SECTION 8.2 THROUGH 8.8: The miscibility of liquids with water The miscibility of two liquids or the solubility of a solid in a liquid depends on the attractive forces that operate between the substances. INTERMOLECULAR FORCES: Dispersion Forces - attractions between all covalent molecules -forces related to electrons and shape of molecule eg. hexane and carbon tetrachloride (nonpolar molecules) are not miscible with water Diple- Dipole Forces - Polar molecules are attracted to one another by permanent charges on neighbouring molecules -based on difference in electronegativity, and shape of molecule eg. acetone - polar molecule DIAGRAM: NaCl (aq) (ION - DIPOLE ATTRACTION) Hydration - when the water molecules surround the ions, this is called hydration. Hydrogen Bonds - an exceptionally strong dipole-dipole attraction in substances with H attached to high electronegative atom (F, O, N) The solubility of solids See text p. 345 eg. water and ethanol water will only mix with other polar liquids with which it can share the same type of intermolecular forces. CAUTION: Like dissolves like does not always work, see p. 344
STRONG ELECTROLYTE: A solid substance which when dissolved is i.e. dissolves to give an electrically conducting eg. STRONG ELECTROLYTIC SOLUTION: -one in which over 50% of a dissolved solute WEAK ELECTROLYTE: A solid substance which when dissolved is WEAK ELECTROLYTIC SOLUTION: -one in which less than 50% of a dissolved solute NON - ELECTROLYTE: A solid which ionizes i.e. a substance which dissolves to give a non-conducting solutions containing eg. A MOLECULAR SOLUTION CONTAINS ONLY NEUTRAL MOLECULES AND AN IONIC SOLUTION CONTAINS IONS. CLASSIFYING COMPOUNDS AS IONIC OR MOLECULAR: IONIC COMPOUNDS -a compound that is made up of a metal and a non-metal is likely to be ionic in solution. eg. FeCl3 (s) Fe 3+ (aq) + 3Cl - (aq) -a compounds made up of polyatomic ions will be ionic in solution. eg. NaH3P2O7 (s) Na + (aq) + H3P2O7 - (aq) MOLECULAR COMPOUNDS -a covalent compound (non metal and non metal), especially organic compounds IONIC vs. MOLECULAR?? 1. RbBr 3. CsNO3 2. CHCl3 4. S8
Dissociation vs. Ionization Dissociation - separation (when a solvent such as water The strength of the ion-dipole attractions with the water cause Ionization - when the ions are produced from eg. acids and bases - Unit 4 SOLUBILITY RULES: SATURATED SOLUTION: There exists a dynamic state of equilibrium between the ions in the solution and the solid. eg. NaCl (s) Na + (aq) + Cl - (aq) For every molecule of NaCl(s) which dissolves, then a Na + (aq) + Cl - (aq) ion recombine to form a molecule of NaCl(s). [Na + ](aq) and [Cl - ](aq) remain constant. UNSATURATED SOLUTION: If unsaturated, then the rate of the forward reaction is greater than the reverse. More NaCl (s) dissolves. [Na + ](aq) and [Cl - ](aq) increase. SUPERSATURATED SOLUTION: A supersaturated solution holds more dissolved solute than it should at that temp. -unstable -reverts back to unsaturated state. -as it does so, the rate of the reverse reaction will be greater.
The effect of temperature on solubility K can only be changed by temperature. THEREFORE: The solubility of a solute can also be changed by changing temperature. Whether or not a solute will be more soluble in hot water depends on the ΔS (entropy) value. As long as the entropy is getting more random (ΔS is positive), then the increasing temp. will increase the solubility. eg. NaCl (s) Na + (aq) + Cl - (aq) Recall GIBB's FREE ENERGY Equation: ΔG = ΔH - TΔS If ΔS is positive, as T increases, -system gets more random -TΔS becomes more negative THEREFORE increasing the temp increases the solubility as long as the ΔS is positive (the system is getting more random) If ΔS is negative, then the system becomes more ordered or less random. eg. dissolving a gas in water: O 2 (g) O 2 (aq) more random less random Recall GIBB's FREE ENERGY Equation: ΔG = ΔH - TΔS If ΔS is negative, as T increases, -TΔS becomes more positive -less spontaneous -less soluble THEREFORE GASES ARE LESS SOLUBLE IN HOT WATER!!!!!
