SOLUBILITY EQUILIBRIUM

Transcription

1 Introduction SOLUBILITY EQUILIBRIUM A. Ionic vs Molecular Solutions 1. Ionic Compounds form Ionic Solutions a) Ionic compounds ( metal + non-metal ) dissolved in water to form Ionic Solutions eg1: dissociation equation AlCl3(s) Al 3+ (aq) + 3Cl - (aq) ionic compound free ions / electrolytes Electrolyte: A substance which dissolves to give an electrically conducting solution containing ions. (e.g. AlCl 3 above) b) Compounds containing polyatomic ions form Ionic solutions. eg: KMnO4 KMnO4(s) K + (aq) + MnO4 - (aq) Polyatomic ion 2. Covalent Compounds form Molecular Solutions a) Covalent Compounds (Non-metal + Non-metal) generally form Molecular solutions. eg. I2(s) I2(aq) C 2 H 2(g) C 2 H 2 (aq) do NOT break up into ions when dissolve: are called non electrolytes b) organic compounds form molecular solutions. eg: C12H22O11(s) C12H22O11(aq) exception: organic acids (they end in the "COOH") What Causes Electrical Conductivity? A liquid must have IONS in it to conduct electrical current Dissolved ions move freely and carry a charge, therefore can transfer electrical current 1

2 SUMMARY 1. Compounds made up of a metal and a non-metal will form ionic solutions. 2. Compounds containing polyatomic ions will form ionic solutions. 3. Compounds containing only non-metals (covalent compounds) will form molecular solutions. 4. Organic compounds (other than those ending in the "COOH" group) will form molecular solutions. 5. Organic acids (organic compounds ending in the "COOH" group) will form partially ionic solutions. (These are ionic but because there are fewer ions formed, these solutions are weak conductors or weak electrolytes.) B. Solubility and Solubility Equilibrium E.g. Put solid ionic solid (crystal) in water: Initia lly,no ions in the water, so rate of dissolving is high no free ions will go back onto the crystal, so rate of precipitation is zero. 1. Ionic Solid Aqueous Solution (dissolving) At first, only the forward reaction is taking place: Eventually, solution fills with free ions and forward rate begins to decrease 2. Ionic Solid Aqueous Solution (precipitation or crystallization) The number of free ions that collide and stick to the crystal increases. This is the reverse reaction: Eventually, reverse rate matches forward rate, and we reach Solubility Equilibrium Ionic Solid CaCO3(s) Aqueous Solution Ca 2+ (aq) + CO3 2- (aq) Solubility Equilibrium: The rate of dissolving = The rate of precipitation A saturated solution is a solution in which there exists a dissolved substance in equilibrium with the undissolved substance Solubility is the equilibrium concentration of a substance in a solution at a given temperature The Solubility of a substance is the maximum amount that can dissolve in a given amount of solution at a given temperature. (the "ability" to dissolve) Solubility increases with temperature 2

3 Molar solubility: when concentration is expressed in moles/l or g/l Hebden # 1-7 WRITING FORMULA, COMPLETE AND NET IONIC EQUATIONS eg. Mix solution of AgNO 3 with a solution of Na 2 CO 3. Formula equation - balanced eqn with all reactants and products: 2AgNO 3 (aq) + Na 2 CO 3 (aq) Ag 2 CO 3 (s) + 2NaNO 3 (aq) Complete ionic equation - balanced eqn showing all ions: 2Ag + - (aq) + 2NO 3 (aq) + 2Na + 2- (aq) + CO 3 (aq) 2Ag + 2- (s) + CO 3 (s) + 2Na + - (aq) + 2NO 3 (aq) - only Ag and CO 3 are reacting and precipitating. How do we know which ions will precipitate? (must look to Ksp value), - the other ions are spectator ions Hebden # 25 Net ionic equation - balanced eqn showing only species involved in the reaction: 2Ag + (aq) + CO 3 2- (aq) Ag 2 CO 3 (s) PREDICTING THE SOLUBILITY OF SALTS 1. Definitions "Low Solubility" = a precipitate will form. "Soluble" = a precipitate won't form. A substance has low solubility if a saturated solution of the substance has a concentration less than 0.1 M 2. Finding the solubility of a compound To find this, we use a Solubility Table(Hebden p.332) 1. Find anion. (column 1) 2. Look for matching cation (column 2) 3. Answer (column 2) IMPORTANT: Cu + Cu +2 Know what an alkali is FeCO 3 CuI PbBr 2 Soluble Low Solubility Hebden # SEPARATING MIXTURES OF IONS BY PRECIPITATION METHODS 3

