SOLUBILITY PRODUCT (K sp ) Slightly Soluble Salts & ph AND BUFFERS (Part Two)

Transcription

1 SOLUBILITY PRODUCT (K sp ) Slightly Soluble Salts & ph AND BUFFERS (Part Two) ADEng. PRGORAMME Chemistry for Engineers Prepared by M. J. McNeil, MPhil. Department of Pure and Applied Sciences Portmore Community College Main Campus

2 LECTURE OBJECTIVES Use the solubility product constant, K sp, concept. Calculate the K sp of compounds. Have an understanding of the common-ion effect. Predict the formation of a precipitate.

3 SOLUBILITY OF IONIC SALTS The solubility of ionic salts varies considerably. Most nitrates are very soluble, while phosphates and may other salts of transition metal ions are often insoluble.

4 SOLUBILITY AND TEMPERATURE The solubility of ionic salts varies considerately with the temperature. Some such as KNO 3 have a wide solubility range. The solubility of NaCl in water varies only slightly. Define saturation, supersaturation and undersaturation.

5 SOLUBILITY Table salt is water soluble, but the addition of more and more salt the dissolution decreases until it ceases. Why? Atomic level - NaCl (ionic bonds) grain will break apart into their ions and ushered away by the water. Result - A net increase in the [Na + (aq)] and [Cl - (aq)] in the solution. With the increase increase increase the amount of ions, eventually no more ions can be added, and it will be pushed back to the solid form of salt. Solution = Solute and Solvent (solid) Breaking apart the solid solute into its ions by dissolving it in solution. Solid form (solute) (on the LHS) Aqueous form in ionic form (on the RHS) An equilibrium is achieved in a solution since the reaction is reversible. The compound thus have its unique equilibrium constant and we call this.

6 THE SOLUBILITY PRODUCT CONSTANT, K sp Many important ionic compounds are only slightly soluble in water and equations are written to represent the equilibrium between the compound and the ions present in a saturated aqueous solution. K sp is way of measuring the solubility of any salt. The solubility product constant, K sp, is the product of the concentrations of the ions involved in a solubility equilibrium, each raised to a power equal to the stoichiometric coefficient of that ion in the chemical equation for the equilibrium. Substances that forms a satuarated solution will have a K sp.

7 SOLUBILITY PRODUCT E.g. Copper(II) sulphate On adding more and more CuSO 4, the crystals - not just sitting there, but The rate of dissolution = rate of crystallization (precipitation) in a solution equilibrium. How do we measure whether the reaction goes to the RHS or LHS = equilibrium constant of the substance being dissolved in water.

8 SOLUBILITY EQUILIBRIUM In saturated solutions dynamic equilibrium exists between undissolved solids and ionic species in solutions. Solids continue to dissolve and ionpairs continue to form solids. The rate of dissolution process is equal to the rate of precipitation. General expression: Always take the solid [LHS] and break it down into its constituent ions [RHS]. M m X n (s) mm n+ (aq) + nx m- (aq) The ions, M n+ (aq) and X m- (aq) will remain constant over time and an equilibrium constant expression can be derived that can describe the solution at a given temperature. Solubility product, K sp = [M n+ ] m [X m- ] n If the molar solubility of the ions, [M n+ (aq)] and [X m- (aq)] are altered, the Ksp value can still be determined just before precipitation. Large Ksp have high solubility (NaCl) Small Ksp have low solubility

9 Example 1 SOLUBILITY EQUILIBRIUM EQUATION AND K sp CaF 2 (s) Ca 2+ (aq) + 2 F - (aq) K sp = [Ca 2+ ][F - ] 2 K sp = 5.3 x 10-9 units Example 2 As 2 S 3 (s) 2 As 3+ (aq) + 3 S 2- (aq) K sp = [As 3+ ] 2 [S 2- ] 3 K sp = 4.4 x units

10 SOLUBILITY PRODUCT CONSTANTS AT 25 o C

11 K sp AND MOLAR SOLUBILITY CALCULATIONS The solubility product constant is related to the solubility of an ionic solute, but K sp and molar solubility - the molarity of a solute in a saturated aqueous solution - are not the same thing. Calculating solubility equilibria fall into two categories: determining a value of K sp from experimental data calculating equilibrium concentrations when K sp is known. SOLUBILITY (SOLID) AND SOLUBILITY PRODUCTS (IONS)

