AQUEOUS EQUILIBRIA. Salts Hydrolysis The Common-Ion Effect Buffer Solutions Acid-Base Titrations Solubility. Brown et al., Chapter 15,

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1 AQUEOUS EQUILIBRIA Salts Hydrolysis The Common-Ion Effect Buffer Solutions Acid-Base Titrations Solubility Brown et al., Chapter 15, CHEM120 Lecture Series One : 2011/01

2 SALTS AND HYDROLYSIS Recall: an acid is a proton donor and a base is a proton acceptor. A salt arises from the reaction of an acid with a base. e.g., the reaction between HBr and Mg(OH) 2 leads to MgBr 2. 2HBr(aq) + Mg(OH) 2 (aq) MgBr 2 (aq) + 2H 2 O(l) When a solid sample of a salt (the solute) is placed in water (the solvent) the salt dissolves to give a solution. NaCl(s) + (aq) Na + (aq) + Cl - (aq) solute solvent solution SALTS dissociate FULLY in water to form ionic solutions and are thus STRONG ELECTROLYTES. CHEM120 Lecture Series One : 2011/02

3 Weak electrolytes do not dissociate fully into ions when in solution. EXAMPLE CH 3 COOH(aq) CH 3 COO - (aq) + H + (aq) What percentage of molecules have dissociated in a mol dm -3 solution of acetic acid? K a (CH 3 COOH) = 1.75 x 10-5 SOLUTION CH 3 COOH(aq) CH 3 COO - (aq) + H + (aq) Initial (M) Change (M) -x +x +x Equili. (M) (0.100 x) x x CHEM120 Lecture Series One : 2011/03

4 Assume that x x 2 = x = % dissociated = ( /0.1) 100% = 1.32% SALTS CAN REACT WITH WATER TO BE EITHER ACIDIC, BASIC OR NEUTRAL. CHEM120 Lecture Series One : 2011/04

5 SALTS OF STRONG ACIDS AND STRONG BASES NaOH + HCl NaCl + H 2 O H 2 O NaCl Na + + Cl - Na + + H 2 O no reaction and Cl - + H 2 O no reaction a solution of NaCl is neutral. CHEM120 Lecture Series One : 2011/05

6 SALTS OF WEAK ACIDS AND STRONG BASES CH 3 COOH + NaOH CH 3 COONa + H 2 O H CH 3 COONa 2 O CH 3 COO + Na + Na + does not react with water. CH 3 COO undergoes hydrolysis CH 3 COO + H 2 O CH 3 COOH + OH ph of sodium acetate solution > 7 (basic). CHEM120 Lecture Series One : 2011/06

7 SALTS OF STRONG ACIDS AND WEAK BASES HCl + NH 3 NH 4 Cl H 2 O NH 4 Cl NH Cl - NH 4+ + H 2 O NH 3 + H 3 O + ph of NH 4 Cl solution < 7 (acidic) CHEM120 Lecture Series One : 2011/07

8 SALTS OF WEAK ACIDS AND WEAK BASES Solution ph depends on relative strengths of weak acid and weak base. HNO 2 + NH 3 NH 4 NO 2 H 2 O NH 4 NO 2 NH 4+ + NO - 2 NH 4+ + H 2 O NH 3 + H 3 O + K a = NO H 2 O HNO 2 + OH - K b = K b < K a solution acidic acidic if K a (cation) > K b (anion) neutral if K a (cation) = K b (anion) basic if K a (cation) < K b (anion) CHEM120 Lecture Series One : 2011/08

9 EXAMPLE Calculate the ph of a mol dm -3 solution of NH 4 Cl at 25 o C. SOLUTION K a = NH 4+ (aq) + H 2 O(l) NH 3 (aq) + H 3 O + (aq) Initial (M) Change (M) -x +x +x Equili. (M) ( x) x x x = From, ph = - log [H 3 O + ], ph = 5.14 CHEM120 Lecture Series One : 2011/09

10 EXAMPLE Calculate the ph of a mol dm -3 solution of sodium acetate. K b = SOLUTION CH 3 COO - (aq) + H 2 O(l) CH 3 COOH(aq) + OH - (aq) Initial (M) Change (M) -x +x +x Equili. (M) (0.154 x) x x x = poh = 5.03, therefore ph = 8.97 CHEM120 Lecture Series One : 2011/10

