Chemistry 1B Summer School 2016: Part Two

Transcription

1 Chemistry 1B Summer School 2016: Part Two Topic 9 Phase diagrams and Solubility 1 Topic 10 Transition Metals and Coordination Chemistry 24 Topic 11 Kinetics 43 Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille, Chemistry, 2012 (John Wiley) ISBN:

2 Topic 9 - Phase diagrams and Solubility Phase diagrams The phase (solid, liquid or gas) of a substance depends on temperature, T and pressure, P. A phase diagram describes the conditions of T & P under which each phase is stable. Question: Indicate the phase change indicated by the following term. The first is given as an example. Melting S L Condensation Freezing Vapourisation Sublimation Deposition Key Terms: Triple Point: The conditions under which three phases (solid, liquid and gas) are at equilibrium. Critical Point: The highest temperature at which it is possible to liquefy a gas. Normal Boiling Point: The boiling point at 1 atmosphere pressure. Normal Melting Point: The melting point at 1 atmosphere pressure. Carbon dioxide Water P P critical point critical point solid liquid solid liquid 1 atm 1 atm triple point gas gas - 78 oc T 0 oc 100 oc T 1

3 [WS 27] Vapour Pressure 2.5 A liquid boils when the pressure of the vapour escaping from the liquid is equal to the pressure exerted on the liquid by its surroundings. The pressure of the vapour escaping is called the vapour pressure. The graph opposite shows how the vapour pressure of water changes with temperature. vapour pressure (atm) temperature ( C) Questions: 1. The normal boiling point of water is 100 C. What is the vapour pressure of water at this temperature? 2. The safety valve on commercial pressure cookers is set to allow a maximum pressure of 2 atm. What is the boiling point of water at this pressure? 3. The air pressure at the highest point in Australia is Mount Kosciuszko (2,228 m) is around 0.75 atm. What is the maximum temperature of a cup of tea on top on the mountain? 4. The graph below shows the low temperature part of the vapour pressure diagram for water (a) How could we make water boil at 25 C? vapour pressure (atm) (b) The pressure on the surface of Mars is about atm. If the temperature on Mars in summer reaches 20 C, what phase would water be in? temperature ( C) 2

4 Supercritical Fluids Critical point is the T & P at which the boundary between liquid & vapour disappears. If sealed tube of CO 2 liquid + vapour heated, liquid-vapour phase boundary disappears at 31 C and 73 atm: Supercritical fluids: substance above critical temperature (T c ). Can behave as solvents dissolving a wide range of substances; also new industrial reaction medium. Environmentally friendly: CO 2 is harmless solvent Easily removed (open valve to release pressure). Removal of caffeine from coffee beans by supercritical CO 2 : removes caffeine from beans but leaves flavour and aroma components. Also used in dry-cleaning & as solvent for chemical industry. [WS 27] Phase Diagram of CO 2 A phase diagram allows us to predict the phase (solid, liquid or gas) of a substance that is stable at any given temperature and pressure. The phase diagram opposite is for CO 2. The phases are divided up by lines, which represent the temperatures and pressures at which the substance is in equilibrium between the phases. 3

5 Questions: 1. Add and label the following points on the diagram and hence predict which phase CO 2 will be in under the stated conditions (a) (b) (c) -80 C and 1 atm pressure 25 C and 1 atm pressure 25 C and 10 atm pressure. 2. Draw a line connecting points (a) and (b). Describe in words what happens to CO 2 as: (i) it is warmed from -80 C to -25 C at 1 atm pressure (ii) it is cooled from 25 C to -80 C at 1 atm pressure 3. Draw a line connecting points (b) and (c). Describe in words what happens to CO 2 as: (i) the pressure is increased from 1 atm to 10 atm at 25 C (ii) the pressure is reduced from 10 atm to 1 atm at 25 C 4. Carbon dioxide sequestration has been suggested as a way of storing CO 2 from burning fossil fuels so that it does not enter the atmosphere. It involves pumping liquid CO 2 into empty oil fields such as the 2 km deep Otway Basin in western Victoria. What is the minimum pressure required to liquefy CO 2? 4