12 CALCULATING K sp FROM MOLAR SOLUBILITY 1. Determine the K sp of copper(ii) bromate. Use the ICE table: Cu(BrO 3 ) 2 Cu 2+ 2BrO 3 - Initial Change - +x +2x Equilibrium - +x +2x Cu 2+ K sp = [Cu 2+ ] [BrO 3- ] K sp = x. 2x K sp = 2x 2 units 12

13 CALCULATING K sp FROM MOLAR SOLUBILITY 2. It is found that 1.2 x 10-3 mol of lead (II) iodide, PbI 2, dissolves in 1.0 dm 3 of aqueous solution at 25 o C. What is the K sp at this temperature? PbI 2 (s) Pb 2+ (aq) + 2 I - (aq) K sp = [Pb 2+ ] [I - ] 2 For every lead iodide that dissolves, there is one lead ion and 2 iodide ions. Therefore, [Pb 2+ ] = 1.2x10-3 units [I - ] = 2.4x10-3 units K sp = (1.2x10-3 ) (2.4x10-3 ) 2 K sp = 6.9 x 10-9 units So we must find the concentrations of each ion and then solve for K sp.

14 CALCULATING MOLAR SOLUBILITY FROM K sp 3. Calculate the concentration of [Ag + ] and [Cl - ] in a saturated solution of silver chloride. The K sp for silver chloride : K sp = 1.8x K sp = [Ag + ] [Cl - ] = 1.8x10-10 Let x = [Ag + ] = [Cl - ] using I.C.E. Substitute into the K sp expression x 2 = 1.8x10-10 x = 1.8x10-10 x = 1.34 x 10-5 units [Ag + ] = [Cl - ] = 1.34 x 10-5 units

15 CALCULATING MOLAR SOLUBILITY FROM K sp 4. Calculate the molar solubility of silver chromate, Ag 2 CrO 4, in water from K sp = 1.1x for Ag 2 CrO 4. Ag 2 CrO 4 (s) 2 Ag + + CrO 2-4 K sp = [Ag + ] 2 [CrO 2-4 ] K sp = 1.1x10-12 Let x = [CrO 2-4 ] 2x = [Ag + ] using I.C.E. Substitute into the K sp expression 4x 3 = 1.1x10-12 x = 3 1.1x10-12 /4 x = [CrO 2-4 ] = 6.5 x10-5 units 2x = [Ag + ] = 1.30 x 10-4 units

16 FACTORS THAT AFFECT SOLUBILITY Temperature Solubility generally increases with temperature; Common ion effect Common ions reduce solubility; Salt effect This slightly increases solubility; ph of solution ph affects the solubility of ionic compounds in which the anions are conjugate bases of weak acids; Formation of complex ion The formation of complex ion increases solubility.

17 THE COMMON ION EFFECT IN SOLUBILITY EQUILIBRIA The precipitate of a solute on addition of another solution which has an ion in common with the solute is referred to as the common ion effect. The solubility of a slightly soluble ionic compound is lowered when a second solute that furnishes a common ion is added to the solution. The common ion effect also affects solubility equilibria. Le Châtelier s principle is followed for the shift in concentration of products and reactants upon addition of either more products or more reactants to a solution.

18 THE COMMON ION EFFECT E.g. Consider the following solubility equilibrium: AgCl(s) Ag + (aq) + Cl - (aq); K sp = 1.6 x ; The solubility of AgCl is 1.3 x 10-5 mol/l at 25 o C. If NaCl is added, equilibrium shifts left due to increase in [Cl - ] and some AgCl will precipitate out. AgCl(s) Ag + (aq) + Cl - (aq); NaCl(s) Na + (aq) + Cl - (aq); For example, if [Cl - ] = 1.0 x 10-2 M, Solubility of AgCl = (1.6 x )/(1.0 x 10-2 ) = 1.6 x 10-8 mol/l (Solubility decreases

19 THE COMMON ION EFFECT BaSO 4 (s) Ba 2+ (aq) + SO 42 (aq) Na 2 SO 4 (s) 2Na + (aq) + SO 42 (aq) The presence of the sodium sulphate increases the concentration of SO 4 2- ions in the mixture. An increase in the concentration of [SO 42 ] (common ion) would cause a decrease in the [Ba 2+ ] ions to keep the solubility product at a constant value. The equilibrium will thus shift to counteract the stress imposed and reduce the solubility of BaSO 4. The addition of the sulphate ions results in the precipitation of the solid BaSO 4.

