CHEMISTRY 1012 CHAPTER 17 ADDITIONAL ASPECTS OF AQUEOUS EQUILIBRIA

CHEMISTRY 1012 CHAPTER 17 ADDITIONAL ASPECTS OF AQUEOUS EQUILIBRIA (Please note that you have to attend lectures to complete this set of notes) Dr A Hart 1

THE COMMON ION EFFECT Consider a 0.10 M solution of acetic acid in water, at equilibrium. CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) Add CH 3 COONa (sodium acetate) i.e. Na + (aq) and CH 3 COO - (aq) ions. According to Le Chatelier s principle in which way is the equilibrium expected to shift? 2

Answer CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) There is an increase in [CH 3 COO - ]. Therefore, equilibrium shifts to Le Chatelier s principle. H + ions are used up and The ph increases (less acidic). The common ion is CH 3 COO - as it is in both acetic acid and sodium acetate. Note: Na + ions are spectator ions. according 3

WHAT DOES THIS MEAN? Acetic acid (CH 3 COOH) is a weak acid and ionises incompletely. Sodium acetate (CH 3 COONa) is a strong electrolyte and dissociates completely. They share a common ion CH 3 COO -. Addition of CH 3 COONa solution causes a shift to the This means that the added acetate ions cause the acetic acid to ionise than it normally would. 4

THE COMMON ION EFFECT What we have just seen is called the common ion effect. We define the common ion effect as follows: Whenever a weak electrolyte and a strong electrolyte, in solution together, contain a common ion, the weak electrolyte ionises less than it would if it were alone in solution. The common ion effect has important applications in buffer solutions. 5

EXAMPLE In a solution containing 1.0 M HF (K a = 7.2 x 10-4 ) and 1.0 M NaF, calculate: a) the [H + ]. b) the percentage dissociation of HF. HF is a weak acid: HF(aq) + H 2 O(l) H 3 O + (aq) + F - (aq) conc at start 1.0 M 0 M 0 M NaF is a strong electrolyte: NaF(aq) Na + (aq) + F - (aq) conc 1.0 M 1.0 M 1.0 M In solution we have HF, F - (common ion), Na + (spectator ion) and H 2 O (very weak acid or base). 6

a) HF(aq) + H 2 O(l) H 3 O + (aq) + F - (aq) Initial 1.0 M 0 1.0 M Change -x M +x M +x M Equilibrium K a = [H + ][F - ] [HF] = (x)(1.0 + x) (1.0 x) Because HF is a weak acid x is expected to be very small. Therefore, (1.0 + x) M 1.0 M (1.0 - x) M 1.0 M K a = (x)(1.0) = 7.2 x 10-4 (1.0) x = [H + ] = x M = 7.2 x 10-4 Assumption valid: (1.0 7.2 x 10-4 ) 1.0 M (1.0 + 7.2 x 10-4 ) 1.0 M 7

b) % dissociation of HF = [H 3 O + ] x 100 [HF] = 7.2 x 10-4 x 100 1.0 = 0.072 % NOTE: The [H 3 O + ] in a 1.0 M HF aqueous solution is known to be 2.7 x 10-2 M and % dissociation of HF is 2.7 %. Therefore NaF decreased the dissociation of HF. 8

BUFFER SOLUTIONS These are solutions that contain a conjugate acid-base pair. These solutions can resist drastic changes in ph when small amounts of acid or base are added. Therefore ph is kept In order for proteins and enzymes to function properly the ph must be kept constant. 9

BUFFER CAPACITY Buffer capacity is the amount of acid or base that a buffer can neutralise before an appreciable ph change occurs. A buffer needs large concentrations of a weak acid and its conjugate base (which is also weak). For the best efficiency of the buffer, they should be in The greater the amount of the conjugate acidbase pair, the is the buffer capacity. 10

HOW DOES A BUFFER WORK? Consider the acetate buffer: CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) weak acid conjugate base ADD A STRONG ACID (e.g. HCl) [H 3 O + ] increases. Equilibrium shifts to the Both CH 3 COO - and [H 3 O + ] are used up. ph remains nearly The solution will act as a buffer when acid is added as long as enough CH 3 COO - ions are present. 11