6 5. Use the following data to draw a phase diagram for hydrogen. Label the solid, liquid and gas regions of the graph. NB the pressure scale is logarithmic. Normal melting point 14 K Normal boiling point 20.3 K Triple point 14 K, 0.07 atm Critical point 33 K, 13 atm P / atm 10 Hydrogen T / K 5

7 Solubility What determines solubility? Cl - Cl - Cl - Na + ClṈa+ Cl - Na + Cl - Cl - Na + Cl- Na + Na + Cl - Cl - Cl - + H O H H O H H H O OH 2 H 2 O Na + OH 2 H 2 O OH 2 + HOH HOH O H H HOH Cl - HOH HOH Solubility Equilibria Solubility - 'how much dissolves?' In solubility SI data book, the solubility is quoted as the grams of solute that dissolves in 100 ml (100 g) of water. A note on some units: 1. mass% = (grams of solute/grams of solution) x 100 e.g. 5.0 g of sugar in a cup (250 ml) = 2 % 2. molarity = moles solute/litre of solution e.g. [sugar] = (50 g/342 g mol -1 x 0.25 L) = M 3. ppm = mass% x this is 1 mg of a solute in 1 L of solution Saturated solution where the maximum amount of solid is dissolved, Some solid is undissolved so equilibrium between solution and some undissolved solid. solute + solvent solution 6

8 Solubility Equilibrium Constant, K sp K sp is the solubility equilibrium constant (Note: K so may also be used and is equivalent) K sp only involves molecular solutes - solubility and solubility equilibrium constant are the same. It only involves the concentration of the resulting product as the reactants are pure solids. As all solids are slightly soluble there will always be some ions in solution. K sp = [products] / [reactants] = [sugar(aq)] / [sugar(s)] = [sugar(aq)] / 1 (by definition [solid] = 1) e.g. for sucrose at 25 C K sp 6 M (solubilty in water is 2000 gl -1, mw = ) Solubility Product, K sp For ionic solutes - solubility is not equal to solubility equilibrium constant. The solubility product constant is the equilibrium constant for a slightly soluble (or nearly insoluble) ionic compound: MX(s) M + (aq) + X - (aq) K sp = [M + ][ X - ] Solubility is how much MX dissolves, in this case = [M + ] = [X - ]; equilibrium constant is the solubility product. At equilibrium the product of ion concentrations is a constant at each temperature. If the concentration of one ion increases the concentration of the other must decrease to keep equilibrium product constant. Example: How much AgCl dissolves in water at 298 K? K sp (AgCl) = 1.6 x M 2 every x moles of AgCl produces x moles of Ag + and x moles of Cl - 7

9 AgCl (s) Ag + (aq) + Cl - (aq x x x K sp = [Ag + ][Cl - ] = 1.6 x M 2 = [x][x] = x 2 x = 1.3 x 10-5 M (solubility in water) Question: What is the solubility of AgI at 298 K? (K sp (AgI) = 8 x M 2 ) [WS 31] Table 1. The results of adding solid Mg(OH) 2 to a litre of water at 25 C: Total amount of Mg(OH) 2 added (g) (moles) Concentration of ions in solution [Mg 2+ (aq)] [OH - (aq)] / /M M Mass of Mg(OH) 2 that does not dissolve (g) [Mg 2+ (aq)][o H - (aq)] Questions: 1. What is the maximum number of moles of Mg(OH) 2 that can be dissolved in a litre of water? 2. What is the maximum value of the expression [Mg 2+ (aq)][oh - (aq)] 2? 8