20 THE COMMON ION EFFECT CALCULATIONS 1. What is the solubility of Ag 2 CrO 4 in 0.10 M K 2 CrO 4? K sp = 1.1x10-12 for Ag 2 CrO 4. Let 2x = Ag + and CrO 4 2- = 10-1 (4x 2 )(10-1 ) = 1.1 x x 2 = 1.1 x / x 2 = 1.1 x x = 1.66 x 10-6 [Ag + ] = 2x = 3.32 x 10-6 Comparison of solubility of Ag 2 CrO 4 In pure water: 6.5 x 10-5 M In 0.10 M K 2 CrO 4 : 1.7 x 10-6 M The common ion effect!!

21 DETERMINING WHETHER PRECIPITATION OCCURS Q ip is the ion product reaction quotient and is based on initial conditions of the reaction. Q ip is ion product expressed in the same way as K sp for a particular system. To predict if a precipitation occurs: Q ip can then be compared to K sp!!!!!!!! - Precipitation should occur if Q ip > K sp (precipitate). - Precipitation cannot occur if Q ip < K sp (no precipitate). - A solution is just saturated if Q ip = K sp (unsaturated solution). Q sp and Q ip are termed the ion product expressed in the same way as K sp for a particular system.

22 PREDICTING PRECIPITATION 1. The concentration of calcium ion in blood plasma is M. If the concentration of oxalate ion is 1.0x10-7 M, do you expect calcium oxalate to precipitate? K sp = 2.3x10-9. CaC 2 O 4 Ca 2+ (aq) + C 2 O 4 2- (aq) [Ca 2+ ] [C 2 O 4 2- ] = 2.3x10-9 Three steps: 1. Determine the initial concentrations of ions. (Put both in the K sp formula) 2. Evaluate the reaction quotient Q ip. 3. Compare Q ip with K sp. If Q ip > K sp, then it will precipitate.

23 PREDICTING PRECIPITATION In applying the precipitation criteria, the effect of dilution when solutions are mixed must be considered. E.g ml of M Pb(NO 3 ) 2 is added to 30.0 ml of 0.10 M NaCl. Predict if precipitate of PbCl 2 will form. (Ksp for PbCl 2 = 1.6 x 10-5 ) Calculation: [Pb 2+ ] = (20.0 ml x M)/(50.0 ml) = M [Cl - ] = (30.0 ml x 0.10 M)/(50.0 ml) = M Q sp = [Pb 2+ ][Cl - ] 2 Q sp = (0.010 M)(0.060 M) 2 Q sp = 3.6 x 10-5 Q ip can then be compared to K sp!!!!!!!! To predict if a precipitation occurs: - Precipitation should occur if Q ip > K sp (precipitate). Q sp > K sp precipitate of PbCl 2 will form. - Precipitation cannot occur if Q ip < K sp (no precipitate). - A solution is just saturated if Q ip = K sp (unsaturated solution).

24 PRACTICAL APPLICATIONS OF SOLUBILITY EQUILIBRIA Qualitative Analyses Isolation and identification of cations and/or anions in unknown samples Synthesis of Ionic Solids of commercial interest. Selective Precipitation based on K sp Separation and identification of cations, such as Ag +, Ba 2+, Cr 3+, Fe 3+, Cu 2+, etc. can be carried out based on their different solubility and their ability to form complex ions with specific reagents, such as HCl, H 2 SO 4, NaOH, NH 3, and others. Separation and identification of anions, such as Cl -, Br -, I -, SO 4 2-, CO 3 2-, PO 4 3-, etc., can be accomplished using reagents such as AgNO 3, Ba(NO 3 ) 2 under neutral or acidic conditions.