ADD A STRONG BASE (e.g. NaOH) CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) The OH - ions from the added base will react with the H + ions of the acid. [H 3 O + ] decreases. Equilibrium shifts to the Some CH 3 COOH dissociates and more [H 3 O + ] ions are formed. The ph is kept nearly constant. The solution will act as a buffer provided enough is present. Acetic acid by itself cannot act as a buffer. Additional acetate ions (conjugate base of acetic acid) must be added. 12

BLOOD BUFFERS Carbonic acid-bicarbonate buffer H 2 CO 3 /HCO - 3 Phosphate buffer HPO 2-4 /H 2 PO - 4 Protein buffer haemoglobin/oxyhaemoglobin 13

EXAMPLE : UNCONTROLLED DIABETES/KETOSIS Uncontrolled diabetes results in acidosis where the plasma becomes more acidic (ph drops). This affects the following equilibrium: HCO 3- (aq) + H + (aq) H 2 CO 3 (aq) H 2 O(l) + CO 2 (g) base conjugate acid (bicarbonate ion) (carbonic acid) As ph drops (more H + ions), the equilibrium shifts to the right. More H 2 CO 3 is produced. H 2 CO 3 decomposes into H 2 O and CO 2. 14

Excess CO 2 stimulates the brain so the patient breathes faster (to remove CO 2 from the blood). As the level of CO 2 in the blood drops, breathing returns to normal. 15

CALCULATION PROBLEMS Two types of calculation problems: 1. Work out the ph of a given buffer. 2. Prepare a buffer of a certain ph. 16

EXAMPLE We have a solution of 3.0 mol acetic acid and 2.0 mol sodium acetate in 1.0 L of solution. What is the ph of this buffer solution? (K a CH 3 COOH = 1.8 x 10-5 ) CH 3 COONa CH 3 COO - + Na + 2.0 M 2.0 M 2.0 M CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) start 3.0 M 2.0 M 0 M change x M +x M +xm equilibrium 17

K a = [CH 3 COO - ][H 3 O + ] = 1.8 x 10-5 [CH 3 COOH] = (2.0 + x)(x) = 1.8 x 10-5 (3.0 x) Assume x is very small (2.0)(x) = 1.8 x 10-5 (3.0) [x]= [H + ] = ph = -log (2.7 x 10-5 ) = 18

HENDERSON-HASSELBACH EQUATION ph = pk a + log [base] [acid] pk a = - log K a [base] = concentration of the salt If [base] = [acid] then log [base] = log 1 = 0 [acid] This is the optimal ph of any buffer. 19

By choosing appropriate components and adjusting their relative concentrations, one can buffer a solution to almost any ph. The previous example with a 1.0 L solution of 3.0 mol acetic acid and 2.0 mol sodium acetate can be solved using the Henderson-Hasselbach Equation: ph = -log (1.8 x 10-5 ) + log (2.0 M) = 4.54 (3.0M) 20

A WEAK BASE/CONJUGATE ACID BUFFER SYSTEM If the buffer system is a weak base/conjugate acid, the Henderson-Hasselbach equation is: poh = pk b + log[acid] [base] [acid] = concentration of the salt pk b = - log K b 21

EXAMPLE Calculate the amount of sodium nitrite that must be added to 250 ml of a 0.50 M solution of nitrous acid to produce a buffer solution with ph 3.50. K a (HNO 2 )= 4.5 x 10-4 In solution we have: NaNO 2 Na + + NO 2 - strong electrolyte HNO 2 + H 2 O H 3 O + + NO 2 - weak acid ph = 3.50 [H + ] = 10-3.50 = 22

HNO 2 + H 2 O H 3 O + + NO 2 - start 0.50 M 0 M x M from NaNO 2 change -3.16 x 10-4 M +3.16 x 10-4 M +3.16 x 10-4 M equilibrium 3.16 x 10-4 M [H 3 O + ][NO 2- ] = K a [NO 2- ] = K a [HNO 2 ] [HNO 2 ] [H 3 O + ] x + 3.16 x 10-4 = (4.5 x 10-4 )(0.50-3.16 x 10-4 ) (3.16 x 10-4 ) x = 0.7116 3.16 x 10-4 x = [ NO 2- ] = [NaNO 2 ] = nnano 2 = cv = 0.71 M x 0.250 L = Therefore of NaNO 2 must be added. 23