4 eg1: A solution contains both Ag + and Ba 2+. How do you separate these cations? Use Solubility Table to find a negative ion which will form a precipitate with one.but not both! if you add Cl- or Br- or I- the precipitate should contain Ag + and the solution should still contained dissolved Ba 2+. ALSO SO4 2-, would be a bad choice for the above example because both would form a precipitate. E.g. 2) A solution may contain Ag +, Ba 2+, and Ni 2+. What ions could be added, and in what order to determine which cations are present? make a precipitate table: Cl - 2- SO 4 S 2- OH - 3- PO 4 Ag + precipitate precipitate precipitate precipitate precipitate Ba 2+ - precipitate - precipitate precipitate Ni precipitate precipitate precipitate Can not use OH - or PO 4 3- b/c they will form precipitate w all three cations Step 1. Add Cl - first because it forms precipitate only with Ag + then filter out the AgCl Ag + is gone, so now only consider Ba 2+ and S 2- Step 2. (Two Possibilities) 2- Can add either SO 4 which will form a precipitate with only Ba 2+ - then filter out the BaSO 4 - only S 2- remains in the solution or S 2- (foms precipitate w only Ni 2+ ) - then filter out the NiS - 2- only SO 4 remains in the solution Note: Anions cannot be added by themselves you should add a spectator cation (Li +, Na +, K + ) at the same time Hebden # REVIEW: DILUTION AND UNITS OF CONCENTRATION 4

5 A. Dilution dilution formula: C 1 V 1 = C 2 V 2 eg1) Mix 5.0mL of Cl 2 solution with 15.0mL of M Br - solution. What is the concentration(c 2 ) of the Br - ion in the new solution? C 2 = C1V1 = V M x 15.0mL 15.0mL + 5.0mL =.0090 M eg2) There are 7.0 x 10 4 mol [Na 2 SO 4 ] in ml of solution. What is the [Na + ]? [Na + ] = x 10 mol Na2SO L B. Units of Concentration x 2Na + 1Na2SO 4 = 1.4 x 10-2 M Na + eg1) 1L of saturated solution contains 1.96g of AgBrO 3 What is the molar solubility of AgBrO 3? [AgBrO 3 ] = 1.96g AgBrO 3 x L 1mol 235.8g = 8.31 x 10-3 M eg2) The molar solubility of BaF 2 is 3.58mol/L. How many grams of BaF 2 exist in 726mL of saturated solution? Grams of BaF 2 = 3.58mol BaF2 L x 175.3g BaF2 mol BaF2 x 0.726L = 456g BaF 2 THE SOLUBILITY PRODUCT (K sp ) K sp = the equilibrium constant (K eq ) for an ionic substance dissolving in water. eg. CaCO3(s) dissolves in water net-ionic eqn: CaCO3(s) Ca2+ (aq) + CO3 2- (aq). Ksp = [Ca 2+ ] [CO3 2- ] ([CaCO3] not included is a solid) [Ca2+] and [CO3 2- ] at equilib are their solubilities Writing K sp expression from the Net-Ionic Equation. 5

