1. 19.1 The common Ion Effect 19.2 Controlling pH : Buffer Solution 19.3 Acid and Base Titration 19.4 Solubility of Salts 19.5 Precipitation Reactions 19.6 Solubility Constant Ch 19Reactivity of Aqueous Solution The common-ion effect to describe the effect on a solution of two dissolved solutes that contain the same ion. The presence of a common ion suppresses the ionization of a weak acid or a weak base.

2. Formation of acidic precipitation.

3. A view inside Carlsbad Caverns, New Mexico

4. The effect of acid rain on marble statuary. 19441994 Location: New York City

5. A forest damaged by acid rain

6. How a buffer works. Buffer with equal concentrations of conjugate base and acid OH - H3O+H3O+ Buffer after addition of H 3 O + H 2 O + CH 3 COOH H 3 O + + CH 3 COO - Buffer after addition of OH - CH 3 COOH + OH - H 2 O + CH 3 COO - Buffer Solution Buffer solution is defined as a mixture of a conjugate acid and base pair. Buffer solution corresponds to approximately 10-90 % titration of a relative flat region so the titration curve. Buffer solution tends to resist changes in pH when an acid or base is added into the system. Buffer solution is commonly used when pH must be maintained at a relative constant or in many biological systems.

7. Sample Problem Calculating the Effect of Added H 3 O + or OH - on Buffer pH PROBLEM:Calculate the 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. Initial Change Equilibrium 0.50 + x 0.50-x - - - 0.500 + x 0.50 +xx - x SOLUTION: CH 3 COOH( aq ) + H 2 O( l ) CH 3 COO - ( aq ) + H 3 O + ( aq )Concentration (M) (a)

8. Calculating the Effect of Added H 3 O + and OH - on Buffer pH continued (2 of 4) [CH 3 COOH] equil ≈ 0.50M[CH 3 COO - ] initial ≈ 0.50M[H 3 O + ] = x K a = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] [H 3 O + ] = x = K a [CH 3 COO - ] [CH 3 COOH] = 1.8  10 -5 M Check the assumption:1.8x10 -5 /0.50 X 100 = 3.6x10 -3 % CH 3 COOH(aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l)Concentration (M) Before addition Addition After addition (b) [OH - ] added = 0.020 mol 1.0 L soln = 0.020M NaOH 0.50- - - - -0.020- 0.4800.52

9. 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. CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq)Concentration (M) Initial Change Equilibrium 0.48- - x 0.48 -x - - 0.520 x + x 0.52 + x [H 3 O + ] = 1.8  10 -5 0.48 0.52 = 1.7  10 -5 pH = 4.77 CH 3 COO - (aq) + H 3 O + (aq) CH 3 COOH(aq) + H 2 O (l)Concentration (M) Before addition Addition After addition (c)[H 3 O + ] added = 0.020 mol 1.0L soln = 0.020M H 3 O + 0.50- - - - -0.020- 0.4800.52

10. 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. CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq)Concentration (M) Initial Change Equilibrium 0.52- - x 0.52 -x - - 0.480 x + x 0.48 +x [H 3 O + ] = 1.8  10 -5 0.48 0.52 = 2.0  10 -5 pH = 4.70 The Henderson-Hasselbalch Equation pH = pK a + log [base] [acid]

11. Buffer Capacity and Buffer Range Buffer capacity is the ability to resist pH change. Buffer range is the pH range over which the buffer acts effectively. 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. Buffers have a usable range within ± 1 pH unit of the pK a of its acid component. pH = pK a + log [base] [acid] 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.

12. Sample Problem Preparing a Buffer SOLUTION: PROBLEM:An environmental chemist needs a carbonate buffer of pH 10.00 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 4.7  10 -11. PLAN: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 3 2- ][H 3 O + ] [HCO 3 - ] pH = 10.00; [H 3 O + ] = 1.0x10 -10 4.7x10 -11 = [CO 3 2- ](0.20) 1.0x10 -10 [CO 3 2- ] = 0.094M moles of Na 2 CO 3 = (1.5L)(0.094mols/L)= 0.14 = 15 g Na 2 CO 3 0.14 moles 105.99g mol

13. Fig. 1 Curve for a strong acid-strong base titration

14. Fig. 2 Curve for a weak base-strong acid titration Titration of 40.00mL of 0.1000M NH 3 with 0.1000M HCl pH = 5.27 at equivalence point pK a of NH 4 + = 9.25

15. Fig. 3 Curve for a weak acid-strong base titration Titration of 40.00mL of 0.1000M HPr with 0.1000M NaOH [HPr] = [Pr - ] pH = 8.80 at equivalence point pK a of HPr = 4.89 methyl red

16. pK a = 7.19 pK a = 1.85 Curve for the titration of a weak polyprotic acid. Titration of 40.00mL of 0.1000M H 2 SO 3 with 0.1000M NaOH