If you were asked for the mass of NaNO 2 needed to produce a ph 3.50 in a 0.5 M solution of nitrous acid: m (NaNO 2 ) = nm = 0.178 mol x 69.0 g mol -1 = 24

EXAMPLE a) Why is a 1.2 M solution of formic acid, HCOOH, not a buffer solution? b) What would the possible buffer system be? a) weak acid conjugate base HCOOH + H 2 O H 3 O + + HCOO - start 1.2 M 0 M 0 M change - x M + x M + x M equilibrium x M Because x is small, not enough available to sustain a buffering system. is 25

b) What would the possible buffer system be? Unless there is an extra source of ions, we cannot claim to have a buffer. One needs more conjugate base in the form of a salt. Adding (sodium formate ) to the formic acid would create a buffer system. 26

EXAMPLE If we mix 100 ml 1 M HCl and 200 ml 1 M NH 3 do we have a buffer solution? n HCl = cv = 0.1 x 1 = 0.1 mol n NH 3 = cv = 0.2 x 1 = 0.2 mol from weak base from strong acid HCl + NH 3 NH 4 Cl NH 4 + + Cl - start 0.1 mol 0.2 mol 0 mol 0 mol end 0 0.1 mol 27

weak acid conjugate base NH 4 + + H 2 O NH 3 + H 3 O + unreacted start 0.1 mol 0.1 mol 0 mol change - x mol + x mol + x mol equilibrium We can assume x is small because NH 4 + is a Therefore there are nearly equal amounts of the weak acid (NH 4+ ) and the conjugate base (NH 3 ). Therefore we have a A buffer solution can be created by the reaction of a strong acid (HCl) with excess weak base (NH 3 ) to form the conjugate acid (NH 4+ ) of that base. 28

EXAMPLE A buffer solution contains 0.50 M sodium lactate (C 3 H 5 O 3 Na) and 0.50 M lactic acid (HC 3 H 5 O 3 ). a) What is the ph of the buffer solution? b) What is the ph after adding 0.10 mol HCl to one litre of buffer? c) What is the ph after adding 0.10 mol KOH to one litre of buffer? (K a lactic acid = 1.4 x 10-4 ) 29

a) weak acid conjugate base (added) HC 3 H 5 O 3 (aq) + H 2 O C 3 H 5 O 3 - (aq) + H 3 O + (aq) start 0.50 M 0.50 M 0 M change - x M + x M + x M equilibrium K a = [C 3 H 5 O - 3 ][H 3 O + ] = 1.4 x 10-4 [HC 3 H 5 O 3 ] (assume x <<< 0.50 M) (0.50)x = 1.4 x 10-4 (0.50) [H + ] = x = 1.4 x 10-4 M ph = -log(1.4 x 10-4 ) = 30

b) HCl is a strong acid, therefore we are effectively adding 0.10 mol of H + (and Cl - ) to 1 litre of buffer. The H + reacts with C 3 H 5 O - 3 (the conjugate base). C 3 H 5 O - 3 (aq) + H + (aq) HC 3 H 5 O 3 (aq) start 0.50 M 0.10 M 0.50 M end 0.40 M 0 M 0.60 M Recalculate the ph of the buffer using new concentrations: HC 3 H 5 O 3 (aq) + H 2 O C 3 H 5 O 3 - (aq) + H 3 O + (aq) start 0.60 M 0.40 M 0 M change - x M + x M + x M equilibrium 31

K a = [C 3 H 5 O - 3 ][H 3 O + ] = 1.4 x 10-4 [HC 3 H 5 O 3 ] (assume x is very small) (0.40)x = 1.4 x 10-4 (0.60) [H + ] = x = 2.1 x 10-4 M ph = -log(2.1 x 10-4 ) = Therefore there is only a small change in ph ( ). 32

c) KOH is a strong base, therefore we are effectively adding The OH - reacts with HC 3 H 5 O 3. HC 3 H 5 O 3 (aq) + OH - (aq) C 3 H 5 O - 3 (aq) + H 2 O(l) start 0.50 M 0.10 M 0.50 M end 0.40 M 0 M 0.60 M Recalculate the ph of the buffer using new concentrations: HC 3 H 5 O 3 (aq) + H 2 O C 3 H 5 O 3 - (aq) + H 3 O + (aq) start 0.40 M 0.60 M 0 M change - x M + x M + x M equilibrium 33