6 Steps 1. Write out equilib eqn showing the salt dissolving 2. Write out the equilib expression (leave out the solid) eg1: PbCl2(s) Pb2+ (aq) + 2Cl- (aq) K sp = [Pb 2+ ] [Cl - ] 2 eg2: Ag2SO4(s) 2Ag + (aq) + SO4 2- (aq) K sp = [Ag + ] 2 [SO4 2- ] Definitions: Solubility: the amount of a substance required to make a saturated solution Molar solubility: the molar concentration of a saturated solution Solubility product: the Ksp value obtained when the concentrations of the ions are multiplied together Hebden # CALCULATING SOLUBILITY & ION CONCENTRATION E.g.1: What is the molar solubility (mol/l) of a saturated AgBrO 3(aq) solution if 1 L contains 1.96 g of AgBrO 3? [AgBrO 3(aq) ] = 1.96 g AgBrO 3 x 1 mol AgBrO 3 = 8.31 x10-3 M 1L AgBrO g AgBrO 3 E.g. 2: Express the molar solubility of PbI 2 in g/l. Molar solubility = 1.37 x 10-3 Solubility in g/l = 1.37 x 10-3 mol PbI 2 x g PbI 2 = g/mol 1 L PbI 2 1 mol PbI 2 Read examples on p. 80 Use Molar mass Hebden # 8-20 CALCULATIONS WITH K sp Calculating Solubility given K sp. e.g. Calculate molar solubility of AgCl. K sp for AgCl = 1.8 x Step 1. Pick a variable and use mole ratios to find other concs: -s +s +s AgCl(s) Ag+ (aq) + Cl- (aq) Step 2. Write out solubility expression. 6

7 K sp = [Ag + ] [Cl - ] Step 3. substitute in variables Since: K sp = [Ag+] [C] = [s][s] = s 2 Step 4. Solve for s K sp = s 2 = 1.8 x M s = 1.8 x = x 10-5 M Answer: the molar solubility of AgCl is 1.3 x 10-5 M Calculating K sp, Given Solubility eg. solubility of Ag 2 CrO 4 is 1.31 x 10-4 moles/l. Calculate Ksp. Step1. write net-ionic eqn Ag2CrO4 (s) 2Ag + (aq) + CrO4 2- (aq) Step 2. use mole ratios to find the other concs: [Ag+] = 2 x 1.31 x 10-4 M = 2.62 x 10-4 M [CrO4 2- ] = 1.31 x 10-4 M Step 3. substitute values into K sp equation: Calculating [ion] K sp = [Ag+]2 [CrO4 - ] = (2.62 x 10-4 ) 2 (1.31 x 10-4) = 8.99 x Eg. What is the [ Mg 2+ ] in a saturated solution of Mg(OH) 2? Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) K sp =[ Mg 2+ ][ OH - ] 2 -x x 2x If you are not given Ksp find K sp = [ Mg 2+ ][ OH - ] 2 = (x)(2x) 2 = 4x 3 = 5.6x10-12 it on table of Solubility Product Constants (p. 333) x = 1.1x10-4 Therefore, [ Mg 2+ ] = 1.1x10-4 M Hebden # Predicting Precipitates and Maximum Ion Concentration 7

8 A. Predicting Precipitates When Two Solutions Are Mixed 1. Review The Ksp for CaCO3 is 5.0 x 10-9 Ksp = [Ca2+] [CO3 2- ] The Ksp value from above comes from a saturated solution if you try to add more than 5.0 x 10-9 Ca2+ or CO3 2- a precipitate will form because the solution can t dissolve any more! 2. Mixing two solutions (ions are from different sources) E.g. Mixing solutions of Ca(OH) 2 and Na 2 CO3 resulted in the following concentrations: [Ca2+] = 2.3 x 10-4 M and [CO3 2- ] = 8.8 x 10-2 M. Will a precipitate form? Table shows CaCO3 Ksp = 5.0x Step 1. Calculate "trial Ksp" Trial Ksp = [Ca2+] [CO3 2- ] = (2.3 x 10-4 ) (8.8 x 10-2 ) = x 10-5 Step 2. Compare the Trial Ksp with the real value for Ksp Trial Ksp = x 10-5 Ksp = 5.0 x 10-9 Trial Ksp > Ksp precipitate will form B. Maximum Possible Concentration of an Ion in Solution AgCl(s) Ag + (aq) + Cl - (aq) 4/5 Why? Net ionic equation should be written out If trial Ksp < Ksp, a precipitate will NOT form. Net ionic equation is given you don t have to write it out e.g. 1: Cl- is added to a solution having only Ag+ ([Ag+] = is 1.0 x 10-5 M). What is the maximum [Cl-] before precipitate forms? Ksp = 1.8 x Ksp = [Ag+] [ Cl - ] = 1.8 x [ Cl- ] = 1.8 x = 1.8 x 10-5 M 1.0 x10-5 4/4 or 2/2 e.g. 2: If a 5.0 ml sample of 6.0 x 10-5 M Ag + is added to 10.0 ml of 4.2 x 10-6 Cl -, will a precipitate form? 8

