More Acid and Base Chemistry

Common-ion effect In the last chapter, we calculated the [H3O+] of a 0.123M HClO as 6.02x10-5M. The percent dissociation for this solution would be: 6.02x10 5 x100  0.0489% 0.123 But what would happen if we place 0.123moles of HClO in a 1.000L container that already has 0.100moles of the conjugate base NaClO. Will this change anything? More Acid and Base Chemistry HClO(aq)+ H2O(l)⇌ H3O+(aq) + ClO-(aq) I C E [H O ][ClO ] Ka  3 [HClO] 2.95x10 8  nitial (x)(0.100  x) 0.123 - x 0.123M -x hange quilibirum Assume these “x’s” will be very small 2.95x10 8  (x)(0.100) 0.123 3.63x10 8 x100  0.0000295% 0.123 Ka  [H3O ][ClO ] [HClO] 0.123M (0.100  x)(x) 0.123 - x 2.95x10 8  Assume these “x’s” will be very small 0M +x +x Hydromiun ion already in solution before acid added 0.123M - x (0.100)(x) 0.123 0.100+x ΔpH = -7 HCl adds H3O+ to the solution with its reaction with water. Cl- H+ HCl + H2O → H3O+ + Cl- x pH = 7 Through the reaction, 1 mol of H3O+ is added to the water. pH = 0 x  3.63x10 8 3.63x10 8 x100  0.0000295% 0.123 3.63x10 9  0.100x How would ionization be affected if both H3O+ and ClO- were 0.100M initially? Hint: think about Qa ΔpH = -0.22 NH3 /NH4+ buffer solution NH4+ Both products of dissociation of HClO, ClOand H3O+, caused a decrease in the dissociation of the acid. This is the common ion effect. Note the change in percent dissociation. The percent dissociation decreased because there were already ClO- ions in solution. The reaction did not proceed as far toward the reactants to reach equilibrium. This is the common-ion effect. Common ions in solution decrease dissociation of weak acids and bases. Plain ole water Add HCl quilibirum 2.95x10 8  0.100M -x hange 0.100+x Buffers have the ability to stabilize pH, even with the addition of acids or bases. Buffers are composed of weak conjugate bases and acids in equilibrium. HClO(aq)+ H2O(l)⇌ H3O+(aq) + ClO-(aq) nitial +x x Buffers Common-ion effect What would happen if the solution had a pH of 1.000 before adding 0.123M HClO? I C E Hypoclorite ion already in solution before acid added 0.100M +x 0.123M - x x  3.63x10 8 3.63x10 9  0.100x 6.02x10 5 x100  0.0489% 0.123 0M NH3 Add HCl + NH3 NH3 NH4 NH + 4 NH4+ NH3 NH4+ NH4+ Cl + NH3 NH3 NH4 NH + 4 NH + NH3 - 4 pH = 9.25 pH = 9.03 HCl reacts with NH3 to give more NH4+ ions. H3O+ can only be added to the solution through the NH3/NH4+ equilibrium HCl + NH3 → NH4+ + ClThrough the reaction, only 3.7x10-10 mol of H3O+ is added to the water. Calculating pH of buffer solutions Adding Strong Acid/Base to a Buffer Solution Calculate the pH of a buffer solution in a 1.0L container with 3.0 moles NH 3 of and 5.0 moles of NH4+. Ka for NH4+ is 5.8x10-10. Let’s take our first buffer problem from the last page: “Calculate the pH of a buffer solution in a 1.0L container with 3.0 moles NH3 of and 5.0 moles of NH4+. Ka for NH4+ is 5.8x10-10.” Ka  [H3O ][NH3 ] [NH4  ] 5.8x10 10  (x)(3.0) 5.0 5.8x10 10  NH4+(aq)+ H2O(l)⇌ H3O+(aq) + NH3(aq) (x)(3.0  x) 5.0 - x x  9.7x10 10 pH = -log[H3O+] = -log(9.7x10-10) = 9.01 What happens when we add 0.50L of 2.00M HCl to the solution 5.0M Initial Change Equilibrium 0M -x 5.0M - x 3.0M +x +x x 3.0M+x Because the [NH3] + [NH4+] = 8.0M, this is an 8.0 M buffer First, we need to determine how adding HCl changes the NH3/NH4+ buffer. To do this, we will use a change table. It is best to use moles not M in this table (you will see why later). H+ (aq) + NH3(aq) → NH4+(aq) Note that the HCl (acid) will react