17. Homework Calculate pH for (b), (c), and (d) Sample ProblemCalculating the pH During a Weak Acid- Strong Base Titration PROBLEM:Calculate the pH during the titration of 40.00 mL of 0.1000M propanoic acid (HPr; K a = 1.3  10 -5 ) after adding the following volumes of 0.1000M NaOH: (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 + ] x = 1.1  10 -3 ; pH = 2.96 (a) 0.00mL

18. (b) Before addition Addition After addition 0.04000 0.03000 0.01000 0- - -0 - -- HPr( aq ) + OH - ( aq ) Pr - ( aq ) + H 2 O ( l )Amount (mol) Continued [H 3 O + ] = 1.3x10 -5 0.001000 mol 0.003000 mol = 4.3x10 -6 MpH = 5.37 PROBLEM: Calculate the pH during the titration of 40.00 mL of 0.1000M propanoic acid (HPr; K a = 1.3  10 -5 ) after adding the following volumes of 0.1000M NaOH: Change0.03000-0.03000- K a = [Pr - ][H 3 O + ]/[HPr] [H 3 O + ] = K a [HPr] / [Pr - ] pH = -log [H + ]

19. Calculating the pH During a Weak Acid- Strong Base Titration (c) When 40.00mL of NaOH are added, all of the HPr will be reacted and the [Pr - ] will be (0.004000 mol) (0.004000L) + (0.004000L) = 0.05000M K a x K b = K w K b = K w /K a = 1.0x10 -14 /1.3  10 -5 = 7.7x10 -10 [H 3 O + ] = K w / = 1.6  10 -9 M pH = 8.80 (d) 50.00mL of NaOH will produce an excess of OH -. mol  V of base = (0.1000M)(0.05000L - 0.04000L) = 0.00100mol M = (0.00100) (0.0900L) M = 0.01111 [H 3 O + ] = 1.0  10 -14 /0.01111 = 9.0  10 -11 M pH = 12.05

20. Solubility Rules For Ionic Compounds in Water 1. All common compounds of Group 1A(1) ions (Li +, Na +, K +, etc.) and ammonium ion (NH 4 + ) are soluble. 2. All common nitrates (NO 3 - ), acetates (CH 3 COO - or C 2 H 3 O 2 - ) and most perchlorates (ClO 4 - ) are soluble. 3. All common chlorides (Cl - ), bromides (Br - ) and iodides (I - ) are soluble, except those of Ag +, Pb 2+, Cu +, and Hg 2 2+. 1. All common metal hydroxides are insoluble, except those of Group 1A(1) and the larger members of Group 2A(2)(beginning with Ca 2+ ). 2. All common carbonates (CO 3 2- ) and phosphates (PO 4 3- ) SO 3 2-, SO 4 2-, S 2- are insoluble, except those of Group 1A(1) and NH 4 +. 3. All common sulfides are insoluble except those of Group 1A(1), Group 2A(2) and NH 4 +. Soluble Ionic Compounds Insoluble Ionic Compounds

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

22. Sample ProblemDetermining K sp from Solubility PROBLEM: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 (net ionic equation); find moles of dissociated ions; convert solubility to M and substitute values into solubility product constant expression. SOLUTION: PbF 2 (s) Pb 2+ (aq) + 2F - (aq) K sp = [Pb 2+ ][F - ] 2 = 2.6  10 -3 M K sp = (2.6  10 -3 )(5.2  10 -3 ) 2 = 0.64g L soln245.2g PbF 2 mol PbF 2 7.0  10 -8

23. Sample ProblemDetermining 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 6.5  10 -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 -Initial Change Equilibrium - - 00 +S+ 2S S2S K sp = (S)(2S) 2 S = = 1.2  10 -2 M Ca(OH) 2 (s) Ca 2+ (aq) + 2OH - (aq)Concentration (M)

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

25. Sample ProblemCalculating the Effect of a Common Ion on Solubility PROBLEM:In Sample Problem above, 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:Ca(OH) 2 (s) Ca 2+ (aq) + 2OH - (aq)Concentration(M) Initial Change Equilibrium - - - 0.100 +S+2S 0.10 + S2S K sp = 6.5x10 -6 = (0.10 + S)(2S) 2 = (0.10)(2S) 2 S << 0.10 S = = 4.0x10 -3 Check the assumption: 4.0% 0.10M 4.0x10 -3 x 100 =

26. Sample ProblemPredicting 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. Br - is the anion of a strong acid. No effect. SOLUTION:(a) PbBr 2 (s) Pb 2+ (aq) + 2Br - (aq) (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 -. Both weak acids serve to increase the solubility of FeS. FeS(s) + H 2 O(l) Fe 2+ (aq) + HS - (aq) + OH - (aq)

27. Sample ProblemPredicting 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.

28. Sample ProblemSeparating 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.