Ionic Equilibria in Aqueous Systems. Dr.ssa Rossana Galassi

Transcription

1 Ionic Equilibria in Aqueous Systems Dr.ssa Rossana Galassi

2 Ionic Equilibria in Aqueous Systems 19.1 Equilibria of Acid-Base Buffer Systems 19.2 Acid-Base Titration Curves 19.3 Equilibria of Slightly Soluble Ionic Compounds 19.4 Equilibria Involving Complex Ions 19.5 Application of Ionic Equilibria to Chemical Analysis

3 Figure 19.1 The effect of addition of acid or base to acid added base added Figure 19.2 an unbuffered solution acid added base added or a buffered solution

4 Table 19.1 The Effect of Added Acetate Ion on the Dissociation of Acetic Acid [CH 3 COOH] initial [CH 3 COO - ] added % Dissociation* ph * % Dissociation = [CH 3COOH] dissoc [CH 3 COOH] initial x 100

5 Figure 19.3 How a buffer works. Buffer after addition of H 3 O + Buffer with equal concentrations of conjugate base and acid Buffer after addition of OH - H 3 O + OH - H 2 O + CH 3 COOH H 3 O + + CH 3 COO - CH 3 COOH + OH - H 2 O + CH 3 COO -

6 Sample Problem 19.1 Calculating the Effect of Added H 3 O + or OH - PROBLEM: Calculate the ph: on Buffer ph (a) of a buffer solution consisting of 0.50M CH 3 COOH and 0.50M CH 3 COONa (b) after adding 0.020mol of solid NaOH to 1.0L of the buffer solution in part (a) (c) after adding 0.020mol of HCl to 1.0L of the buffer solution in part (a) K a of CH 3 COOH = 1.8x10-5. (Assume the additions cause negligible volume changes. PLAN: We know K a and can find initial concentrations of conjugate acid and base. Make assumptions about the amount of acid dissociating relative to its initial concentration. Proceed step-wise through changes in the system. SOLUTION: Concentration (M) (a) CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) Initial Change Equilibrium x 0.50-x x + x x x

7 Sample Problem 19.1 Calculating the Effect of Added H 3 O + and OH - continued (2 of 4) on Buffer ph [H 3 O + ] = x [CH 3 COOH] equil 0.50M [CH 3 COO - ] initial 0.50M K a = [H 3O + ][CH 3 COO - ] [CH 3 COOH] [CH [H 3 O + 3 COOH] ] = x = K a [CH 3 COO - ] Check the assumption: 1.8x10-5 /0.50 X 100 = 3.6x10-3 % = 1.8x10-5 M ph = 4.74 (b) [OH - ] added = mol 1.0L soln = 0.020M NaOH Concentration (M) CH 3 COOH(aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l) Before addition Addition After addition

8 Sample Problem 19.1 Calculating the Effect of Added H 3 O + and OH - on Buffer ph continued (3 of 4) Set up a reaction table with the new values. Concentration (M) Initial Change Equilibrium CH 3 COOH(aq) + H 2 O(l) x x - CH 3 COO - (aq) + H 3 O + (aq) x + x x x 0.48 [H 3 O + ] = 1.8x = 1.7x10-5 ph = 4.77 (c) [H 3 O + ] added = mol 1.0L soln = 0.020M H 3 O + Concentration (M) CH 3 COO - (aq) + H 3 O + (aq) CH 3 COOH(aq) + H 2 O (l) Before addition Addition After addition

9 Sample Problem 19.1 Calculating the Effect of Added H 3 O + and OH - on Buffer ph continued (4 of 4) Set up a reaction table with the new values. Concentration (M) Initial Change Equilibrium CH 3 COOH(aq) + H 2 O(l) x x - CH 3 COO - (aq) + H 3 O + (aq) x + x x x [H 3 O + ] = 1.8x = 2.0x10-5 ph = 4.70