10 3. Is it possible for the value of the expression [Mg 2+ (aq)][oh - (aq)] 2 to be less than the value your answer to Q2? 4. What will happen if the concentrations of Mg 2+ (aq) and OH - (aq) are such that the expression [Mg 2+ (aq)][oh - (aq)] 2 is: (a) less than your answer to Q3? (b) more than your answer to Q3? Solubility Product of 2:1 Salts Solubility product equilibrium constants differ between 1:1 and 2:1 electrolytes Example: For PbCl 2 (1:2 salt) PbCl 2 (s) Pb 2+ (aq) + 2 Cl - (aq) for x moles dissolved x 2x K sp = [Pb 2+ ] [Cl - ] 2 = [x] [2x] 2 units of K sp are M 3 K sp (PbCl 2 ) = 1.6 x 10-5 M 3 = x. (2x) 2 = 4x 3 x = 1.6 x 10-2 M Example: Alternative type of problem: a saturated solution of PbCl 2 at 298 K has 3.2 x 10-2 M Cl - ; what is K sp of PbCl 2? K sp = (1.6 x 10-2 )(3.2 x 10-2 ) 2 = 1.6 x 10-5 M 3 Question: What is the solubility of Ag 2 S? K sp (Ag 2 S) = 6 x M 3 9

11 [WS 31] The solubility product A solution is described as saturated if as much solid has dissolved as is possible so there is an equilibrium: Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) If the reaction is at equilibrium, the expression [Mg 2+ (aq)][oh - (aq)] 2 is the equilibrium constant for the reaction. The concentration of the solid is constant so does not appear in the expression. Clearly, this is only true if solid is present: if there is no solid present, the solution is not saturated and the reaction is not at equilibrium. The equilibrium constant is known as the solubility product and given the symbol K sp : Questions: K sp = [Mg 2+ (aq)][oh - (aq)] 2 1. If x moles of Mg(OH) 2 dissolve in 1.00 L of water, what will [Mg 2+ (aq)] and [OH - (aq)] be in terms of x? [Mg 2+ (aq)] = [OH - (aq)] = 2. K sp of Mg(OH) 2 is at 25 C, using your answer to Q1, work out [Mg 2+ (aq)] and [OH - (aq)]. 3. Using the same approach, write down the formula for K sp in terms of x for salts of the following stoichiometry: (a) MX (b) M 2 X (c) MX 3 4. Work out the molar solubility of the following silver salts at 25 C. (a) AgCl; K sp = (b) Ag 2 CrO 4 ; K sp = (Hint: dissolves to give Ag + (aq) and CrO 4 2- (aq)) Relative Solubility Note It is important to take care when comparing the relative solubility using K sp values. K sp = [cation][anion], if x is the solubility in moles/l then x = (K sp ) 1/2 : AgI K sp = 1.5 x x = (K sp ) 1/2 = 1.2 x 10 8 M CuI K sp = 5.0 x x = (K sp ) 1/2 = 2.2 x 10 6 M CaSO 4 K sp = 6.1 x 10 5 x = (K sp ) 1/2 = 7.8 x 10 3 M 10

12 K sp of CuS is larger, but more Bi 2 S 3 dissolves: CuS K sp = 8.5 x K sp = x 2 x = 9.2 x M Ag 2 S K sp = 1.6 x K sp = (4x 3 ) x = 3.4 x M Bi 2 S 3 K sp = 1.1 x K sp = (4x 2 *27x 3 ) x = 1.0 x M Ion Product, Q To determine whether an equilibrium system will go in the forward or reverse direction requires that we evaluate the reaction quotient, Q. This is the product of ionic concentrations before solubility equilibrium is established. For the reaction: PbCl 2 (s) Pb 2+ (aq) + 2 Cl - (aq); Q C = [Pb 2+ ] i [Cl -1 ] i where i = initial concentration (same expression as equilibrium constant), c in terms of concentration. If Q c > K sp then a precipitate forms If Q c = K sp then the solution is saturated If Q c < K sp then more solute can dissolve Example: 1. Does a precipitate of AgCl(s) form if equal volumes of 1 mm HCl and 1 mm AgNO 3 solutions are mixed? AgCl (s) Ag + (aq) + Cl - (aq) [Ag + ] = 5 x 10-4 M (after mixing) [Cl - ] = 5 x 10-4 M (after mixing) Q = [Ag + ][ Cl - ] = 2.5 x 10-7 M 2 > K sp = 1.6 x M 2 AgCl(s) precipitates, i.e. reaction moves to left toward equilibrium. 11