K a = [C 3 H 5 O 3 - ][H 3 O + ] = 1.4 x 10-4 [HC 3 H 5 O 3 ] (assume x is very small) (0.60)x = 1.4 x 10-4 (0.40) [H + ] = x = 9.3 x 10-5 M ph = -log(9.3 x 10-5 ) = Therefore there is only a small change in ph ( ). 34

TITRATION A base of known concentration is slowly added to a solution of an acid or an acid of known concentration is slowly added to a solution of a base. 35

A ph meter or indicators are used to determine when the solution has reached the, the point at which the stoichiometric amount of acid equals that of base. 36

TITRATION OF A STRONG ACID WITH A STRONG BASE A plot of ph versus volume of acid (or base) added is called a titration curve. Consider adding a strong base (e.g. NaOH) to a solution of a strong acid (e.g. HCl). Before any base is added, the ph is given by the strong acid solution. Therefore, the ph is When base is added, before the equivalence point, the ph is given by the amount of strong acid in excess. Therefore, the ph is less than 7. 37

From the start of the titration to near the equivalence point, the ph goes up slowly. 38

Just before and after the equivalence point, the ph increases rapidly. 39

At the equivalence point, an equal number of moles of HCl and NaOH have reacted and the solution contains only their salt, NaCl and water. At the equivalence point the ph = 7 (neutral). 40

As more base is added, the increase in ph again levels off. The ph after the equivalence point is determined by the amount of excess base. HCl 41

TITRATION OF A WEAK ACID WITH A STRONG BASE Unlike in the previous case, the conjugate base of the acid affects the ph when it is formed. The ph at the equivalence point will be Phenolphthalein is commonly used as an indicator in these titrations. 42

CH 3 COOH CH 3 COOH + CH 3 COONa CH 3 COONa CH 3 COONa + NaOH 43

Before the end point, every mole of strong base added reacts completely with the equivalent amount of weak acid, forming a solution that contains the salt of the weak acid and a progressively diminishing amount of weak acid. At the equivalence point the solution is a solution of the salt of the After the equivalence point the solution contains the salt of the weak acid as well as a progressively increasing amount of hydroxide ions. 44

TITRATION OF A WEAK BASE WITH A STRONG ACID The ph at the equivalence point in these titrations is less than 7. Methyl red is the indicator of choice. weak base (weak base) strong base 45

Before the end point, every mole of strong acid added reacts completely with the equivalent amount of weak base, forming a solution that contains the salt of the weak base and a progressively diminishing amount of weak base. At the equivalence point the solution is a solution of the salt of the After the equivalence point the solution contains the salt of the weak base as well as a progressively increasing amount of hydrogen ions. 46

EFFECT OF K a ON TITRATION CURVES With weaker acids, the initial ph is higher and ph changes near the equivalence point are more subtle. 47

TITRATIONS OF POLYPROTIC ACIDS In these cases there is an equivalence point for each dissociation. 48

CALCULATIONS RELATING TO ACID-BASE TITRATIONS Calculate the ph at the equivalence point of a titration in which 20.00 cm 3 of 0.10 M acetic acid is titrated with 0.080 M sodium hydroxide. (K a for acetic acid = 1.8 10-5 ) The chemical reaction involved in the titration is: CH 3 COOH + NaOH CH 3 COONa + H 2 O weak acid strong base goes to completion mole ratio 1 1 1 1 49

We first need to calculate the total volume at the end of the titration: n acetic acid initially = cv = (20 x 10-3 ) dm 3 0.10 mol dm -3 = 1 mol of CH 3 COOH reacts with 1 mol of NaOH n NaOH required to reach the equiv. pt. = 0.0020 mol nch 3 COONa formed = 0.0020 mol V NaOH reqd. to reach the equivalence point: c = n V = n = 0.0020 mol V c 0.080 mol dm -3 = 0.025 dm 3 = V total at end of titration= 20 + 25 = 50

At the equivalence point the solution obtained will consist of 0.0020 mol of sodium acetate and will have a volume of 45 cm 3. [CH 3 COONa] = 0.0020 mol = 0.044 M. 45 x 10-3 dm 3 At the equivalence point we have a 0.044 M solution of sodium acetate. CH 3 COONa(aq) CH 3 COO - (aq) + Na + (aq) 0.044 M 0.044 M 0.044 M CH 3 COO - is the conjugate base of a weak acid, CH 3 COOH, therefore it gives a basic solution. 51