10 The Henderson-Hasselbalch Equation HA + H 2 O H 3 O + + A - K a = [H 3 O + ] [A - ] [HA] K [H 3 O + ] = a [HA] [A - ] - log[h 3 O + ] = - log K a + log [A - ] [HA] ph = pk a + log [base] [acid]

11 Buffer Capacity and Buffer Range Buffer capacity is the ability to resist ph change. The more concentrated the components of a buffer, the greater the buffer capacity. The ph of a buffer is distinct from its buffer capacity. A buffer has the highest capacity when the component concentrations are equal. Buffer range is the ph range over which the buffer acts effectively. Buffers have a usable range within ± 1 ph unit of the pk a of its acid component.

12 The relation between buffer capacity and ph change.

13 Preparing a Buffer 1. Choose the conjugate acid-base pair. 2. Calculate the ratio of buffer component concentrations. 3. Determine the buffer concentration. 4. Mix the solution and adjust the ph.

14 Sample Problem Preparing a Buffer PROBLEM: An environmental chemist needs a carbonate buffer of ph to study the effects of the acid rain on limsetone-rich soils. How many grams of Na 2 CO 3 must she add to 1.5L of freshly prepared 0.20M NaHCO 3 to make the buffer? K a of HCO 3- is 4.7x PLAN: SOLUTION: We know the K a and the conjugate acid-base pair. Convert ph to [H 3 O + ], find the number of moles of carbonate and convert to mass. HCO 3- (aq) + H 2 O(l) CO 3 2- (aq) + H 3 O + (aq) K a = [CO 2-3 ][1.0x10-10 ] ph = 10.00; [H 3 O + ] = 1.0x x10-11 = (0.20) moles of Na 2 CO 3 = (1.5L)(0.094mols/L) = 0.14 [CO 3 2- ][H 3 O + ] [HCO 3- ] [CO 3 2- ] = 0.094M 0.14 moles g mol = 15 g Na 2 CO 3

15 Colors and approximate ph range of some common acid-base indicators. ph

16 Figure 19.6 The color change of the indicator bromthymol blue. basic acidic change occurs over ~2pH units

17 Curve for a strong acid-strong base titration

18 Curve for a weak acidstrong base titration Titration of 40.00mL of M HPr with M NaOH pk a of HPr = 4.89 ph = 8.80 at equivalence point [HPr] = [Pr - ] methyl red

19 Sample Problem Calculating the ph During a Weak Acid- Strong Base Titration PROBLEM: Calculate the ph during the titration of ml of M propanoic acid (HPr; K a = 1.3x10-5 ) after adding the following volumes of M NaOH: (a) 0.00mL (b) 30.00mL (c) 40.00mL (d) 50.00mL PLAN: The amounts of HPr and Pr - will be changing during the titration. Remember to adjust the total volume of solution after each addition. SOLUTION: (a) Find the starting ph using the methods of Chapter 18. K a = [Pr - ][H 3 O + ]/[HPr] [Pr - ] = x = [H 3 O + ] [Pr - ] = x = [H 3 O + ] x (1.3x10 5 )(0.10) x = 1.1x10-3 ; ph = 2.96 (b) Amount (mol) Before addition Addition After addition HPr(aq) + OH - (aq) Pr - (aq) + H 2 O (l)

20 Sample Problem 19.3 continued Calculating the ph During a Weak Acid- Strong Base Titration [H 3 O + ] = 1.3x mol mol = 4.3x10-6 M ph = 5.37 (c) When 40.00mL of NaOH are added, all of the HPr will be reacted and the [Pr - ] will be ( mol) = M ( L) + ( L) K a x K b = K w K b = K w /K a = 1.0x10-14 /1.3x10-5 = 7.7x10-10 [H 3 O + ] = K w / = 1.6x10-9 M K b x[pr ] ph = 8.80 (d) 50.00mL of NaOH will produce an excess of OH -. mol XS base = (0.1000M)( L L) = mol [H 3 O + ] = 1.0x10-14 / = 9.0x10-11 M M = ( ) (0.0900L) M = ph = 12.05