13 2. What about equal volumes of 2.0 x 10-5 M solutions? [Ag + ] = 1.0 x 10-5 M (after mixing) [ Cl - ] = 1.0 x 10-5 M (after mixing) Q = [Ag + ][ Cl - ] = 1.0 x M 2 < K sol = 1.6 x M 2 AgCl(s) does not precipitate. Question: Lead iodide is an almost insoluble salt with a dense golden yellow colour. (It is used in ornamental work requiring a gold-like colour). The solubility product equilibrium constant is 7.1x10-9 mol 3 /L 3. Calculate the solubility of PbI 2 (assume excess solid). PbI 2 (s) Pb 2+ (aq) + 2I - (aq); K sp =7.1x10-9 Initial: 0 0 mol/l Change: +x +2x mol/l Equilibrium: x 2x mol/l Solve for x and work out how many mol/l of PbI 2 will dissolve. 12

14 [WS 31] As we saw earlier, the solubility product gives the maximum values of the ion concentrations that are allowed. The concentrations can certainly be less but their product cannot be more! If their concentrations such that their product is less than K sp, then more solid can dissolve. If [Mg 2+ (aq)][oh - (aq)] 2 < K sp then more solid can dissolve If their concentrations are such that their product is more than K sp then the concentrations must reduce: precipitation must occur. If [Mg 2+ (aq)][oh - (aq)] 2 > K sp then precipitation must occur The value of the product can thus be used to predict whether dissolution or precipitation can occur. Because of its importance, it is called the ionic product and given the symbol Q sp : Q sp = [Mg 2+ (aq)][oh - (aq)] 2 If Q sp < K sp then dissolution will occur. If Q sp > K sp then precipitation will occur. Questions: ml of 0.12 M NaOH solution with ml of 0.10 M Mg(NO 3 ) 2 are mixed. (a) Assuming that no reaction occurs, what would [Mg 2+ (aq)] and [OH - (aq)] be after mixing? (b) Write down the value of the ionic product, Q sp. (c) Does a precipitate form? 13

15 Common Ion Effect The initial presence of an ion in solution may suppress the solubility of a solid if the solid also contains the same ion the equilibrium constant remains the same. Example: How much AgCl dissolves in 0.1 M HCl? AgCl (s) Ag + (aq) + Cl - (aq) x (0.1+ x) 0.1 K sp = [Ag + ][Cl - ] = 1.6 x M 2 = [x][0.1 +x] 0.1x x = 1.6 x 10-9 M (This is << than 10-5 M in H 2 O) Question: What is the solubility of AgBr in 0.05 M sodium bromide solution at 298 K? (K sp (AgBr) = 5 x M 2 ) 14

16 [WS 32] Le Châtalier s principle states: If a change is imposed on a system, the position of equilibrium will shift in a direction that tends to reduce that change If the concentration of a reactant is increased, the equilibrium responds by producing more products. If the concentration of a product is increased, the equilibrium responds by producing more reactant. PbCl 2 (s) Pb 2+ (aq) + 2Cl - (aq) PbCl 2 is not very soluble in water. The picture shows a test tube containing a saturated solution of lead chloride in contact with a precipitate of solid. The effect on this solubility of adding Pb 2+ (aq) or Cl - (aq) ions from another source is called the common ion effect. Questions: 1. Write down the solubility product expression, K sp, for lead chloride. 2. Sodium chloride dissolves completely to give Na + (aq) and Cl - (aq) ions. If sodium chloride is added to the saturated solution, what would be the effect on the solubility of lead(ii) chloride? (Hint: consider how the equilibrium written above would shift, according to Le Châtelier s principle, when these ions are added). 3. If sodium chloride is added so that [Cl - (aq)] = 0.5 M, rearrange your K sp expression to give [Pb 2+ (aq)]. 15