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26 SELECTIVE PRECIPITATION (Mixtures of Metal Ions) Use a reagent whose anion forms a precipitate with only one or a few of the metal ions in the mixture. Example: Solution contains Ba 2+ and Ag + ions. (Double displacement reaction) Adding NaCl will form a precipitate with Ag + (AgCl), while still leaving Ba 2+ in solution. Read up on limitations of K sp.

27 ph and BUFFERS 1. What is meant by a buffer solution? 2. Explain how buffer solutions control ph? 3. Calculate the ph of a specific buffer. 4. Discuss the importance of buffers in biological systems and industrial processes.

28 BUFFER SOLUTIONS Buffers are solutions which resists changes in ph when small quantities of acid of alkali are added Acidic Buffer (ph < 7) made from a weak acid + its sodium or potassium salt (conjugate base or a salt of the weak acid) ethanoic acid sodium ethanoate Alkaline Buffer (ph > 7) made from a weak base + its chloride (conjugate acid or a salt of the weak base ammonia ammonium chloride

29 USES OF BUFFER SOLUTIONS In biological systems (saliva, stomach, and blood) it is essential that the ph stays constant in order for any processes to work properly. e.g. If the ph of blood varies by 0.5 it can lead to unconsciousness and coma. ph affects the structure and function of enzymes. Many household and cosmetic products need to control their ph values. Buffer solutions counteract the alkalinity of the soap and prevent irritation. Washing powder and eye drops. Standardising ph meters

30 ACTION OF BUFFER SOLUTION It is essential to have a weak acid for an equilibrium to be present so that ions can be removed and produced. The dissociation is small and there are few ions. CH 3 COOH(aq) CH 3 COO (aq) + H + (aq) relative concs. HIGH LOW LOW NB A strong acid can t be used as it is fully dissociated and cannot remove H + (aq) HCl(aq) > Cl (aq) + H + (aq) Adding acid Any H + is removed by reacting with CH 3 COO ions to form CH 3 COOH via the equilibrium. Unfortunately, the concentration of CH 3 COO is small and only a few H + can be mopped up. A much larger concentration of CH 3 COO is required. To build up the concentration of CH 3 COO ions, sodium ethanoate is added, which dissociates completely. CH 3 COO Na + (aq) > CH 3 COO (aq) + Na + (aq) relative concs. LOW HIGH HIGH

31 ACTION OF BUFFER SOLUTION We have now got an equilibrium mixture which contains a high concentration of both the undissociated weak acid, CH 3 COOH(aq), and its conjugate base, CH 3 COO (aq). CH 3 COOH(aq) CH 3 COO (aq) + H + (aq) relative concs. HIGH LOW LOW CH 3 COO Na + (aq) CH 3 COO (aq) + Na + (aq) relative concs. LOW HIGH HIGH This adds OH ions Adding alkali These react with the small concentration of H + ions: H + (aq) + OH (aq) H 2 O(l) Removal of H + from the weak acid equilibrium means that more CH 3 COOH will dissociate to form ions to replace those being removed. CH 3 COOH(aq) CH 3 COO (aq) + H + (aq) As the added OH ions remove the H + from the weak acid system, the equilibrium moves to the right to produce more H + ions. (There needs to be a large concentration of undissociated acid molecules to be available)

32 BUFFER SOLUTIONS - IDEAL CONC. The concentration of a buffer solution is important If the concentration is too low, there won t be enough CH 3 COOH and CH 3 COO to cope with the ions added. Summary For a buffer solution one needs... large [CH 3 COOH(aq)] - for dissociating into H + (aq) when alkali is added weak acid (equilibrium shifts to the right) large [CH 3 COO (aq)] - for removing H + (aq) as it is added conjugate base (equilibrium shifts to the left) `This situation can t exist if only acid is present; a mixture of the acid and salt is used. The weak acid provides the equilibrium and the large CH 3 COOH(aq) concentration. The sodium salt provides the large CH 3 COO (aq) concentration. One uses a WEAK ACID + its SODIUM OR POTASSIUM SALT