21 Figure 19.9 Titration of 40.00mL of M NH 3 with M HCl pk a of NH 4+ = 9.25 Curve for a weak basestrong acid titration ph = 5.27 at equivalence point

22 Curve for the titration of a weak polyprotic acid. pk a = 7.19 pk a = 1.85 Titration of 40.00mL of M H 2 SO 3 with M NaOH

23 Ion-Product Expression (Q sp ) and Solubility Product Constant (K sp ) For the hypothetical compound, M p X q At equilibrium Q sp = [M n+ ] p [X z- ] q = K sp

24 Sample Problem 19.4 Writing Ion-Product Expressions for Slightly Soluble Ionic Compounds PROBLEM: Write the ion-product expression for each of the following: PLAN: SOLUTION: (a) Magnesium carbonate (c) Calcium phosphate (b) Iron (II) hydroxide (d) Silver sulfide Write an equation which describes a saturated solution. Take note of the sulfide ion produced in part (d). (a) MgCO 3 (s) Mg 2+ (aq) + CO 3 2- (aq) K sp = [Mg 2+ ][CO 3 2- ] (b) Fe(OH) 2 (s) (c) Ca 3 (PO 4 ) 2 (s) (d) Ag 2 S(s) Fe 2+ (aq) + 2OH - (aq) 3Ca 2+ (aq) + 2PO 4 3- (aq) 2Ag + (aq) + S 2- (aq) K sp = [Fe 2+ ][OH - ] 2 K sp = [Ca 2+ ] 3 [PO 4 3- ] 2 S 2- (aq) + H 2 O(l) HS - (aq) + OH - (aq) K sp = [Ag + ] 2 [HS - ][OH - ] Ag 2 S(s) + H 2 O(l) 2Ag + (aq) + HS - (aq) + OH - (aq)

25 Table 19.2 Solubility-Product Constants (K sp ) of Selected Ionic Compounds at 25 0 C Name, Formula Aluminum hydroxide, Al(OH) 3 Cobalt (II) carbonate, CoCO 3 Iron (II) hydroxide, Fe(OH) 2 Lead (II) fluoride, PbF 2 Lead (II) sulfate, PbSO 4 Mercury (I) iodide, Hg 2 I 2 Silver sulfide, Ag 2 S Zinc iodate, Zn(IO 3 ) 2 K sp 3 x x x x x x x x 10-6

26 Sample Problem 19.5 Determining K sp from Solubility PROBLEM: (a) Lead (II) sulfate is a key component in lead-acid car batteries. Its solubility in water at 25 0 C is 4.25x10-3 g/100ml solution. What is the K sp of PbSO 4? (b) When lead (II) fluoride (PbF 2 ) is shaken with pure water at 25 0 C, the solubility is found to be 0.64g/L. Calculate the K sp of PbF 2. PLAN: Write the dissolution equation; find moles of dissociated ions; convert solubility to M and substitute values into solubility product constant expression. SOLUTION: (a) PbSO 4 (s) Pb 2+ (aq) + SO 2-4 (aq) K sp = [Pb 2+ ][SO 4 2- ] 4.25x10-3 g 1000mL mol PbSO 4 = 1.40x10-4 M PbSO 4 100mL soln L 303.3g PbSO 4 K sp = [Pb 2+ ][SO 2-4 ] = (1.40x10-4 ) 2 = 1.96x10-8

27 Sample Problem 19.5 Determining K sp from Solubility continued (b) PbF 2 (s) Pb 2+ (aq) + 2F - (aq) K sp = [Pb 2+ ][F - ] g mol PbF 2 L soln 245.2g PbF 2 = 2.6x10-3 M K sp = (2.6x10-3 )(5.2x10-3 ) 2 = 7.0x10-8