CH 3 COO - (aq) + H 2 O(l) CH 3 COOH(aq) + OH - (aq) start 0.0444 M 0 M 0 M change -x +x +x equilibrium (0.044 x) M x M x M Note: We are given K a for CH 3 COOH, we need to calculate K b for CH 3 COO - using K a x K b = K w. K b = 1.0 x 10-14 = 5.6 x 10-10 1.8 x 10-5 K b = [CH 3 COOH][OH - ] = x 2 = 5.6 x 10-10 [CH 3 COO - ] (0.044 x) 52

Assume x to be small (0.044 - x) M 0.044 M x = [OH - ] = 0.044 x 5.6 x 10-10 x = [OH - ] = Test assumption: 0.044 (5.0 x 10-6 ) = 0.044 valid. We were not asked for [OH - ], we were asked for the ph. Therefore, we use the ion product of water: [H 3 O + ][OH - ] = 1.0 x 10-14 [H 3 O + ] = 1.0 x 10-14 = 5.0 x 10-6 ph = -log(2.0 x 10-9 ) = 53

ANOTHER CALCULATION Calculate the ph of a solution formed when 25 cm 3 of 0.25 M NaOH is added to 50 cm 3 of 0.20 M acetic acid. K a for acetic acid = 1.8 10-5. The NaOH added will react with acetic acid (CH 3 COOH) to form the appropriate amount of the salt sodium acetate (CH 3 COONa) in solution. CH 3 COOH + NaOH CH 3 COONa + H 2 O 54

nch 3 COOH = cv = 0.20 M x 50 x 10-3 L = 0.0100 mol nnaoh = cv = 0.25 M x 25 x 10-3 L = 0.00625 mol CH 3 COOH + NaOH CH 3 COONa + H 2 O Mol start 0.0100 mol 0.00625 mol 0 mol Mol end 0.0100-0.00625 = 0 mol Conc end 0.00375 75 x 10-3 = 0 M 0.00625 75 x 10-3 = Note: For greater accuracy we do not round up to the correct number of significant figures in the middle of this calculation. 55

We see that after reaction between the NaOH and the acetic acid, we end up with a solution that contains appreciable concentrations of both a weak acid and its conjugate base. This is a buffer solution. Its ph can be calculated by use of the Henderson-Hasselbach equation: ph = pk a + log[base] [acid] = -log1.8 10-5 + log 0.0833 0.0500 = 4.74 + 0.22 = 56

ACID-BASE INDICATORS The acid-base indicators are themselves weak organic acids (HIn). Like any other weak acid, the indicator dissociates as follows: HIn(aq) + H 2 O(l) In - (aq)+ H 3 O + (aq) red (acidic) blue (basic) They exhibit a certain colour when the proton is attached and a different one when it is absent. Example: Phenolphthalein is colourless in its HIn form and pink in its In - form. 57

THE ph RANGES OF COLOUR CHANGE OF SOME ACID-BASE INDICATORS 58

K a OF ACID-BASE INDICATORS HIn(aq) + H 2 O(l) In - (aq)+ H 3 O + (aq) Each indicator HIn will have its own acid dissociation constant K a. K a = [H 3 O + ][In - ] [HIn] ph = pk a + log [In - ] [HIn] 59

ph AT THE MIDPOINT OF AN INDICATOR S COLOUR TRANSITION RANGE When an indicator is at the midpoint of its colour transition range, then: [HIn] = [In] log [In - ] = 0 [HIn] The midpoint of an indicator s colour transition range occurs when the ph of the solution is equal to the pk a of the indicator. 60

SOLUBILITY EQUILIBRIA Up to now we have dealt with homogeneous acid-base equilibria (all species in the same phase). We will now consider equilibria which involve the dissolution or precipitation of ionic compounds. The compounds are usually slightly soluble. 61

THE SOLUBILITY OF COMMON SALTS Before: Used solubility rules to predict whether a substance was going to be soluble in water or whether a precipitate would form. Now: Use solubility equilibria to predict quantitatively the amount of a given compound that will or will not dissolve. 62