33 ACTION OF ALKALINE BUFFERS Alkaline buffer Very similar but is based on the equilibrium surrounding a weak base; AMMONIA NH 3 (aq) + H 2 O(l) OH (aq) + NH 4+ (aq) relative concs. HIGH LOW LOW but one needs ; a large conc. of OH (aq) to react with any H + (aq) added a large conc of NH 4+ (aq) to react with any OH (aq) added There is enough NH 3 to act as a source of OH but one needs to increase the concentration of ammonium ions by adding an ammonium salt. Use AMMONIA (a weak base) + AMMONIUM CHLORIDE (one of its salts)

34 CALCULATING THE ph OF AN ACIDIC BUFFER SOLLUTION Whenever you are asked to calculate the ph of a buffer solution, ALWAYS USE THE Henderson-Hasselbach equation ph = pk a + log 10 [salt]/[acid] Calculate the ph of a buffer whose [HA] is 0.1 mol dm -3 and [A ] of 0.1 mol dm -3. The K a of the weak acid HA is 2 x 10-4 mol dm -3

35 CALCULATING THE ph OF AN ACIDIC BUFFER SOLLUTION (Alternative calculation to Henderson-Hasselbalch equation use) Question One Calculate the ph of a buffer whose [HA] is 0.1 mol dm -3 and [A ] of 0.1 mol dm -3. The K a of the weak acid HA is 2 x 10-4 mol dm -3 K a = [H + (aq)] [A (aq)] [HA(aq)]

36 CALCULATING THE ph OF AN ACIDIC BUFFER SOLLUTION Calculate the ph of a buffer whose [HA] is 0.1 mol dm -3 and [A ] of 0.1 mol dm -3. K a = [H + (aq)] [A (aq)] [HA(aq)] re-arrange [H + (aq)] = [HA(aq)] x K a [A (aq)]

37 CALCULATING THE ph OF AN ACIDIC BUFFER SOLLUTION Calculate the ph of a buffer whose [HA] is 0.1 mol dm -3 and [A ] of 0.1 mol dm -3. K a = [H + (aq)] [A (aq)] [HA(aq)] re-arrange [H + (aq)] = [HA(aq)] x K a [A (aq)] from information given [A ] = 0.1 mol dm -3 [HA] = 0.1 mol dm -3

38 Calculating the ph of an acidic buffer solution Calculate the ph of a buffer whose [HA] is 0.1 mol dm -3 and [A ] of 0.1 mol dm -3. K a = [H + (aq)] [A (aq)] [HA(aq)] re-arrange [H + (aq)] = [HA(aq)] x K a [A (aq)] from information given [A ] = 0.1 mol dm -3 [HA] = 0.1 mol dm -3 If the K a of the weak acid HA is 2 x 10-4 mol dm -3. [H + (aq)] = 0.1 x 2 x 10-4 = 2 x 10-4 mol dm

39 CALCULATING THE ph OF AN ACIDIC BUFFER SOLLUTION Calculate the ph of a buffer whose [HA] is 0.1 mol dm -3 and [A ] of 0.1 mol dm -3. K a = [H + (aq)] [A (aq)] [HA(aq)] re-arrange [H + (aq)] = [HA(aq)] x K a [A (aq)] from information given [A ] = 0.1 mol dm -3 [HA] = 0.1 mol dm -3 If the K a of the weak acid HA is 2 x 10-4 mol dm -3. [H + (aq)] = 0.1 x 2 x 10-4 = 2 x 10-4 mol dm ph = - log 10 [H + (aq)] = 3.699

40 CALCULATING THE ph OF AN ACIDIC BUFFER SOLLUTION Question Two Calculate the ph of the solution formed when 500cm 3 of 0.1 mol dm -3 of weak acid HX is mixed with 500cm 3 of a 0.2 mol dm -3 solution of its salt NaX. K a = 4 x 10-5 mol dm -3. Henderson-Hasselbach equation ph = pk a + log 10 [salt]/[acid] ph = 4.699

41 ASSIGNMENT Discuss the importance of buffers in biological and industrial processes. Include reference to blood buffer systems such as hydrogencaronale, phosphate and amino acid systems (zwitterions), enzyme catalysed reactions and the food processing industry.