28 Sample Problem 19.6 Determining Solubility from K sp PROBLEM: Calcium hydroxide (slaked lime) is a major component of mortar, plaster, and cement, and solutions of Ca(OH) 2 are used in industry as a cheap, strong base. Calculate the solubility of Ca(OH) 2 in water if the K sp is 6.5x10-6. PLAN: Write out a dissociation equation and K sp expression; Find the molar solubility (S) using a table. SOLUTION: Ca(OH) 2 (s) Ca 2+ (aq) + 2OH - (aq) K sp = [Ca 2+ ][OH - ] 2 Concentration (M) Initial Change Equilibrium Ca(OH) 2 (s) Ca 2+ (aq) + 2OH - (aq) 0 0 +S + 2S S 2S K sp = (S)(2S) 2 S = 6.5x = 1.2x10x -2 M

29 Table 19.3 Relationship Between K sp and Solubility at 25 0 C No. of Ions Formula Cation:Anion K sp Solubility (M) 2 MgCO 3 1:1 3.5 x x PbSO 4 1:1 1.6 x x BaCrO 4 1:1 2.1 x x Ca(OH) 2 1:2 5.5 x x BaF 2 1:2 1.5 x x CaF 2 1:2 3.2 x x Ag 2 CrO 4 2:1 2.6 x x 10-5

30 The effect of a common ion on solubility CrO 4 2- added PbCrO 4 (s) Pb 2+ (aq) + CrO 4 2- (aq) PbCrO 4 (s) Pb 2+ (aq) + CrO 4 2- (aq)

31 Sample Problem Calculating the Effect of a Common Ion on Solubility PROBLEM: In Sample Problem 19.6, we calculated the solubility of Ca(OH) 2 in water. What is its solubility in 0.10M Ca(NO 3 ) 2? K sp of Ca(OH) 2 is 6.5x10-6. PLAN: Set up a reaction equation and table for the dissolution of Ca(OH) 2. The Ca(NO 3 ) 2 will supply extra [Ca 2+ ] and will relate to the molar solubility of the ions involved. SOLUTION: Concentration(M) Ca(OH) 2 (s) Ca 2+ (aq) + 2OH - (aq) Initial Change S +2S Equilibrium S 2S K sp = 6.5x10-6 = ( S)(2S) 2 = (0.10)(2S) 2 S << 0.10 S = 6.5x10 6 Check the assumption: 4 = 4.0x x10-3 x 100 = 4.0% 0.10M

32 Test for the presence of a carbonate.

33 Sample Problem 19.8 Predicting the Effect on Solubility of Adding Strong Acid PROBLEM: Write balanced equations to explain whether addition of H 3 O + from a strong acid affects the solubility of these ionic compounds: (a) Lead (II) bromide (b) Copper (II) hydroxide (c) Iron (II) sulfide PLAN: Write dissolution equations and consider how strong acid would affect the anion component. SOLUTION: (a) PbBr 2 (s) Pb 2+ (aq) + 2Br - (aq) No effect. Br - is the anion of a strong acid. (b) Cu(OH) 2 (s) Cu 2+ (aq) + 2OH - (aq) OH - is the anion of water, which is a weak acid. Therefore it will shift the solubility equation to the right and increase solubility. (c) FeS(s) Fe 2+ (aq) + S 2- (aq) S 2- is the anion of a weak acid and will react with water to produce OH -. FeS(s) + H 2 O(l) Fe 2+ (aq) + HS - (aq) + OH - (aq) Both weak acids serve to increase the solubility of FeS.

