodium are very commonex: soluble salt containing CO32- would be Na2CO34)For a articular cation needed, strongly suggested that you choose NO3- to combine with your cationex: soluble salt containg Fe3+ would be Fe(NO3)3 Heb (12) p. 84 #24Ex-1 Write a chemical formula equation, a complete ionic equation and a net ionic equation for the reaction which occurs when 0.2 M solutions of Pb(NO3)2(aq) and FeCl3(aq) are mixed.

a) Set up an ion boxCation Anion ProductsPb+2 NO3

- Fe(NO3)3 (aq)

Fe3+ Cl- PbCl2(s)b) Find the products and indicate if there are soluble or notc) Chemical Formula Equation: balanced eqn, use "(s)" to show ppt

3Pb(NO3)2(aq) + 2FeCl3(aq) → 3PbCl2(s) + 2Fe(NO3)3 (aq) d)Complete Ionic Equation: 3Pb2+(aq) + 6NO3-(aq) + 2Fe3+(aq) + 6Cl-(aq) → 3PbCl2(s) + 2Fe3+(aq) + 6NO3-(aq) e)Net Ionic Equation: Pb2+(aq) +2Cl-(aq) → PbCl2(s)Heb (12) p.87 #25

Ex-2 Write the net ionic equation for the reaction between sodium hydroxide and hydrochloric acid

1) Set up an ion boxCation Anion ProductsNa+ OH- H2O (l)H+ Cl- NaCl(aq)

2) Chemical Formula Equation:

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NaOH(aq) + HCl(aq) → NaCl (aq) + H2O(l)

3) Complete Ionic Equation: Na+(aq) + OH+(aq) + H+(aq) + Cl-(aq) → Na+(aq) + Cl-(aq) + H2O(l)

4) Net Ionic Equation: OH-(aq) + H+(aq) → HOH(l) (no ppt formed)

Ex-3 Write the net ionic equation for the reaction between zinc metal and hydrochloric acid

1) Set up an ion boxCation Anion ProductsZn H2 (g)H+ Cl- ZnCl2(aq)

2) Chemical Formula EquationZn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

3) Complete Ionic Equation:Zn(s) + 2H+ (aq) + 2Cl- (aq) → 2Zn2+(aq) + 2Cl- (aq) + H2(g)

4) Net Ionic Equation:Zn(s) + 2H+ aq) → Zn2+ aq) + H2(g)

Problems: Sol Bkt p.4-6, p.9-10#6-10, p.12#19, p.22-23#1,2

Calculating the Concentrations of Ions in SolutionEx-1:What is the molar concentration of chloride ions in 0.25 M AlCl3(aq)

AlCl3(s) → Al3+(aq) + 3Cl-(aq)1 mol 1 mol 3 mol

[Cl-] = 3 X [AlCl3] = 3 X 0.25 M = 0.75 Mor using factor label method:[Cl-] = 0.25 mol AlCl3 X 3 mol Cl- = 0.75 M

1 L 1 mol AlCl3

Ex-2: What is the concentration of each type of ion in a solution made by mixing 50.0 mL of 0.240 M AlBr3 and 25.0 mL of 0.300 M CaBr2?When AlBr3 and CaBr2 dissolve in water, they undergo dissociation:AlBr3 → Al3+ + 3Br- and CaBr2 → Ca2+ + 2Br-

1)Perform a dilution calculation on each of the starting solutionsC1V1 = C2V2

C1=initial concentrationV1=initial volumeC2=final concentrationV2=final volumeV2: 50.0 mL + 25.0 mL = 75.0 mL

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C2= dilution concentration = C1V1 V2

[AlBr3]dil = 0.240 M X 50.0 mL = 0.160 M 75.0 mL

[CaBr2 ]dil= 0.300 M X 25.0 mL = 0.100 M 75.0 mL

2)For each dissociation equation, indicate each ion's concentrationAlBr3 → Al3+ + 3Br- 0.160 M 0.160 M 0.480MCaBr2 → Ca2+ + 2Br- 0.100 M 0.100 M 0.200MThe mixture has 2 sources of Br-, which are added to give a total [Br-] :

[Br-]total =0.480 + 0.200 = 0.680 M

In summary:[Al3+] = 0.160 M [Ca2+] = 0.100 M [Br-]= 0.680 MProblems: Sol Bkt p.11#15-18, p.14-15#4-10, p.24, p.31-33

Molar Solubility ProblemsSolubility: concentration of a dissolved ion in a saturated solutionConcept Map Bkt p.5,6