17 4. If K sp = at 25 o C, what is [Pb 2+ (aq)] in 0.5 M NaCl? 5. What is the effect of adding extra PbCl 2 (s) to the test tube? (Be careful!) 6. What is the solubility of PbCl 2 (s) chloride (in mol L 1 ) in 1.00 M magnesium chloride solution? (Hint: what is [Cl - (aq)] for a 1.00 M magnesium chloride solution?) Solubilities of Hydroxides Example: 1. What is solubility of Zn(OH) 2 in water at 298 K? K sp = 1.0 x M 3 Zn(OH) 2 Zn OH - x dissolves x 2x K sp = (x)(2x) 2 = 4x 3 = 1.0 x M 3 x = 6.3 x 10-6 M 2. What is the ph of a saturated solution of Zn(OH) 2? [OH - ] = 2x = 2 (6.3 x 10-6 M) = 1.2 x 10-5 M poh = 4.9 ph = poh = 9.1 Hydroxides are generally insoluble - except Group 1 (alkali metals) Ca(OH) 2 and Ba(OH) 2. Divalent hydroxides are moderately insoluble; trivalent hydroxides are exceedingly insoluble. 16

18 Example: What is solubility of Fe(OH) 3 in water at 298 K? K sp = 1.0 x M 4 Fe(OH) 3 Fe OH - x dissolves x 3x K = [Fe 3+ [OH - ] 3 = [x] [3x] 3 = 27 x 4 x = 1.4 x M ph and Hydroxide Solubility Sometimes it is necessary to account for other reactions aqueous ions might undergo. For example, if the anion is the conjugate base of a weak acid, it will react with H 3 O +. Terefore the solubility will be affected by ph. Example: 1. What are solubilities of Zn(OH) 2 and Fe(OH) 3 at ph 7? Zn(OH) 2 : K sp = 1.0 x M3 [Zn 2+ ] = [x] = 0.1 M = [Zn 2+ ][OH - ]2 = [x ] [10-7 ] 2 Fe(OH) 3 : K sp = 1.0 x M 4 = [x] [10-7 ] 3 [Fe 3+ ] = [x] = 1.0 x M 1. At what ph is Fe(OH) 3 soluble to 0.1 M? K sp = 1.0 x M 4 = [0.1] [x] 3 x 3 = 1.0 x M 3 x = 4.6 x M poh = 12.3 and ph = 1.7 ph dependence of solubilities can be calculated and plotted - ph control can be used to separate metal ions. 17

19 Application: Al(OH) 3 is used to clarify drinking water - how much Al (in ppm) remains in water at ph 6.6? K sp = [Al 3+ ][OH - ] 3 = 3 x M 4 at ph 6.6, [OH - ] = = 4 x 10-8 M [Al 3+ ] = 3 x / (4 x 10-8 ) 3 = 5 x M and 5 x mol L -1 x 27 g mol -1 x 10 3 mg L -1 = 1 x 10-7 mg L -1 (ppm) Example: Consider the following equilibrium is calcium oxalate likely to be more soluble in acidic solutions?: CaC 2 O 4 (s) Ca 2+ (aq) + C 2 O 4 2- (aq) Because the oxalate ion is conjugate to a weak acid (HC 2 O 4 ), it will react with H 3 O +. C 2 O 4 2- (aq) + H 3 O + (aq) HC 2 O 4 - (s) + H 2 O (l) C 2 (OOH) 2 à C 2 (OOH)OO - (pk a = 1.23) C 2 (OOH)OO - à C 2 (OOH) 2 ` (pk a = 4.12) 2- According to Le Chatelier s principle, as C 2 O 4 ion is removed by the reaction with H 3 O +, more calcium oxalate dissolves. Therefore, calcium oxalate should be more soluble in acidic solution (low ph) than in pure water. Case Study 1: Kidney Stones and Calcium Oxalate The presence of a common ion reduces the solubility of the salt. In some cases considerably. - eg Ca and Kidney Stones Each salt contributes the same cation Ca 2+. The effect of the calcium ion provided by the calcium chloride is to make calcium oxalate less soluble than it would be in pure water CaC 2 O 4 (s) Ca 2+ (aq) + C 2 O 2-4 (aq) K sp of Ca(C 2 O 4 ) = 1.96 x 10 8 M 2 = [Ca 2+ ][C 2 O 2 4 ] 18