11 COMMON ION EFFECT CH 3 CO 2 H + H 2 O H 3 O + + CH 3 CO 2 What happens if sodium acetate is added? Can be explained by Le Chatelier s Principle. When a reactant containing a given ion is added to an equilibrium mixture that already contains that ion, the position of equilibrium shifts away from forming more of it. CHEM120 Lecture Series One : 2011/11

12 EXAMPLE What is the ph of a solution made by adding 0.30 mol acetic acid (HC 2 H 3 O 2 ) and 0.15 mol of sodium acetate (NaC 2 H 3 O 2 ) to enough water to make 1.0 dm 3 of solution? K a = 1.8 x 10-5 SOLUTION CH 3 COOH H + + CH 3 COO - Initial 0.30 M M Change -x M +x M +x M Equilibrium (0.30 x) M x M ( x) M CHEM120 Lecture Series One : 2011/12

13 x = M = [H + ] ph = -log( ) = 4.44 EXERCISE FOR THE IDLE MIND Practice Exercises: p 571, p 572 and Exercise p 602 CHEM120 Lecture Series One : 2011/13

14 BUFFERS A buffer solution is a solution whose ph remains essentially constant despite the addition of a small amount of either acid or base. Must have an acidic component to react with added OH - and a basic component to react with added H 3 O + ion. Typical buffer is acetic acid (CH 3 COOH) and sodium acetate (CH 3 COONa) CH 3 COOH + H 2 O CH 3 COO - + H 3 O + acid 1 base 1 CHEM120 Lecture Series One : 2011/14

15 Some common buffer solutions are: NaHCO 3 + Na 2 CO 3 Brønsted acid conjugate base NaH 2 PO 4 + Na 2 HPO 4 Brønsted acid conjugate base CHEM120 Lecture Series One : 2011/15

16 ph < 7.00 Two components a weak acid and the salt of a weak acid. WEAK ACID + SALT OF A WEAK ACID CH 3 COOH + CH 3 COO - Na + acetic acid sodium acetate The weak acid is there to react with any OH - ions which might be introduced into the system. CH 3 COOH(aq) + OH - (aq) CH 3 COO - (aq) + H 2 O(l) Result - to convert a strong base (OH - ) into a weak base (CH 3 COO - ). Although the ph will increase due to more base being present, the increase will be relatively small as the acetate ion is a weak base (K b = ). CHEM120 Lecture Series One : 2011/16

17 The salt of the weak acid is there to react with any H 3 O + ions which might be introduced into the system. CH 3 COO - (aq) + H 3 O + (aq) CH 3 COOH(aq) + H 2 O(l) Result - a strong acid (H 3 O + ) has been converted into a weak acid (K a (CH 3 COOH) = ). The ph of the solution will decrease but only by a small amount. This buffer solution can be described by the following equilibrium: CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) CHEM120 Lecture Series One : 2011/17

18 EXAMPLE Calculate the ph of a solution which contains mol dm -3 of CH 3 COOH and mol dm -3 of sodium acetate. SOLUTION We first determine the ph using a method described before. CH 3 COOH(aq) H + (aq) + CH 3 COO - (aq) Initial M M Change -x M +x M +x M Equilibrium ( x) M x M ( x) M CHEM120 Lecture Series One : 2011/18

19 x = M = [H + ] ph = -log( ) = 4.83 CHEM120 Lecture Series One : 2011/19

20 EXAMPLE Calculate the ph of the solution which contains mol dm -3 CH 3 CH 2 COO - and mol dm -3 CH 3 CH 2 COOH solution. (pk a CH 3 CH 2 COOH = 4.88). SOLUTION CH 3 CH 2 COOH + H 2 O CH 3 CH 2 COO - + H 3 O + Initial (M) Change (M) -x +x +x Equilibrium ( x) ( x) x CHEM120 Lecture Series One : 2011/20

21 x = mol dm -3 ph = -log( ) = 4.79 CHEM120 Lecture Series One : 2011/21

22 ph > 7.00 Weak base and the salt of a weak base. WEAK BASE + SALT OF A WEAK BASE NH 3 + NH 4 Cl ammonia + ammonium chloride The weak base will take care of any added H 3 O +. NH 3 (aq) + H 3 O + (aq) NH H 2 O(l) The salt of a weak base will react with any added OH - NH OH - (aq) NH 3 (aq) + H 2 O(l) and the strong base gives way to a weak base (NH 3 ). CHEM120 Lecture Series One : 2011/22