CaF2(aq) ⇌ Ca2+(aq) + 2F-(aq)-s s 2sSol Bkt p.9#11 ÷MMRoad Map: s(g/L) → s(mol/L)

You need to use a road mapWhen in doubt, mole it out!!Use two triangles for mole and concentration:

xg xg: mass in grams (g) n: number of mol

n MM MM: molar mass in g/mol

n n: number of mol

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[ ]: concentration in mol/L or Molar(M) [ ] V V: volume in litres

Sol Bkt p.14#11 X V X MMRoad Map : s(M) ----→ n(mole) ------→xg(g)Solubility Bkt p.10#11-14, p.15#11-13

Solubility ProductCaF2(s) ⇌ Ca2+(aq) + 2F-(aq) (Solubility Product Equation)salt CaF2(s) : only slightly soluble in waterKsp =[Ca2+][F-]2 Ksp : Solubility Product constantspecial symbol given to Keq when dealing with dissociation rxn of saltData Bkt p. 4, 5

Ksp value:large value: more soluble saltsmall value: less soluble salt

Note: solubility and solubility product are different concepts:solubility: amount of a substance to make saturated soln(molar) solubility: amount of ionic compound dissolved in a solnsolubility product: Ksp value is obtained when [ ]s of ions in saturated soln are multipled together

Heb p.92 # 41

CaF2(s) ⇌ Ca2+(aq) + 2F-(aq)Ksp =[Ca2+][F-]2 = 1.46 X 10 -10

Case [Ca2+] [F-] [Ca2+][F-]21 4.46 X 10-3 1.81 X 10-4 1.46 X 10-10 2 1.21 X 10-3 3.47 X 10-4 1.46 X 10 -103 3.32 X 10-4 6.64 X 10-4 1.46 X 10 -104 9.65 X 10-5 1.23 X 10-3 1.46 X 10 -10

observations:1)Ksp : constant value2)values [F-] increase, values [Ca2+] decrease3)case #3: [F-] is double [Ca2+]process of dissolving CaF2 gaves ratio 1:2

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CaF2(s) ⇌ Ca2+(aq) + 2F-(aq)1 mol 1 mol 2 mol

2 situations:1)Saturated soln made by separately pouring Ca2+ and 2F- into soln, we have separate values for [Ca2+] and [F-]. These values are placed directly into the Ksp expression2)Saturated soln made by dissolving CaF2(s). Once we know the molarity of dissolved CaF2 , we are assured that:

[Ca2+] = [CaF2 ]dissolved [F-] = 2 X [CaF2 ]dissolved

suppose that [Ca2+] = s Ksp = x • 2s2 Ksp = 4s3 Note:Strongly suggested that for EVERY problem involving Ksp you:

.write out equilibrium equation showing dissolving of salt

.write out solubility product expressionProblems: p.20#10-11, p.25#1-3, p.27#7, p.28#10, p.35-37#26-32

Separating Mixtures of Ions by Precipitation MethodsQualitative Analysis Device a procedure for separating a mixture of Ba2+ and Pb2+ by precipitating them individually.To determine which elements or ions are present in a substance

SO42- S2- Cl- Br- I-Ba2+ ppt --- --- --- ---Pb2+ ppt ppt ppt ppt ppt

Add Na2S or NaCl or NaBr or NaINa2S + Pb2+ PbS(s)and Ba2+ is still in solutionThen we could add Na2SO4 or Na2CO3 or Na3PO4to precipitate the barium ions

Cl - SO 42- S 2- OH - PO 43-Ag+ ppt ppt ppt ppt pptSr2+ --- ppt --- --- ppt.Assume that aqueous soln contains one or both cations Ag+ and Sr2+ ..Find out some anion which could form ppt with only one of the cation.If Cl-, S2-, or OH- is added and ppt formed with sample, only Ag+ is present

To separate Ag+ from Sr2+ :

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i)added Cl- to sample, any Ag+ present ppt ii)filter off the ppt (or use a centrifuge) iii)add SO42- or PO43- to the remaining soln to ppt any Sr2+

Example: A solution contains one or more of Ag+ ,Ba2+ and Ni2+. What ions could be added, and in what order, to determine which of these cations are present?