THE SOLUBILITY PRODUCT (K sp ) Consider the following equilibrium: CaF 2 (s) Ca 2+ (aq) + 2F - (aq) Since this equilibrium describes the dissolution of a solid, we call the equilibrium constant the solubility product or K sp. K sp = 63

WHAT IS K sp? K sp is the product of the molar concentrations of the ions involved in the equilibrium, each raised to the power of its coefficient in the equilibrium reaction. Molar concentrations of the ions are in mol L -1. 64

SOLUBILITY OF SLIGHTLY SOLUBLE SALTS For slightly soluble salts it is more usual to describe their solubility using K sp. Examples: AgCl(s) Ag + (aq)+ Cl - (aq) K sp = [Ag + ][Cl - ] = 1.8 x 10-10 Ca 3 (PO 4 ) 2 (s) 3Ca 2+ (aq) + 2PO 4 3- (aq) K sp = [Ca 2+ ] 3 [PO 4 3- ] 2 = 2.0 x 10-29 (often implicated in kidney stones) A small K sp value means solubility. 65

DIFFERENCE BETWEEN SOLUBILITY AND SOLUBILITY PRODUCT Solubility of a substance is the quantity of a substance that dissolves to form a saturated solution. Often expressed as g L -1 or mol L -1 The solubility product (K sp ) is an equilibrium constant of the equilibrium between an ionic solid and its saturated solution. The solubility of a substance can change as concentrations of other solutes change. However, K sp has only one value for a given solute at any specific temperature. 66

LIMITATIONS TO SOLUBILITY PRODUCTS There are limitations to solubility products because: we assume that ionic compounds fully dissociate when they dissolve and this is not always true. we tend to ignore other equilibria that may occur simultaneously in solution e.g. acidbase equilibria. 67

EXAMPLE At 25 C, 0.00188 g of silver chloride dissolve in water to give 1.00 L of a saturated solution. What is the K sp? n AgCl = m = 0.0188 g = M 143.4 gmol -1 AgCl(s) Ag + (aq)+ Cl - (aq) Mole ratio 1 : 1 : 1 [Ag + ] = [Cl - ] = n = 1.31 x 10-5 mol = V 1.00 L K sp = [Ag + ][Cl - ] = (1.31 x 10-5 mol L -1 ) 2 = 68

EXAMPLE The K sp for CaF 2 is 3.9 x 10-11 at 25 C. What are the concentrations of Ca 2+ and F - ions in a saturated solution of CaF 2? CaF 2 (s) Ca 2+ (aq) + 2F - (aq) x M x M 2x M K = [Ca 2+ ][F - ] 2 = 3.9 x 10-11 (x)(2x) 2 = 3.9 x 10-11 4x 3 = 3.9 x 10-11 x = [Ca 2+ ] = [F - ] = = [CaF 2 ] in solution is 69

PRECIPITATION AND SOLUBILITY PRODUCT Given the concentration of substances in a reaction mixture, will a precipitate form? We need to compare the values. Ion product Q = [initial ion] x [initial ion] y and the K sp The ion product expression has the form of the K sp expression each concentration is raised to the power of the number of ions in the ionic formula. However, the concentrations are the initial concentrations not the equilibrium concentrations like in K sp. Example: For Mg(OH) 2 Q = 70

precipitation occurs until Q = K sp 71 WILL A PRECIPITATE FORM? Q < K sp the solution is unsaturated, therefore no precipitation occurs. Solid dissolves until Q = K sp Q = K sp the solution is saturated. Equilibrium exists. Therefore, no precipitation occurs. Q > K sp the solution is supersaturated and

EXAMPLE Will a precipitate of lead(ii) chloride form when you add a solution of Pb(NO 3 ) 2 and NaCl to water to give a solution with 0.050 M Pb 2+ and 0.10 M Cl -? (K sp (PbCl 2 ) = 1.6 x 10-5 ) Pb(NO 3 ) 2 (aq) + 2NaCl(aq) PbCl 2 (s) Pb 2+ (aq) + 2Cl - (aq) PbCl 2 (s) + 2NaNO 3 (aq) Q = [Pb 2+ ][Cl - ] 2 = (0.050)(0.1) 2 = Because precipitation will occur until Q = K sp and equilibrium is reached. 72