23 The equilibria which determines the final ph of the buffer will be NH H 2 O(l) NH 3 (aq) + H 3 O + (aq) If instead we had written the equilibrium as then NH 3 (aq) + H 2 O(l) NH OH - (aq) CHEM120 Lecture Series One : 2011/23

24 EXAMPLE Calculate the ph of a solution prepared from ml of mol NH 3 and ml of mol NH 4 Cl (K a (NH 4+ ) = ). SOLUTION NH H 2 O(l) NH 3 (aq) + H 3 O + (aq) Initial (mol) Change (mol) -x +x +x Equili(mol) ( x) ( x) x Equili(M) ( x) ( x) x CHEM120 Lecture Series One : 2011/24

25 x = mol dm -3 ph = -log( ) = 9.04 EXERCISE FOR THE IDLE MIND Practice Exercise p 575and Exercise15.14, p 603 CHEM120 Lecture Series One : 2011/25

26 What happens to the ph of a buffer solution on adding base or acid? When an acid is added to a buffer solution, it is consumed by the weak base present. When a base is added to a buffer solution, it reacts fully with the weak acid present. EXAMPLE Consider the addition of 0.20 cm 3 of a 0.50 mol dm -3 HNO 3 solution to a buffer solution consisting of 50 cm mol dm -3 aqueous CH 3 CO 2 H (pk a = 4.77) and 50 cm mol dm -3 mol dm -3 Na[CH 3 CO 2 ]. Assume that the addition of HNO 3 causes minimal increase in the total volume and it remains at 100 cm 3. CHEM120 Lecture Series One : 2011/26

27 Before (mol) CH 3 COO - (aq) + H + (aq) CH 3 COOH(aq) After (mol) CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) Initial (mol) Change (mol) -x +x +x Equili(mol) ( x) ( x) x Equili(M) ( x) ( x) x CHEM120 Lecture Series One : 2011/27

28 x = mol dm -3 ph = -log( ) = 4.73 CHEM120 Lecture Series One : 2011/28

29 EXERCISE FOR THE IDLE MIND 1. A buffer is made by adding mol HC 2 H 3 O 2 and mol of NaC 2 H 3 O 2 to enough water to make 1.0 dm 3 of solution (ph of the buffer is 4.74). (a) Calculate ph after mol NaOH is added. (b) Calculate ph after mol HCl is added to original solution. (K a = 1.8 x 10-5 ) 2. Calculate the ph of a buffer solution of 0.50 mol dm -3 HF and 0.45 mol dm -3 F - before and after addition of 0.01 mol NaOH to 1.0 dm 3 of the buffer. (K a of HF = 6.8 x 10-4 ) CHEM120 Lecture Series One : 2011/29

30 MAKING UP A BUFFER SOLUTION OF SPECIFIED ph EXAMPLE (Sample Exercise 15.4 p 575) How many moles of NH 4 Cl must be added to 2.0 dm 3 of 0.1 M NH 3 to form a buffer whose ph is 9.00? (Assume that the addition of NH 4 Cl does not change the volume of the solution: K b = ) SOLUTION NH 3 (aq) + H 2 O(l) NH OH - (aq) poh = = 5.00 [OH - ] = M CHEM120 Lecture Series One : 2011/30

31 Amount of NH 4 Cl must be added = (0.18 mol dm -3 ) (2.0 dm 3 ) = 0.36 mol EXERCISE FOR THE IDLE MIND Practice Exercise p 576 and Exercise15.18, p 603 CHEM120 Lecture Series One : 2011/31

32 ACID-BASE TITRATIONS Titration - one solution of known concentration is used to determine the concentration of another solution. Indicators mark the endpoint of the titration by changing colour. HIn(aq) + H 2 O(l) H 3 O + (aq) + In - (aq) acid colour base colour Various indicators that cover a wide range of ph values. CHEM120 Lecture Series One : 2011/32