9 C 2 = C 1 V 1 V 2 [Ag + ] = (6.0 x 10-5 )( 5.0 ml) = 2.0 x 10-5 M 15.0 ml Four Steps dilution trial Ksp comparison to Ksp statement [Cl - ] = (4.2 x 10-6 ) )( 5.0 ml) = 2.8 x 10-6 M 15.0 ml Q = [Ag + ][Cl - ] = [2.0 x 10-5 M][ 2.8 x 10-6 M] = 5.6 x Q = 5.6 x < Ksp = 1.8 x a precipitate will/will not form e.g. 3: If 0.1 M KCl is added dropwise to a beaker containing 0.10 M Ag + and 0.10 M Pb 2+, which precipitate would form first? Show all equations and calculations. We can calculate the minimum [Cl - ] needed to start the precipitation of AgCl and PbCl 2 by using the Ksp values. AgCl Ksp = 1.8 x PbCl 2 Ksp = 1.2 x 10-5 AgCl(s) Ag + (aq) + Cl - (aq) PbCl 2(s) Pb 2 + (aq) + 2 Cl - (aq) Ksp = [Ag + ] [Cl - ] 1.8 x = [0.10 M] [Cl - ] [Cl - ] = 1.8 x = 1.8 x 10-9 M 0.10 M Ksp = [Pb 2 + ] [Cl - ] x 10-5 = [0.10 M] [Cl - ] 2 [Cl - ] = (1.2 x 10-5 ) = M ( 0.10 M) As 1.8 x 10-9 M < M AgCl will precipitate out first. Hebden # Solubility and Titrations 9

10 Sample Solution - A solution of unknown conc. Standard Solution - solution of known conc used in a titration. Equivalence Point The point where moles of reactant in sample solution have all reacted with moles of the reactant added from standard solution Titration the process of adding a standard solution to a sample solution until the equivalence point is reached, in order to determine the [sample solution] Indicator substance which does something (i.e. changes colour, form a precipitate) to show equivalence point The purpose of a titration is to find the conc of an unknown solution. E.g.: What is the concentration of a Cl - solution which is in the range 0.10 and 5.0 M? Use CrO 4 2- as an indicator and force out a precipitate with Ag M Ag + Note: the flask should be swirled throughout the procedure and the titrant should be added slowly??? M Cl - and M CrO 4 2- White These 2 equilibria exist in the flask AgCl (s) Ag + + Cl - Ksp = [Ag + ][ Cl - ] = 1.8 x Red Ag 2 CrO 4(s) 2Ag + + CrO 4 2- Ksp = [Ag + ] 2 [ CrO 4 2- ] = 1.1 x Recall Ksp tells us when a precipitate will form. Therefore we can use the Ksp values to determine what concentrations of Ag + will form a precipitate first 10

11 Ksp = [Ag + ][Cl - ] [Ag + ] = Ksp [Cl - ] [Ag + ] needed for a precipitate 0.10 M Cl M Cl M CrO 4 [Ag + ] = Ksp [Cl - ] [Ag + ] = Ksp [Cl - ] [Ag + ] 2 = Ksp [CrO 4 2- ] =1.8 x = 1.8 x 10-9 M Ag = 1.8 x = 3.6 x M Ag = 1.1 x = 1.0 x 10-5 M Ag Therefore 5.0 M Cl - would precipitate out first followed by 0.10 M Cl - and finally M CrO 4 2- Note: if CrO 4 2- did not precipitate last then it could not be used as an indicator Visual Observations (while slowly adding Ag + with swirling) 1. WHITE precipitate forms = AgCl(s) take biuret reading A 2. RED precipitate forms = Ag 2 CrO 4 (s) take biuret reading B 3. when red precipitate has formed almost all of Cl - has reacted A B A B = ml of 0.10 Ag + needed to precipitate out Cl - Tion Calculate moles Ag + And since moles Ag + = moles Cl - and we know the initial volume of Cl - solution Finally we can calculate [Cl - ] E.g. 2: What is the concentration of [Cl - ] in a 25.0 ml sample of sea water? To get to the equivalence point 26.8 ml of M AgNO 3 was added. Ag + + Cl - AgCl (s) Moles Cl - = L AgNO 3 x mol AgNO 3 x 1 mol AgCl x 1 mol Cl L AgNO 3 1 mol AgNO 3 1 mol AgCl = mol Cl - [Cl - ] = mol Cl - = M Cl - Hebden # L Cl - Removing Pollution and Hardness from Water by Precipitation Methods PROBLEM: Waste-water from a mine has [Sr 2+ ] = 0.005M, but according to law, [Sr 2+ ] cannot exceed 1.0x10-5 M. 11