with the NH3 (base) in the solution. The strong acid or base will always react completely. Remember, 100% dissociation. But instead of another I.C.E. table, we can use… Before: 1.0mole 3.0moles 5.0moles Change: -1.0mole -1.0mole +1.0mole After: 2.0moles 6.0moles 0mole pKa = -log Ka = -log(5.8x10-10) = 9.24 •Change in acid/base concentrations by dissociation negligible. Henderson – Hasslebalch Formula [conjugatebase] pH  p K a  log [conjugateacid] [acid/base]initial ≈ [acid/base]equilibrium pH  9.24  log Let’s do the same calculation above, but using H-H pH  9.24  log [conjugatebase] pH  p K a  log [conjugateacid] 3.0M 5.0M pH=9.02 2.0mol 6.0mol pH=8.76 Since both NH3 and NH4+ are in the same volume, we can use moles instead of concentration. If you divide both HA and A- by the same volume, the ratio stays the same, 1 A More Realistic Buffer Solution Problem Calculate the pH of a buffer solution in a 100.mL container with 0.100M of each component of a benzoic acid / sodium benzoate buffer. Ka of benzoic acid is 6.4x10-5. [conjugatebase] pH  p K a  log [conjugateacid] pH  4.19  log 0.100M 0.100M pH  4.19 What is the pH after adding 10.00mL of 0.500M NaOH? OH- (aq) HA(aq) → A- (aq) Before: 0.00500mol 0.0100mol 0.0100mol Change: -0.00500mol -0.00500mol +0.00500mol After: 0.00500mol 0.0150mol + 0mol AA- Since there are still appreciable amounts of both HA and A-, it still a buffer solution. Thus, H-H can be used AHA Buffer Solution pH  4.19  log 0.0150mol 0.00500mol pH  4.67 A More Realistic Buffer Solution Problem A- OH- What is the pH after adding a total of 20.00mL of 0.500M NaOH to the original (100.0mL 0.100M each) buffer solution? Ka 6.4x10-5. A- 0.0200L NaOH AA- A- A- A- A- HA A- HA H+ HA ClNa+ Na+ Cl- 0.500mol NaOH  0.0100mol NaOH 1 L NaOH OH- (aq) 0.1000L HA HA(aq) → A- (aq) Before: 0.0100mol 0.0100mol 0.0100mol Change: -0.0100mol -0.0100mol +0.0100mol 0mol 0.0200mol After: + 0mol AA- 0.100mol HA  0.0100mol HA 1 L HA AA- Is this even a buffer solution any more? NO! Only A-! Then I cannot use H-H! Weak Base Solution 0.0200mol A  0.167M A 0.02000L  0.1000L F- (aq)+ H2O(l)⇌ OH-(aq) + HF(aq) Initial Change Equilibrium 0.167M -x 00.167M - x 0M 0M +x +x x x x2  1.6 x10 10 0.167 x  5.2 x10 6 [OH - ]  x  5.2 x10 6 pH  14  pOH  14  ( log 5.2 x10-6 ) pH  8.72 A More Realistic Buffer Solution Problem What is the pH after adding a total of 30.00mL of 0.500M NaOH to the original (100.0mL 0.100M each) buffer solution? OH- (aq) + HA(aq) → A- (aq) Before: 0.0150mol 0.0100mol 0.0100mol Change: -0.0100mol -0.0100mol +0.0100mol 0mol 0.0200mol After: 0.00500mol 0.00500mol OH  0.0385M OH 0.1000L  0.03000 pOH  -log 0.0385  1.415 AA- A- AOH- Which base should have a greater effect on pH? How a buffer reactions to addition of strong base or strong acid Assume we have a buffer with 1M total buffer strength. Let’s start with 0.50M of each HA and A-. adding acid Strong Base Solution adding base pH  14.000  1.415 pH  12.585 Mol of conjugate base in 1mol total buffer [conjugatebase] pH  p K a  log [conjugateacid] Note that our theoretical buffer has a Ka of 1x10-7 and thus a pKa of 7. 