34 Sample Problem Predicting Whether a Precipitate Will Form PROBLEM: A common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when 0.100L of 0.30M Ca(NO 3 ) 2 is mixed with 0.200L of 0.060M NaF? PLAN: Write out a reaction equation to see which salt would be formed. Look up the K sp valus in a table. Treat this as a reaction quotient, Q, problem and calculate whether the concentrations of ions are > or < K sp. Remember to consider the final diluted solution when calculating concentrations. SOLUTION: CaF 2 (s) Ca 2+ (aq) + 2F - (aq) K sp = 3.2x10-11 mol Ca 2+ = 0.100L(0.30mol/L) = 0.030mol [Ca 2+ ] = 0.030mol/0.300L = 0.10M mol F - = 0.200L(0.060mol/L) = 0.012mol [F - ] = 0.012mol/0.300L = 0.040M Q = [Ca 2+ ][F - ] 2 = (0.10)(0.040) 2 = 1.6x10-4 Q is >> K sp and the CaF 2 WILL precipitate.

35 Cr(NH 3 ) 6 3+, a typical complex ion.

36 The stepwise exchange of NH 3 for H 2 O in M(H 2 O) NH 3 M(H 2 O) NH 3 M(H 2 O) 3 (NH 3 ) 2+ M(NH 3 ) 4 2+

37

38 Sample Problem Calculating the Concentration of a Complex Ion PROBLEM: An industrial chemist converts Zn(H 2 O) 4 2+ to the more stable Zn(NH 3 ) 4 2+ by mixing 50.0L of M Zn (H 2 O) 4 2+ and 25.0L of 0.15M NH 3. What is the final [Zn (H 2 O) 4 2+ ]? K f of Zn(NH 3 ) 4 2+ is 7.8x10 8. PLAN: SOLUTION: Write the reaction equation and K f expression. Use a reaction table to list various concentrations. Remember that components will be diluted when mixed as you calculate final concentrations. It is obvious that there is a huge excess of NH 3 and therefore it will drive the reaction to completion. Zn(H 2 O) 4 2+ (aq) + 4NH 3 (aq) Zn(NH 3 ) 4 2+ (aq) + 4H 2 O(l) K f = [Zn(H 2 O) 4 [Zn(NH 3 ) 2+ 4 ] 2+ ] initial = (50.0L)(0.0020M) = 1.3x10-3 M [Zn(H 2 O) 2+ 4 ][NH 3 ] L [NH 3 ] initial = (25.0L)(0.15M) 75.0L = 5.0x10-2 M

39 Sample Problem Calculating the Concentration of a Complex Ion continued Since we assume that all of the Zn(H 2 O) 4 2+ has reacted, it would use 4 times its amount in NH 3. [NH 3 ] used = 4(1.3x10-3 M) = 5.2x10-3 M [Zn(H 2 O) 4 2+ ] remaining = x(a very small amount) Concentration(M) Zn(H 2 O) 4 2+ (aq) + 4NH 3 (aq) Zn(NH 3 ) 4 2+ (aq) + 4H 2 O(l) Initial Change Equilibrium 1.3x x ~(-1.3x10-3 ) ~(-5.2x10-3 ) ~(+1.3x10-3 ) - x 4.5x x K f = [Zn(NH 3 ) 4 2+ ] [Zn(H 2 O) 4 2+ ][NH 3 ] 4 = 7.8x10 8 = (1.3x10-3 ) x(4.5x10-2 ) x = 4.1x10-7 M

40 Sample Problem Calculating the Effect of Complex-Ion Formation on Solubility PROBLEM: In black-and-white film developing, excess AgBr is removed from the film negative by hypo, an aqueous solution of sodium thiosulfate (Na 2 S 2 O 3 ), through formation of the complex ion Ag(S 2 O 3 ) Calculate the solubility of AgBr in (a) H 2 O; (b) 1.0M hypo. K f of Ag(S 2 O 3 ) 2 3- is 4.7x10 13 and K sp AgBr is 5.0x PLAN: Write equations for the reactions involved. Use K sp to find S, the molar solubility. Consider the shifts in equilibria upon the addition of the complexing agent. SOLUTION: AgBr(s) Ag + (aq) + Br - (aq) K sp = [Ag + ][Br - ] = 5.0x10-13 (a) S = [AgBr] dissolved = [Ag + ] = [Br - ] K sp = S 2 = 5.0x10-13 ; S = 7.1x10-7 M (b) AgBr(s) Ag + (aq) + Br - (aq) Ag + (aq) + 2S 2 O 3 2- (aq) AgBr(s) + 2S 2 O 3 2- (aq) Ag(S 2 O 3 ) 2 3- (aq) Br - (aq) + Ag(S 2 O 3 ) 2 3- (aq)