Cl - SO 42- S 2- OH - PO 43-Ag+ ppt ppt ppt ppt pptBa2+ --- ppt --- ppt pptNi2+ --- --- ppt ppt ppti)Only Ag+ ppt with Cl-.Add Cl- to see if any Ag+ present and filter off any pptIf no ppt, Ag+ is absentii)No Ag+ in soln

SO 42- S 2- OH - PO 43-Ba2+ ppt --- ppt pptNi2+ --- ppt ppt pptii)Add SO42- or S2-: if SO42- added: any Ba2+ pptif S2- added: any Ni2+ pptExperimental procedure:Step 1: To 1 mL soln which might contain Ag+, Ba2+ or Ni2+ , add few drops of 1M NaCl solnno ppt: Ag+ absent, proceed to step 2 ppt: Ag+ present. Filter off and discard the ppt. Proceed to step 2 to test the rest of soln.

Step 2: To soln from step 1, add few drops of 1M Na2SO4 soln.no ppt: Ba2+ absent. Proceed to step 3ppt: Ba2+ present. Filter off and discard the ppt. Proceed to step 3 to test the rest of soln.

Step 3: To soln from step 2, add few drops of 1M NaOHno ppt: Ni2+ absent.ppt: Ni2+ present

Heb(12) p. 90#26-39Problems: Sol Bkt p.17-18#1-5, p.21, Topic#16

Predicting Whether a Precipitate Will Form (Trial Ion Product; T.I.P)Concept Map Bkt p.7-8

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If Ag+(aq) and Cl-(aq) are mixed, you have to find out if there is sufficient [ ] of both ions to be at equilibirum:

AgCl(s) ⇌ Ag+(aq) + Cl-(aq) Ksp = [Ag+] [Cl-] = 1.8 X 10 -10

Define Ion product or trial ion product(TIP) as:Q = [Ag+]start [Cl-]start

Q = product of ion [ ]s which actually exist in soln.what we have

Ksp = product of ion [ ]s required to establish a solubility equilibrium.what we need to form saturated soln

3 possible outcomes:Case 1: Q < Ksp

We have less that what we need to form a saturated solnConclusion: ppt can't form

Case 2: Q = KspWhat we have is just equal to what we need to form saturated solnConclusion: barely saturated soln

Case 3: Q > KspWhat we have is greater that what we need to form a saturated solnConclusion: ppt forms

Examples p. 97-98, Exercises p.98 #56-69Problems T.I.P.: Sol Bkt p.37#33-34, Topic #9

Applying Solubility Principles to Chloride TitrationsConcept Map Bkt p.9-10Titration: process in which a measured amount of soln is reacted with a known volume of another soln (One of the solns has unknown [ ] ) until a desired equivalence point (or stoichiometric point) is reached.

purpose of titration: to find [ ] of a particular soln

[ Cl-] determined by doing a titration where Ag+ are added to chloride ions until a ppt of AgCl is formed

Ag+(aq) + Cl-(aq) → AgCl(s)

To signal that the titration is complete, an indicator is usedchromate ions are added

Suppose you use a burette to add a Ag+ 0.10 M soln of into a beaker contai-ning 0.10 M Cl- and 0.010 M chromate ion, CrO42-. The slow addition of Ag+

steadily increase the [Ag+] in the beaker, starting from 0.0M.Both Cl- and CrO42- eventually form a ppt with the added Ag+ .

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AgCl(s) ⇌ Ag+ + Cl- Ksp = [Ag+] [ Cl-] = 1.8 X 10-10whiteAg2CrO4(s) ⇌ 2Ag++ CrO42- Ksp = [Ag+] 2[ CrO42-] = 1.1 X 10-12 red yellow[Ag+] = Ksp = 1.8 X 10-10 = 1.8 X 10-9 M (for AgCl)

[ Cl-] 0.10 [Ag+] = Ksp = 1.1 X 10-12 = 1.0 X 10-5 M (for Ag2CrO4)

√[ CrO42-] √0.010 .1st ppt to form is AgCl(s): smaller [Ag+] is required .when all Cl- has reacted and removed as a ppt, next drop of Ag+will form ppt of Ag2CrO4(s) (red colour)

.point n Ag+ = n Cl-Ksp = [Ag+] [ Cl-] = constant value

steady decrease of Cl- and steady increase of Ag+ .[Ag+] has risen from 0.0M (initially) to 1.0 X 10-5 M (where ppt of Ag2CrO4),[ Cl-] has dropped to:

[ Cl-] = Ksp = 1.8 X 10-10 = 1.8 X 10-5 M [Ag+] 1.0 X 10-5

At this point, % Cl- left is: 1.8 X 10-5 M (when ppt of Ag2CrO4) = X 100% = 0.018%

0.10 m(initially)It means that 99.98 % Cl- has reacted, which agrees with the fact that virtually all Cl- has reacted and been removed.