33 COLOURS AND APPROXIMATE ph RANGE OF SOME COMMON ACID-BASE INDICATORS ph CHEM120 Lecture Series One : 2011/33

34 TITRATION CURVES Strong Acid/Strong Base Weak Acid/Strong Base CHEM120 Lecture Series One : 2011/34

35 PRECIPITATION REACTIONS SOLUBILITY What happens when we add a solution of a salt to a solution of another salt? If we add the two following salts to each other KNO 3 (aq) + NaCl(aq)? Some students might write KNO 3 (aq) + NaCl(aq) KCl(aq) + NaNO 3 (aq) If we write it out more fully Initial: K + (aq) + NO 3- (aq) + Na + (aq) + Cl - (aq) Final: K + (aq) + Cl - (aq) + Na + (aq) + NO 3- (aq) CHEM120 Lecture Series One : 2011/35

36 A reaction will only happen between salts in solution if ions are removed from solution. One of the ways ions can be removed from solution is through formation of a precipitate. Salts containing Group 1 metal ions, NH 4+, nitrates and acetates are soluble. Chlorides, bromides and iodides are soluble except those of Ag +, Pb 2+ and Hg Sulfates are soluble except those of Pb 2+, Hg 2 2+, Sr 2+ and Ba 2+. Ag 2 SO 4 and CaSO 4 are only slightly soluble. Carbonates, phosphates and sulfites are insoluble (except those of group 1 and NH 4+ ). Sulfides (S 2- ) are insoluble except those of Groups 1 and 2 and NH 4+. CHEM120 Lecture Series One : 2011/36

37 If, therefore, we do the following reaction BaCl 2 (aq) + Na 2 SO 4 (aq)? This time a reaction will take place as when the Ba 2+ ions meet the SO 4 2- ions in solution they will form a precipitate and be removed from solution. BaCl 2 (aq) + Na 2 SO 4 (aq) BaSO 4 (s) + 2NaCl(aq) When will a precipitate form? If Q = [Ba 2+ ][SO 4 2- ] is greater than K sp If Q K sp, precipitation occurs until Q = K sp Q = K sp, equilibrium exists (saturated solution) Q K sp, solid dissolves until Q = K sp CHEM120 Lecture Series One : 2011/37

38 SPARINGLY SOLUBLE SALTS Consider the following reaction: BaSO 4 (s) + (aq) Ba 2+ (aq) + SO 4 2- (aq) The equilibrium constant for this type of equation is known as the solubility product, K sp. K sp = [Ba 2+ ][SO 4 2- ] For BaSO 4, K sp = If the value of K sp is very small (as for BaSO 4 ), then that means the ionic product ([Ba 2+ ] [SO 2-4 ] in this case), is very small which in turn means there are very few ions in solution. Salts which have a K sp value (i.e. those which are not fully soluble in water) are said to be sparingly-soluble. CHEM120 Lecture Series One : 2011/38

39 EXAMPLE (Practice Exercise p 586) A saturated solution of Mg(OH) 2 in contact with undissolved solid is prepared at 25 C. The ph of the solution is found to be Assuming that Mg(OH) 2 dissociates completely in water and that there are no other simultaneous equilibria involving the Mg 2+ or OH - ions in the solution, calculate the K sp for this compound. SOLUTION The equilibrium equation and the expression for K sp are Mg(OH) 2 (s) Mg 2+ (aq) + 2 OH - (aq) K sp = [Mg 2+ ][OH - ] 2 CHEM120 Lecture Series One : 2011/39

40 poh = = 3.83 [OH - ] = M Now [Mg 2+ ]/[OH - ] = ½, [Mg 2+ ] = ½[OH - ] = M K sp = [Mg 2+ ][OH - ] 2 = ( )( ) 2 = CHEM120 Lecture Series One : 2011/40

41 MOLAR SOLUBILITIES The molar solubility of a salt in water is calculated from the K sp value. EXAMPLE Calculate the molar solubility of AgCl at 25 C, given the relevant K sp value is SOLUTION For every 1 mole of AgCl that dissolves, 1 mole of Ag + and 1 mole of Cl - results. Therefore, [AgCl] = [Ag + ] = [Cl - ] If we can calculate the molar concentration of either ion then we will have solved the problem. CHEM120 Lecture Series One : 2011/41