2 Strong Acid / Strong Base When does a buffer stop “buffering”? Titration problems (determining the pH at different points of a titration) are very similar to buffer questions. Use change tables, and see what’s left. 10:1 1:1 base:acid 1:10 No longer buffering Determine the pH during of10.00mL of 0.0100M HCl before the titration. pH  -log(0.0100M) Strong Acid Solution - + H+ Cl H Cl Cl- H+ + Cl H What type of solution is this? pH  2.000 Determine the pH during a titration after 1.50mL of 0.0500M NaOH has been added to 10.00mL of 0.0100M HCl. OH- (aq) H+ (aq) → + H2O Mol of conjugate base in 1mol total buffer 1.00x10-4 mol Before: 7.50x10-5 mol Change: A good rule of thumb: A buffer is relatively effective ± 1 pH point of the acid’s pKa. After that point the buffer is exghausted and can no longer buffer in that direction. -7.50x10-5 mol -7.50x10-5 + Cl- Cl H Na+ Na+ Cl Cl- H+ mol 2.5x10-5 mol 0mol After: 2.5x10-5 mol H   0.0022M H  0.01000L  0.00150L pH  -log(0.0022M) Strong Acid / Strong Base Strong Acid Solution pH  2.66 Strong Acid / Strong Base Determine the pH during a titration after 2.00mL of 0.0500M NaOH has been added to 10.00mL of 0.0100M HCl. Determine the pH during a titration after 2.50mL of 0.0500M NaOH has been added to 10.00mL of 0.0100M HCl. OH- (aq) + H+ (aq) → 1.0x10-4 mol 1.0x10-4 mol Change: -1.0x10-4 -1.0x10-4 After: mol OH- (aq) H2O Before: Na+ Cl- mol Na+ 0mol 0mol ClCl- Before: Na+ ClNa+ Change: -1.00x10-4 H2O 1.00x10-4 mol -1.00x10-4 mol Na+ Cl- mol OHNa+Cl- Cl- Na+ 2.5x10-5 mol OH  0.0020M OH 0.01000L  0.00250L Neutral Solution Cl- 0mol 2.5x10-5 mol After: pH  7.00 H+ (aq) → + 1.25x10-4 mol Na+ Strong Base Solution pOH  -log(0.0020M)  1.70 pH  14.00  1.70 pH  12.30 Acid-Base Titrations Acid-Base Titrations Strong Acid/Strong Base Weak Acid/Strong Base Adding Base to Acid Equivalence Point: The amount of base added equals the amount of acid present before the titration. Eq. Pt. The acid/base reaction is complete. For strong acid/strong base titration the pH is >7. After Equivalence Point: The amount of base added equals the amount of acid present before the titration. ½ volume of equivalence point pH = pKa The acid/base reaction is complete. For strong acid/strong base titration the pH is 7. During Before After [HA] = [A-] During (buffer zone) Before Vol of NaOH added (mL) Vol of NaOH added (mL) - + H+ Cl H Cl Cl- H+ + Cl H Before Titration add NaOH + Cl- Cl H Na+ Na+ Cl Cl H+ During Titration Na+ Cladd NaOH Na+ ClCl- Na+ ClNa+ Equivalence Point Na+ Cladd NaOH ClNa+ Cl OH- add NaOH Na+ClNa+ Past Equivalence Point HF HF HF HF Before Titration Na+ F- HF F- Na+ HF During Titration add NaOH Na+ F- Na+ F- Na+ F- F- Na+ Equivalence Point add NaOH Na+ F- F- Na+ F- Na+ Na+ OH F Na+ Past Equivalence Point 3 Acid-Base Titrations Acid-Base Titrations Strong Acid/Weak Base Strong Acid/Strong Base Before Adding Acid to Base Before ½ volume of equivalence point pH = 14-pKb During Equivalence Point: The amount of acid added equals the amount of base present before the titration. The acid/base reaction is complete. For strong acid/strong base titration the pH is 7. OHNa+ OH OH- OH- Na+ Na+ Na+ add HCl Before Titration Cl- + OHNa + OHCl- Na Na+ Na+ After Na+ Cl- add HCl Na Cl+ Cl- During Titration Na+ ClNa+ Na+ Cl- add HCl H+ Cl- Na+ Cl- ClNa+ Cl+ Na Past Equivalence Point Equivalence Point Eq. Pt. The acid/base reaction is complete. For strong acid/weak base titration the pH is <7. add HCl NH3 NH3 ClNH3 NH3 NH3 Before Titration After add HCl NH4+ Cl- NH 3 NH4+ NH4+ During Titration NH4+ ClNH4+ Cl- add HCl ClNH4+ Cl- NH4+ ClNH4+ Cl- Cl- NH4+ H+ NH4+ Past Equivalence Point Equivalence Point Polyprotic acid Titrations Strong Acid / Weak Base Titrations Determine the pH during of10.00mL of 0.0100M HF before the titration. Ka = 6.3x10-4 During (buffer zone) [B]= [BH+] Equivalence Point: The amount of acid added equals