FACTORS AFFECTING SOLUBILITY The solubility of a substance is affected by: Temperature Presence of common ions * ph of the solution ** Presence of complexing agents 73

FACTORS OF INTEREST * The common ion effect In general the solubility of a slightly soluble salt decreases in the presence of a common ion. ** ph of a solution The solubility of slightly soluble salts containing basic anions increases as [H + ] increases i.e. as the ph is lowered. 74

COMMON ION EXAMPLE The K sp for CaF 2 is 3.9 x 10-11 at 25 C. What are the concentrations of Ca 2+ and F - ions if we dissolve CaF 2 in a 1.00 M solution of KF? We expect less CaF 2 to dissolve as predicted by Le Chatelier s principle because there are common F - ions. CaF 2 (s) Ca 2+ (aq) + 2F - (aq) KF K + + F - x M x M 2x M 1.00 M 1.00 M Total [F - ] = (2x + 1.00) M 75

K = [Ca 2+ ][F - ] 2 = 3.9 x 10-11 x <<< 1.00 M (x)(2x + 1.00) 2 = 3.9 x 10-11 x(1.00) 2 = 3.9 x 10-11 x = [Ca 2+ ] = [F - ] = 2(3.9 x 10-11 M) = From dissolved solid Therefore: [CaF 2 ] in solution (in KF solution) = 3.9 x 10-11 M (less) [CaF 2 ] in solution (in pure water) = 2.1 x 10-4 M 76

ph OF SOLUTION EXAMPLE Calculate the molar solubility of Fe(OH) 2 when buffered at ph 7.0. (K sp Fe(OH) 2 = 7.9 x 10-16 ) Fe(OH) 2 (s) Fe 2+ (aq) + 2OH - (aq) The [OH - ] is set by the ph of the solution ph = 7.0 K sp = [Fe 2+ ][OH - ] 2 poh = 7.0 and [OH - ] = 1.0 x 10-7 M [Fe 2+ ] = K sp = 7.9 x 10-16 = [OH - ] 2 (1.0 x 10-7 ) 2 [Fe(OH) 2 ] = [Fe 2+ ] = 77

PREVIOUS CALCULATION AT ph 10.0 ph = 10.0 poh = 4.0 and [OH - ] = 1.0 x 10-4 M K sp = [Fe 2+ ][OH - ] 2 [Fe 2+ ] = K sp [OH - ] 2 [Fe(OH) 2 ] = [Fe 2+ ] = The solubility of Fe(OH) 2 decreased, from 7.9 x 10-2 M to 7.9 x 10-8 M, as the ph 78

PRACTICAL EXAMPLES OF ph AFFECTING SOLUBILITY TOOTH DECAY AND FLUORIDATION Tooth enamel consists of mainly a mineral called hydroxyapatite, Ca 10 (PO 4 ) 6 (OH) 2. Tooth cavities are caused when acids, from the action of bacteria on mainly sugars, dissolve tooth enamel. Ca 10 (PO 4 ) 6 (OH) 2 (s) + 8H + (aq) 10Ca 2+ (aq) + 6HPO 4 2- (aq) + 2H 2 O(l) 79

Fluoride ion, present in drinking water, toothpaste and other sources, can react with hydroxyapatite to form fluoroapatite, Ca 10 (PO 4 ) 6 F 2. This mineral, in which F - has replaced OH -, is much more resistant to attack by acids because it is less soluble in acid than hydroxyapatite. Tooth decay is reduced. 80

ANTACIDS LIKE MILK OF MAGNESIA (Mg(OH) 2 ) Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) Mg(OH) 2 is insoluble in water and milk of magnesia is a white suspension of solid Mg(OH) 2. Mg(OH) 2 dissolves in the stomach because: the H + ions in the stomach react with OH - ions removing them from solution. the equilibrium shifts to the right increasing the solubility of Mg(OH) 2. The increase in [OH - ] combats excess acidity in the stomach. 81

FORMATION OF LIMESTONE CAVES The formation of the huge limestone caves is a result of increased solubility of carbonates in an acidic solution. Carbon dioxide dissolved in ground water makes it acidic and the solubility of CaCO 3 increases, eventually producing huge caverns. As the CO 2 escapes to the air, the ph of the dripping water increases. The solubility of calcium carbonate decreases, resulting in it being deposited, forming stalagmites and stalactites. END OF CHAPTER 82