12 SOLUTION: add CO 3 2- to waste water to remove Sr 2+ as SrCO 3. Hazardous SrCO 3(s) Sr 2+ + CO 3 2- Question: What [CO 3 2- ] is needed in the waste-water to lower [Sr 2+ ] to1.0x10-5 M? K sp of SrCO 3 = [Sr 2+ ][CO 3 2- ] = 5.6x10-10 (from Ksp table) Want [Sr 2+ ] = 1.0x10-5 M Therefore, [CO 3 2- ] = 5.6x x10-5 = 5.6 x10-5 M Hard Water Water is hard if too much Ca 2+ or Mg 2+ is present. Acid rain dissolves natural limestone (CaCO 3 ), and ions flow into water sources. CaCO 3(s) + 2H + (aq) Ca 2+ (aq) + H 2 O (l) + CO 2(g) + heat (Same happens to MgCO 3(s) ) In hot pipes and hot pots: the equilib is forced left evaporation leaves the ions and solid behind Read p.103 Use your head Talk to your partner pots and pipes get that white solid stuck on them. How to soften the water: Add an ion that will form a precipitate with Ca 2+ and Mg 2+ (eg. C 17 H 22 COO - ) Boil the water and force equilib left (precipitate out the ions) Filter out the precipitate Hebden # The Common Ion Effect and Altering Solubility recall Le Chetaliers Principle: 12

13 When a system at equilibrium is disturbed, the equilibrium will shift so as to counteract the disturbance. E.g. 1-3 Saturated solution: CaCO3(s) Ca 2+ (aq) + CO3 2- (aq) 1. The Common Ion Effect Disturbance Dissociation Ion effect Shift Amount of solid add CaCl2 CaCl2 (s) Ca 2+ (aq) + 2Cl - (aq [Ca 2+ ] left increases add K2CO3 K2CO3 (s) 2K + (aq) + CO3 2- (aq) [ CO3 2- ] left increases 2. Decreasing Solubility Added compound Ions Effect on Solubility of CaCO3(s) Reason for effect Ca(NO3)2 Ca 2+ NO3 - decrease equilib shifts left to decrease [Ca 2+ ] KNO3 K + NO3 - none neither ion effects equilib CaCO3 Ca 2+ CO3 2- none Same substance Notice: the equilib shifts when a substance that has a common ion is added. This is the common ion effect. The added compound should not have low solubility because their dissolved ions are needed for the above reaction. 3. Increasing Solubility a) By Adding Acid CaCO3(s) Ca 2+ (aq) + CO3 2- (aq) 13

14 decrease [Ca 2+ ] or [ CO3 2- ] to increase solubility of CaCO3(s) how to decrease? Add acid (H + ) CaCO3(s) Ca 2+ (aq) + CO3 2- (aq) H + (from acid) H2CO3(aq) decomposes H2O(l) + CO2(g) Overall effect: [CO3 2- ] and equilib shifts right Solubility of CaCO3(s) is increased. b) By Forming Another Precipitate AgCl(s) Ag + (aq) + Cl - (aq) Add something that forms a precipitate with Ag + or Cl - Sulphide forms a precipitate with Ag + : AgCl(s) Ag + (aq) + Cl - (aq) + S 2- (S 2- is added) Overall effect: [ Ag + ] and equilib shifts right Solubility of AgCl(s) is increased. Could also use Pb 2+ to precipitate Cl - ions Ag2S(s) Hebden # Solubility Test Coming Soon 14