Summary: When n(Cl-) used in rxn = n(Ag+) added, the titration has arrived at "Equivalence Point" or "Stoichiometric Point"

Don't worry about the chromate indicator; it is simply regarded as being in the background to allow the accurate detection of the equivalence point of the titration.

Examples Heb p.100-101Exercises Heb p.101 #70-75Problems on Titrations: Sol Bkt p.23#3, p.28#11, p.39-40, Topic#11

Solubility-Maximum & Minimum AmountMaximum amount needed to be soluble with a mixtureMinimum amount to start precipitationMethod:(i)Find out what is the precipitate(ii) Write the solubility equation(iii) Write the solubility expressionNB: Both concentrations come from different solutions/compoundsProblems: Sol Bkt p.26#5,6 p.27#8, Topics #14-15

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Common Ion Effect and Other Ways to Alter the Solubility of a SaltConcept Map Bkt p.11

Solubility of ionic compound, such as AgCl, in Pure Water can't change.If pure water is not used to dissolve AgCl(s), it is possible to Increase or decrease the solubility(not the solubility product) by applying Le Chatelier's Principle to the equilibrium

Meaning of Increasing or Decreasing the Solubility of a Salt:AgCl(s) → Ag+(aq) + Cl-(aq) dissolving rxn Ag+(aq) + Cl-(aq) → AgCl(s) crystallization rxn

Both increasing and decreasing the solubility of a salt work on the basis of Le Chatelier's principle:changing the [ ] of the dissolved ions in the equilibrium

AgCl(s) ⇌ Ag+(aq) + Cl-(aq)will shift the equilibrium either to the:

.solid side(decreasing solubility of solid)

.ions side (increasing solubility of solid)In other words, "Increasing and Decreasing the solubility" refers to the solid salt, but the changes in solubility are accompanied by altering the [ ] of the dissolved ions.

A. Decreasing the Solubility of a SaltIf we increase concentration of the ions in soln, Le Chatelier's Principlesays that equilibrium shift to use up the added ions and to cause more solid to form Ex: AgCl(s) ⇌ Ag+(aq) + Cl-(aq)increase [Ag+]:

AgCl(s) Ag (aq) + Cl (aq)+ -

increase [Cl-]:

AgCl(s) Ag (aq) + Cl (aq)+ -

Adding Ag+ and Cl- , a ppt of AgCl forms. Ppt reduces [Ag+] and [Cl-]. It prevents added soln from containing large [ ]s of both ions.

Common Ion Effect: lowering of the solubility of a salt by adding a second salt which has one ion in common with the first salt.

Ex: In a mixture of AgNO3 and AgCl:AgCl ⇌ Ag+ + Cl-

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AgNO3 → Ag+ + NO3-

Ag+ :ion in common

Ex: In a mixture of AgCl and NaCl:AgCl ⇌ Ag+ + Cl-NaCl → Na+ + Cl-

Cl-:ion in common

Common ion effect frequently used in chemistry to prevent a particular salt from dissolving to any greater extent and to force a particular dis-solved ion to leave a solution.

Ex: In the industrial Solvay Process for making sodium hydrogen carbonate (NaHCO3 or baking soda) a saturated soln of NaHCO3 is produced in solution.In order to increase the yield of NaHCO3 the solid produced, a saturated solution of NaCl is added.

NaCl(s) → Na+ + Cl-

solubility of NaCl (6.1 M) > solubility of NaHCO3 (0.82 M)large [Na+] will remove all the HCO3- in soln and form the desiredNaHCO3(s).

B. Increasing the Solubility of a SaltIf we decrease the [ion]s in soln, the equlbrm will shift so as to dissolve more of the solid and bring the [ion]s back up again.

Ex: AgCl(s) ⇌ Ag+(aq) + Cl-(aq)1)decrease [Ag+] by adding some ion which ppte the Ag+ present.looking at Solubility Table: SO42-, S2-, OH-, CO32-, ppt with Ag+ Let's choose S2- and add soluble Na2S (Na+ acts as a spectator)

adding S2- lowers [Ag+] by precipitating Ag+ as Ag2S. This decrease in shifts the equilibrium involving AgCl(s) as to dissolve more AgCl. Net result of adding is to increase solubility of AgCl(s)

2)Decrease the [Cl-] by adding some ion which precipitates the Cl- present. Either Pb2+ or Cu+ will form ppt with Cl-. We will choose Pb2+ in form of soluble salt Pb(NO3)2 . (NO3- acts as a spectator) Adding Pb2+ lowers the [Cl-] by precipitating the Cl- as PbCl2(s).