Using LeChatelier to explain the effect ENDO rxns: kj + solute (s, l, g) solution (aq) EXO rxns: solute (s, l, g) solution (aq) + kj In light of this, gases are less soluble in hot water because in general, the dissolving process for most gases is an exothermic reaction. See lake example on p. 346. Definitions: Recrystallization - see p. 346. The effect of pressure on solubility Using LeChatelier to explain the effect O 2 (g) O 2 (aq) increase pressure: O 2 (g) O 2 (aq) decrease pressure: See example on p. 347
SECTION 8.3: Solid Hydrates, p. 350 Eg. CuSO 4 5H 2 O TEXT BOOK TOPICS THAT YOU SHOULD KEEP UP WITH, READ, AND KNOW FOR A THEORY QUESTION ON YOUR UNIT 3 TEST: DRY CLEANING SOLVENTS - see p. 345 COLLOIDS AND BEAUTY PRODUCTS SOAPS AND DETERGENTS THE POTASH INDUSTRY CALCIUM AND YOUR HEALTH HARD WATER RICHES ON THE SEA FLOOR
Mrs. Toombs Notes on HARD WATER
DUAL EQUILIBRIUM: This is a situation where reaction rate (left and right) for two reactions are in balance simultaneously. AgCH3COO (s) Ag + (aq) + CH3COO - (aq) + H + (aq) + OH - (aq) HOH (l) HCH3COO Equilibrium Stoichiometry -one solute added to a solvent until saturation occurs Equilibrium Non -Stoichiometry -multiple equilibrium sharing ions (common ion) A 3 B 5 3A 5+ + 5B 3- -equilibrium having had a s 3s 5s LeChatelier shift Ksp and s are useful -equilibrium established by mixing solutions -[ion] related by stoichiometric ratio -only Ksp useful -[ion] multiples of "s"
COMMON ION EFFECT: A common ion is a charged particle shared by solutes in a solution. For example, if a solution contains AgCl and Ag 2 CrO 4, the common ion is Ag +. A common ion must be the same species and the same charge. (eg. Fe 2+ and Fe 3+ do not qualify). The existence of a common ion has two important consequences: -the [ion] will normally be non-stoichiometric. -the solubility of each solute will be significantly less, decreasing the solubility To increase the solubility, add an ion which precipitates either the cation or anion present. example: solubility of AgCl in pure water: AgCl (s) Ag + (aq) + Cl - (aq) I x --------- -------- R x s s E x s s no contaminant stoichiometric Ksp = s 2 = 1.6 x 10-10 s = 1.3 x 10-5 M. s is the concentration which saturates a solution in pure water. solubility of AgCl in 0.50 M Ag 2 CrO 4 : AgCl (s) Ag + (aq) + Cl - (aq) I x 0.50 M -------- R x +y +y E x (0.50 + y) M (y) M y does not equal s common ion contaminant non-stoichiometric Ksp can still be calculated. s is not useful here
IN CLASS PRACTICE: SAMPLE CALCULATIONS DUAL EQUILIBRIUM: Answer these questions on a separate piece of paper! 1. Consider a solution in which AgCl and Ag 2 CrO 4 are in dual equilibrium. [Cl - ] = 4.0 x 10-7 M. Find [Ag + ] and [CrO 4 2- ]. 2. Fe (OH) 2 and FeS are in simultaneous equilibrium in solution. Find all [ion] if [OH - ]= 6.0 x 10-6 M. 3. Find [CO 3 2- ] in Al2 (CO 3 ) 3 (aq) saturated solution if [Al 3+ ] = 5.0 x 10-6 M. 4. Ba(NO 3 ) 2 added to 0.0100 M KI, 0.0010 M Na 2 SO 4. What precipitates first? What is the [ion] when the second ppt. commences? YOU CAN NOW COMPLETE ALL OF THE CALCULATIONS and QUESTIONS on the Ca(OH) 2 lab! Another look at PRECIPITATION QUESTIONS 1. If we mix 300. ml of 0.075 M Na2SO4 and 200. ml of 0.045 M Ba (NO3)2, will a precipitate form? 2. Equal volumes of 0.040 M Cu2SO4 and 0.00060M NaI are mixed. Will a precipitate form? 3. 200. ml of 0.000060 M of Al2(SO4)3 are mixed with 300. ml of 0.00010 M Pb (NO3)2. Does a precipitate form? 4. What mass of solute must be added to 3.00 L of a solution of SrSO 4 in order for a precipitate to form?