42 K sp = [Ag + ] [Cl - ] = But [Ag + ] = [Cl - ] and therefore, [Ag + ] = ( ) ½ = mol dm -3 [AgCl] = mol dm -3 EXAMPLE (Practice Exercise p 587) The K sp for LaF 3 is What is the solubility of LaF 3 in water in moles per litre. SOLUTION LaF 3 (s) La 3+ (aq) + 3F - (aq) Let [La 3+ ] = x, [F - ] = 3x CHEM120 Lecture Series One : 2011/42

43 K sp = [La 3+ ][F - ] 3 = (x)(3x) 3 = x 4 = x 4 = x = M [LaF 3 ] = M EXERCISE FOR THE IDLE MIND Exercise15.37, p 604 CHEM120 Lecture Series One : 2011/43

44 COMMON-ION EFFECT Consider a case where we have both NaCl and AgCl dissolved in the same body of water. Let s imagine that the dissolution of NaCl, being freely soluble, results in a chloride ion concentration of mol dm -3. How will this affect the solubility of AgCl? K sp = [Ag + ] [Cl - ] = The chloride ion concentration has two inputs the total concentration (0.100 mol dm -3 ) arises from the NaCl plus the Cl - ions from AgCl. Let us set the molar solubility of [AgCl] as S mol dm -3 then CHEM120 Lecture Series One : 2011/44

45 [AgCl] = [Ag + ] = S mol dm -3 [Cl - ] = ( S ) mol dm -3 Now we can substitute these concentrations into the K sp expression. K sp = [Ag + ] [Cl - ] = (S)( S) = (S)(0.100) Since >> S S = S = [Ag + ] = [AgCl] = mol dm -3 CHEM120 Lecture Series One : 2011/45

46 EXAMPLE Calculate the molar solubility of PbI 2 (K sp = ) in a solution containing mol dm -3 iodide ions. SOLUTION PbI 2 (s) Pb 2+ (aq) + 2I - (aq) [PbI 2 ] = [Pb 2+ ] = S mol dm -3 [I - ] = ( S ) mol dm -3 Now we can substitute these concentrations into the K sp expression. K sp = [Pb 2+ ] [I - ] 2 = (S)( S) 2 = (S)(0.214) 2 since >> S CHEM120 Lecture Series One : 2011/46

47 (0.214) 2 S = S = mol dm -3 [Pb 2+ ] = [PbI 2 ] = mol dm -3 THE EFFECT OF ph ON MOLAR SOLUBILITIES The solubility of any substance whose anion is basic is affected by the ph of the solution. Consider the following: Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) If we decrease the ph of the solution, say by adding an acidic buffer, then what will happen to the solubility of Mg(OH 2 )? CHEM120 Lecture Series One : 2011/47

48 By adding an acid we increase [H + ] and consequently decrease [OH - ] as K w must be maintained. If we decrease the amount of OH - (aq) in the system the position of equilibrium will shift to the right. If the position of equilibrium moves to the right then the concentration of Mg 2+ increases. As [Mg(OH) 2 ] = [Mg 2+ ], the molar solubility of magnesium hydroxide increases as the ph is lowered. THE MOLAR SOLUBILITY OF SPARINGLY-SOLUBLE HYDROXIDES INCREASES AS THE ph IS DECREASED. CHEM120 Lecture Series One : 2011/48

49 EXAMPLE Calculate the solubility of Zn(OH) 2 (K sp = ) in (a) pure water (b) a buffer at ph = 7.00 and (c) a buffer at ph = SOLUTION Zn(OH) 2 (s) Zn 2+ (aq) + 2OH - (aq) (a) Let [Zn 2+ ] = x, [OH - ] = 2x [Zn 2+ ] [OH - ] 2 = (x)(2x) 2 = x = [Zn(OH) 2 ] = [Zn 2+ ] = mol dm -3 CHEM120 Lecture Series One : 2011/49

50 (b) At ph = 7.00, [OH - ] = [Zn 2+ ] [ ] 2 = [Zn 2+ ] = mol dm -3 [Zn(OH) 2 ] = [Zn 2+ ] = mol dm -3 (c) At ph = 11.00, [OH - ] = [Zn 2+ ] [ ] 2 = [Zn 2+ ] = mol dm -3 [Zn(OH) 2 ] = [Zn 2+ ] = mol dm -3 CHEM120 Lecture Series One : 2011/50