41 Sample Problem continued K overall = K sp x K f = Calculating the Effect of Complex-Ion Formation on Solubility [Br - ][Ag(S 2 O 3 ] 2 3- [AgBr][S 2 O 3 2- ] 2 = (5.0x10-13 )(4.7x10 13 ) = 24 Concentration(M) AgBr(s) + 2S 2 O 3 2- (aq) Br - (aq) + Ag(S 2 O 3 ) 2 3- (aq) Initial Change - -2S +S +S Equilibrium S S S K overall = S 2 (1.0-2S) 2 = 24 S 1.0-2S = (24) 1/2 S = [Ag(S 2 O 3 ) 2 3- ] = 0.45M

42 The amphoteric behavior of aluminum hydroxide. 3H 2 O(l) + Al(H 2 O) 3 (OH) 3 (s) Al(H 2 O) 3 (OH) 3 (s) Al(H 2 O) 3 (OH) 4- (s) + H 2 O(l)

43 Sample Problem Separating Ions by Selective Precipitation PROBLEM: A solution consists of 0.20M MgCl 2 and 0.10M CuCl 2. Calculate the [OH - ] that would separate the metal ions as their hydroxides. K sp of Mg(OH) 2 = is 6.3x10-10 ; K sp of Cu(OH) 2 is 2.2x PLAN: Both precipitates are of the same ion ratio, 1:2, so we can compare their K sp values to determine which has the greater solubility. It is obvious that Cu(OH) 2 will precipitate first so we calculate the [OH - ] needed for a saturated solution of Mg(OH) 2. This should ensure that we do not precipitate Mg(OH) 2. Then we can check how much Cu 2+ remains in solution. SOLUTION: Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) K sp = 6.3x10-10 Cu(OH) 2 (s) Cu 2+ (aq) + 2OH - (aq) K sp = 2.2x10-20 [OH - ] needed for a saturated Mg(OH) 2 solution = K sp [Mg 2 ] 6.3x = 5.6x10-5 M

44 Sample Problem Separating Ions by Selective Precipitation continued Use the K sp for Cu(OH) 2 to find the amount of Cu remaining. [Cu 2+ ] = K sp /[OH - ] 2 = 2.2x10-20 /(5.6x10-5 ) 2 = 7.0x10-12 M Since the solution was 0.10M CuCl 2, virtually none of the Cu 2+ remains in solution.

45 Centrifuge Centrifuge The general procedure for separating ions in qualitative analysis. Add precipitating ion Add precipitating ion

46 Centrifuge Centrifuge Centrifuge Centrifuge A qualitative analysis scheme for separating cations into five ion groups. Add 6M HCl Acidify to ph 0.5; add H 2 S Add NH 3 /NH 4 + buffer(ph 8) Add (NH 4 ) 2 HPO 4

47 Centrifuge Centrifuge Centrifuge Extra: A qualitative analysis scheme for Ag +,Al 3+,Cu 2+, and Fe 3+ Step 1 Add NH 3 (aq) Step 2 Add HCl Step 3 Add NaOH Step 4 Add HCl, Na 2 HPO 4 Step 5 Dissolve in HCl and add KSCN

48 A view inside Carlsbad Caverns, New Mexico

49 Formation of acidic precipitation.

50 A forest damaged by acid rain

51 The effect of acid rain on marble statuary Location: New York City