NaHCO (s) Na + HCO + -33

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This decrease in [Cl-] shifts the equilibrium involving AgCl(s) so as to dissolve more AgCl. Net result of adding Pb2+ is the increase the solubility of AgCl(s).

Heb p.108 #81-86

Problems: Topics #12-13

Removing Pollution and Hardness From Water By Precipitation Methods

A.Removing Metal Ion Pollutants By Precipitation MethodsCd2+, Hg2+, Pb2+ interfere with chemical rxns of biological systems and are toxic to organisms that are ingesting them.toxicity of heavy metals proportional to their [ ]s.

Example:Waste water in the"tailings" pond of a mining operation had a cadmium ion concentration of about 0.005 M. Before discharging the waste water into an adjacent river, the mine has to lower the [Cd2+ ] to at most 1.0 X 10-5 M. What [OH-] would be required to bring the [ Cd2+] to acceptable values? Ksp = 5.3 X 10-15 for Cd(OH)2

Cd(OH)2 ⇌ Cd2+ (aq) + 2OH-(aq); Ksp = [ Cd2+] [OH-]2 = 5.3 X 10-15

[OH-]2 = Ksp = 5.3 X 10-15 = 5.3 X 10-10

[ Cd2+] 1.0 X 10-5[OH-] = 2.3 X 10-5 M

B. Hardness in Water: Where It Comes From And How To Get Rid Of ItOrigin of Hard Water 1)Several deposits of limestone(CaCO3) in Canada (Ex: Canadian Shield).acid rain or acids found in humus form soil react with limestone and dissolve it.

CaCO3(s) + 2H+(aq) ⇌ Ca2+(aq) + CO2 (g) + H2O(l) + heat

2) CO2 in the atmosphere dissolves in water to produce carbonic acid which reacts with limestone.

CaCO3(s) + CO2(g) + H2O(l) ⇌ Ca2+(aq) + 2HCO3-(aq) + heat

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These 2 natural ways of dissolving limestone contain noticeable amounts of Ca2+ ion.

Stalactites(at the root) and stalagmites (on the floor) formation in limestone cavern:.acidic water penetrates cracks in rocks and reaches limestone deposit below surface .acid eats away volumes of limestone creating a cavern.water from surface drips from cavern roof producing Ca2+ and HCO3-ions..air in cavern has low [ CO2(g)] since most CO2(g) reacts with limestone.water evaporates as droplets of soln fall from ceiling, above equilibrium proceeds according to LeChatelier's principle:

Ca2+(aq) + 2HCO3-(aq) ⇌ CaCO3(s) + CO2(g) + H2O(l)ppt of CaCO3(s) forms stalactites and stalagmites

Getting Rid of Hardness in Water3 effects of hard water:1)water with bitter taste due to presence of Ca2+ and/or Mg2+.2)white deposit left when water evaporates or is heated

Ca2+(aq) + 2HCO3-(aq) + heat ⇌ CaCO3(s) + CO2(g) + H2O(l).rock-hard deposits

.clog steam pipes

.interfere with heating coils in kettles3)hardness inhibits cleaning action of soaps

.cleaning ingredient in soap is a stearate ion which ppt with Ca2+ and/or Mg2+

Ksp = 1 X 10-12 for Ca(C17H35COO)2Ksp = 1 X 10-12 for Mg(C17H35COO)2

calcium or magnesium stereate causes a gray-white ppt often found around bath tub after one has taken a bath.

How can Ca2+ or Mg2+ be removed from hard water?By softening the water when adding washing soda (Na2CO3).Solubility Table shows that CaCO3 and MgCO3 have low solubility in water

.adding CO32- ions precipitates unwanted Ca2+ or Mg2+.

2 types of hard water:.temporarily hard: contains HCO3-

.such ion can be decomposed by heat:Ca2+(aq) + 2HCO3-(aq) + heat ⇌ CaCO3(s) + CO2(g) + H2O(l).hardness removed by boiling water

.permanently hard: doesn't contain HCO3-

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.only way to get rid of Ca2+ or Mg2+ is to ppt with washing sodaHeb, p. 104 #76-80Problems: Topic #17

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