the amount of base present before the titration. Since there are multiple protons to remove from a polyprotic acid, there are multiple equivalence points – one for each proton. Determine the pH during a titration after 1.50mL of 0.0500M NaOH has been added to 10.00mL of 0.0100M HF. The the pH of the equivalence point of an amphoteric species (on that can either donate or accept an proton) Determine the pH during a titration after 2.00mL of 0.0500M NaOH has been added to 10.00mL of 0.0100M HF. pHof amphotericspecies p K an  p K an1 2 Determine the pH during a titration after 2.50mL of 0.0500M NaOH has been added to 10.00mL of 0.0100M HF. ⇌ H3PO4 H2PO4- HPO42- pKa2 = 7.20 pKa3 = 12.38 Br- Ag+ and Ksp = 5.4x10-13 How do we deal with sparingly soluble salts that do not have cations and anions in a 1:1 ratio? Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq) OHMg2+ OH- So, how do we determine the concentration of Ag+ and Br- in the solution? Ksp = 1.2x10-11 Initial Change Equilibrium 0M 0M +x +2x x 2x The solubility of AgBr is represented by the equation: AgBr(s) ⇌ Ag+(aq) + Br-(aq) Since solids do not appear in any K expression… Initial Change Equilibrium 0M 0M +x +x [Mg2+] = x = 1.4x10-4 M [OH-] = 2x = (2)1.44x10-4 M = 2.9x10-4M x2  5.4x10 13 PO43- Solubility Product – Ksp (more complicated example) In CHM151, we talked about ionic compounds that were “insoluble” in water. In reality, those compounds do dissolve, just a very. very small amount. Ksp = [Ag+][Br-] 3- ⇌ pKa1 = 2.12 Solubility Product – Ksp For example, based on our rules, AgBr is insoluble. How some very small amount does dissociate into ions. The represent this we have another K expression called Ksp. 2- ⇌ Ksp = [Mg2+][OH- ]2 (x)(2x)2 =1.2x10-11 (x)(4x2) = 4x3 = 1.2x10-11 x3 1.2x10 11 4 x = 1.44x10-4M x2  5.4x10 13 x x x  7.3x10 7 M [Ag+] = [Br-] = 7.3x10-7M This is called the “molar solubility” 4 Solubility Product – Ksp Remember LeChatelier’s Principle holds that the addition of a solid should not shift equilibrium. Br- Ag+ Add solid AgBr Br- Does this hold true? YES! Assuming there is already some solid in the solution, the concentration of ions does not change. Ag+ Factors that effect solubility of sparingly soluble salts Common-ions pH (acid/base reactions OH- Ag+ Add water Br- Br- Ag+ Initially, as the water is added, the concentration of ions is decreased. Based on LeChatelier’s Principle the reaction should shift toward the product’s side. BrAg+ Thus, in the end, more AgBr is dissolved and the concentration of ions is the same. In all fours cases, the [Ag+] 7.3x10-7 M and [Br-] is 7.3x10-7 M Fe2+ OH- Common ions will decrease solubility of a sparingly soluble salt. Same as common ion effect. Acids will increase the solubility of a sparingly soluble salts with a basic ion (OH-, F- ,etc.) through an acid/base reaction. Add HCl Cl- Fe2+ Cl- Equilibrium reestablished OH- Fe2+ OHCl- Fe2+ Cl- Formation of Complex ions (Lewis acid/base reactions OHCu2+ OH- Add a lot of NH3 Common ion of a sparingly soluble salt Let’s say we have 0.50L of 0.50M NaOH. Into this solution we drop in a large chunk of Mg(OH) 2. Determine the concentrations of all ions in solution at equilibrium. Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq) Initial Na+ OH- Add Mg(OH)2 Change Equilibrium 0M 0.50M +x +2x x 0.50+2x Ksp = 1.2x10-11 Ksp = [Mg2+][OH- ]2 1.2x10-11 = (x)(0.50)2 = 0.25x x 1.2x10 11  4.8x10 11 0.25 [Mg2+] = x = 4.8x10-11 M Na+ OH- [OH-] = 0.50+2x = 0.50 +(2)4.8x10-11 ≈ 0.50M 5