SOLUBILITY SKULLDUGGERY FOR SUPERIOR STRATEGISTS 1- Will a precipitate form if 10. ml of 0.010 M NaCl are mixed with 30. ml of 0.020 M Ag 2 SO 4? AgCl has a Ksp value of 1.6 x 10-10. 2- What concentration of SO 4-2 must exist to produce a precipitate in a solution of 1.0x 10-4 M Ra(NO 3 ) 2? RaSO 4 has a Ksp = 4.0 x 10-11. 3- A solution contains [Br - ] = 2.00 x 10-5 M and [Cl - ] = 1.00 x 10-2 M. If for AgCl, Ksp = 1.56 x10-10 and for AgBr, Ksp = 3.25 x 10-13, what precipitates first as Ag + is added? What will be the concentrations of bromide, chloride and silver ions when the second precipitate starts forming? 4- A solution is saturated with both AgCl and Ag 2 CrO 4. If AgCl Ksp = 1.1 x 10-10 and Ag 2 CrO 4 Ksp = 1.7 x 10-12, determine all ion concentrations if [CrO 4-2 ] = 1.0 x 10-3 M. ANSWERS 1- Ksp (1.6 x 10-10 ) < Trial Ksp (7.5 x 10-5 ) so a ppt forms 2- [SO 4-2 ] = 4.0 x 10-7 M 3- For AgCl ppt to form, we need [Ag + ] > 1.56 x 10-8 M; for AgBr ppt to form, we need [Ag + ] > 1.63 x 10-8 M; since the [Ag + ] limit for AgCl will be reached first as Ag + is added, AgCl will precipitate first. The second ppt is AgBr and will commence when: [Ag + ] = 1.63 x 10-8 M (as calculated) [Br - ] = 2.0 x 10-5 M (as given) and [Cl - ] = 9.6 x 10-3 M when AgBr starts to precipitate. 4-[Ag + ]=4.1x10-5 M and [Cl - ] =2.7x10-6 M
Worksheet 3.3: More Solubility Puzzles 1- A saturated solution of silver bromate is prepared by adding 5.00 g of silver nitrate to a 2.5 x 10-2 M solution of NaBrO3(aq). What is the volume of solution produced? 2- Will a precipitate form if 750. ml of 2.0 x 10-2 M MgCl2(aq) is mixed with 2.50 L of 0.010 M Cs2CO3(aq)? 3- What volume of 0.050 M CaCl2(aq) can you add to 2.0 L of 0.025 M Na2SO4(aq) without causing a precipitate? 4-250. ml of saturated chromium (III) carbonate solution is allowed to react with zinc. When all the chromium ions have reacted to produce solid chromium metal, it was found that 1.75 g of zinc were required for the process. Determine the ion product of chromium (III) carbonate. 5- What volume of 0.025 M HNO3(aq) would be needed to neutralise 150. ml of saturated magnesium hydroxide solution? 6- Up to 15.0 g of barium chloride can be dissolved in 2.5 L of Al2(SO4)3(aq) solution without occurrence of a precipitate. Find the concentration of the aluminium sulphate solution. 7- Find the solubility of lead (II) chloride in a 0.100 M solution of NaCl(aq). 8- A solution contains 3.5 x 10-3 M IO3 - (aq) and 6.4 x 10-4 M CO3-2 (aq). Identify the precipitate that will form first upon addition of Ag + (aq) to the solution. Find the concentration of the ions when the second precipitate starts forming. 9- Draw a qualitative analysis scheme showing how you would separate a mixture of Mg +2, Pb +4, and Sr +2 ions. 10- Draw a qualitative analysis scheme showing how you would separate a mixture of Cr +3, Cu +, and Ba +2 ions. 11- Draw a qualitative analysis scheme showing how you would separate a mixture of SO4-2, S -2, and Br - ions.
Worksheet 3.3: More Solubility Puzzles KEY 1- Volume of solution produced = 14 L. 2- Yes, a precipitate will form. 3- At saturation, the concentrations of the calcium and sulphate ions will be given by: [Ca +2 (aq)] = 0.050M (x L / (2.0 + x)l) [SO4-2 (aq)] = 0.025 M (2.0 L / (2.0 + x) L) Substituting the above in the Ksp expression for CaSO4 and solving for x gives a volume of CaCl2(aq) of 31 L. 4- Ksp of chromium (III) carbonate = 6.28 x 10-6. 5- Volume of 0.025 M HNO3(aq) required for neutralisation = 1.3 x 10-3 L. 6- Find the concentration of barium ions from the dissolved barium chloride. Using the barium sulphate Ksp, find the maximum sulphate concentration allowed without precipitate formation. Then convert to aluminium sulphate concentration. [Al2(SO4)3(aq)] = 1.3 x 10-9 M. 7- Solubility of PbCl2 in 0.100 M NaCl(aq) = 1.2 x 10-4 M. 8- For AgIO3(s) precipitation, we need [Ag + (aq)] > 9.1 x 10-6 M. For Ag2CO3(s) precipitation, we need [Ag + (aq)] > 1.1 x 10-4 M. The first precipitate formed will be AgIO3(s). The second ppt is Ag2CO3(s) and will commence when: [Ag + (aq)] = 1.1 x 10-4 M (as calculated), [CO3-2 (aq)] = 6.4 x 10-4 M (as given) and, [IO3 - (aq)] = 2.9 x 10-4 M when Ag2CO3(s) starts to precipitate. 9- Draw a qualitative analysis scheme showing how you would separate a mixture of Mg+2,Pb+4,andSr+2 ions. Add SO4-2 (as Na2SO4) to precipitate the strontium ions, then add S -2 (as Na2S) to precipitate the lead (IV) ions, then add CO3-2 (as Na2CO3) to precipitate the magnesium ions. 10- Draw a qualitative analysis scheme showing how you would separate a mixture of Cr+3,Cu+,andBa+2 ions. Add I - (as NaI) to precipitate the copper ions, then add add S -2 (as Na2S) to precipitate the chromium (III) ions, then add CO3-2 (as Na2CO3) to precipitate the magnesium ions. 11- Draw a qualitative analysis scheme showing how you would separate a mixture of SO4-2,S -2,and Br - ions. Add Ca +2 (as Ca(NO3)2) to precipitate the sulphate ions, then add Fe +3 (as Fe(NO3)3) to precipitate the sulphide ions, then add Pb +2 (as Pb(NO3)2) to precipitate the bromide ions.