20 Addition of Ca 2+ solution. pushes the equilibrium to the left causing calcium oxalate to precipitate from Originally it was thought that consumption of too much calcium could promote the development of calcium kidney stones. However, current evidence suggests that the consumption of low-calcium diets is actually associated with a higher overall risk for the development of kidney stones. This is perhaps related to the role of calcium in binding ingested oxalate in the gastrointestinal tract. As the amount of calcium intake decreases, the amount of oxalate available for absorption into the bloodstream increases; this oxalate is then excreted in greater amounts into the urine by the kidneys. In the urine, oxalate is a very strong promoter of calcium oxalate precipitation, about 15 times stronger than calcium. The formation of renal stones is related to the urine ph (range ). Patients being treated for renal calculi are frequently given diets or medications to change the ph of the urine so that kidney stones will not form. Calcium phosphate, calcium carbonate, and magnesium phosphate stones develop in alkaline urine; when this occurs, the urine is kept acidic. Uric acid, cystine, and calcium oxalate stones precipitate in acidic urine; in this situation, the urine should be kept alkaline or less acidic than normal 19

21 [WS32] Metal hydroxides dissolve to give metal ions and hydroxide ions. For example, Fe(OH) 3 (s) Fe 3+ (aq) + 3OH - (aq) The position of the equilibrium (i.e. the solubility) is very sensitive to ph since this controls [OH - (aq)]. All forms of life depend on iron and the concentration of iron in the oceans and elsewhere is one of the primary factors limiting the growth rates of the most basic life forms. One reason for the low availability of iron(iii) is the insolubility of Fe(OH) 3 which has a K sp of only 1 x Questions: 1. Write down the expression for the solubility product, K sp, for Fe(OH) The ph of the oceans is currently Use this to work out [OH - (aq)]. 3. If x moles of Fe(OH) 3 dissolve in 1.00 L of water, [Fe 3+ (aq)] = x mol L -1. Use your answers to Q1 and Q2 to work out x. 4. If the amount of CO 2 in the atmospheres increases, the ph of the oceans will decrease due to the equilibrium below. What will happen to [Fe 3+ (aq)]? CO 2 (g) + H 2 O(l) HCO H 3 O + (aq) 20

22 Thermodynamics of Solubility Temperature dependence The solubility of most substances increases with T, some show little change (e.g. NaCl), for a few solubility decreases with increase in T. solute + solvent solution ± heat if dissolution exothermic, K decreases with T endothermic, K increases with T Heat of solution this is a small difference between two large quantities (lattice energy and hydration energy): G hyd = H hyd - H lat - T S hyd Enthalpy of Hydration The smaller the ionic radius, the more energy is released by surrounding an ion with water molecules. The more highly charged the ion, the more energy is released. ΔH hyd - Always favours dissolution Lattice Energy Energy released when crystal is formed from free ions H latt ca H latt cation charge x anion charge/cation radius + anion radius H latt q 2 /r ΔH latt Always opposes dissolution Entropy of Hydration Solid à Dissolved ions ΔS hyd Almost always favours dissolution Na + (g) + Cl - (g) -LE = 786 kj mol -1 ΔHhyd = kj mol-1 NaCl (s) Na + (aq) + Cl - (aq) ΔHsol = + 3 kj mol -1 21

23 Blackman Fig Case Study 2: Cave Formation Limestone - mainly calcium carbonate, CaCO 3 CaCO 3 dissolves in CO 2 -containing ground-water: CO 2 (g) CO 2 (aq) CaCO 3 (s) + CO 2 (aq) + H 2 O(l) Ca 2+ (aq) + 2HCO 3 (aq) Dissolution of the limestone produces limestone caves. Stalactites & Stalagmites The reverse of these processes occurs as water saturated with calcium hydrogencarbonate drips from the ceiling of the cave. 22

24 Loss of CO 2 to the atmosphere: CO 2 (aq) CO 2 (g) drives the precipitation of calcium carbonate: 2 HCO 3 - (aq) + Ca 2+ (aq) CaCO 3 (s) + H 2 O(l) + CO 2 (aq) 23