Ionic Equilibria in Aqueous Solutions n Equilibria of

Ionic Equilibria in Aqueous Solutions n Equilibria of Acid-Base Buffer Systems Ø 9/29/2020 Buffer l An Acid-Base Buffer is a species added to an solution to minimize the impact on p. H from the addition of [H 3 O+] or [OH-] ions l Small amounts of acid or base added to an unbuffered solution can change the p. H by several units l Since p. H is a logarithmic term, the change in [H 3 O+] or [OH-] can be several orders of magnitude 1

Ionic Equilibria in Aqueous Solutions n The Common Ion Effect Ø Buffers work through a phenomenon known as the: Common Ion Effect 9/29/2020 Ø The common ion effect occurs when a given ion is added to an “equilibrium mixture of a weak acid or weak base that already contains that ion Ø The additional “common ion” shifts the equilibrium away from its formation to more of the undissociated form; i. e. , the acid or base dissociation decreases Ø A buffer must contain an “acidic” component that can react with the added OH- ion, and a “basic” component than can react with the added [H 3 O+] 2

Ionic Equilibria in Aqueous Solutions The buffer components cannot be just any acid or base Ø The components of a buffer are usually the conjugate acid-base pair of the weak acid (or base) being buffered Ex: Acetic Acid & Sodium Acetate Acetic acid is a weak acid, slightly dissociated Ø CH 3 COOH(aq) + H 2 O(l) CH 3 COO-(aq) + H 3 O+(aq) Acid 9/29/2020 Base Conjugate Base Acid 3

Ionic Equilibria in Aqueous Solutions Ø Ex: Acetic Acid & Sodium Acetate (Con’t) l Now add Sodium Acetate (CH 3 COONa), a strong electrolyte (Acetate ion is the conjugate base) CH 3 COOH(aq) + H 2 O(l) H 3 O+(aq) + CH 3 COO-(aq) (added) 9/29/2020 l The added Acetate ions shift the equilibrium to the left forming more undissociated acetic acid l This lowers the extent of acid dissociation, which lowers the acidity by reducing the [H 3 O+] and increasing the p. H l Similarly, if Acetic Acid is added to a solution of Sodium Acetate, the Acetate ions already present act to suppress the dissociation of the acid 4

Ionic Equilibria in Aqueous Solutions When a small amount of [H 3 O+] is add to acetic acid/acetate buffer, that same amount of acetate ion (CH 3 COO-) combines with it, increasing the concentration of acetic acid (CH 3 COOH). The change in the [HA]/[A-] ratio is small; the added [H 3 O+] is effectively tied up; thus, the change in p. H is also small 9/29/2020 5

Ionic Equilibria in Aqueous Solutions n 9/29/2020 Essential Features of a Buffer Ø A buffer consists of high concentrations of the undissociated acidic component [HA] and the conjugate base [A-] component Ø When small amounts of [H 3 O+] or [OH-] are added to the buffer, they cause small amounts of one buffer component to convert to the other component 6

Ionic Equilibria in Aqueous Solutions Ex. When a small amount of H 3 O+ is added to an Acetate buffer, that same amount of CH 3 COOcombines with it, increasing the amount of undissociated CH 3 COOH tying up potential [H 3 O+] Similarly, a small amount of OH- added combines with undissociated CH 3 COOH to form CH 3 COO- & H 2 O tying up potential [OH-] In both cases, the relative changes in the amount of buffer components is small, but the added [H 3 O+] or [OH-] ions are tied up as undissociated components; thus little impact on p. H 9/29/2020 7

Ionic Equilibria in Aqueous Solutions n 9/29/2020 The equilibrium perspective Ø The [H 3 O+] (p. H) of the solution depends directly on the buffer-component concentration ratio Ø If the ratio [HA]/[A-] goes up, [H 3 O+] goes up (p. H down) Ø If the ratio [HA]/[A-] goes down, [H 3 O+] goes down ([OH-] goes up) and p. H increases (less acidic) Ø When a small amount of a strong acid is added, the increased amount of [H 3 O+] reacts with a stoichiometric amount of acetate ion from the buffer to form more undissociated acetic acid 8

Ionic Equilibria in Aqueous Solutions n The Henderson-Hasselbalch Equation Ø For any weak acid, HA, the dissociation equation and Ka expression are: The key variable that determines the [H 3 O+] is the ratio of acid species (HA) to base species (A-) Note the change in sign of “log” term and inversion of acid-base terms 9/29/2020 9

Ionic Equilibria in Aqueous Solutions n 9/29/2020 Buffer Capacity Ø A buffer resists a p. H change as long as the concentration of buffer components are large compared with the amount of strong acid or base added Ø Buffer Capacity is a measure of the ability to resist p. H change Ø Buffer capacity depends on both the absolute and relative component concentrations l Absolute - The more concentrated the components of a buffer, the greater the buffer capacity l Relative – For a given addition of acid or base, the buffer-component concentration ratio changes less when the concentrations are similar than when they are different 10

Ionic Equilibria in Aqueous Solutions n Ex. Consider 2 solutions Ø Solution 1 – Equal volumes of 1. 0 M HAc and 1. 0 M Ac - Ø Solution 2 – Equal volumes of 0. 1 M HAc and 0. 1 M Ac - Ø Same p. H (4. 74) but 1. 0 M buffer has much larger buffer capacity (Con’t) 9/29/2020 11

Ionic Equilibria in Aqueous Solutions Ex. Buffer #1 [HA] = [A-] = 1. 000 M Add 0. 010 mol of OH- to 1. 00 L of buffer [HA] changes to 0. 990 M 9/29/2020 [A-] changes to 1. 010 M 12

Ionic Equilibria in Aqueous Solutions Buffer #2 [HA] = 0. 250 M [A-] = 1. 75 M Add 0. 010 mol of OH- to 1. 00 L of buffer [HA] changes to 0. 240 M [A-] changes to 1. 760 M Buffer-component concentration ratio is much larger when the initial concentrations of the components are very different 9/29/2020 13

Ionic Equilibria in Aqueous Solutions n A buffer has the highest capacity when the component acid / base concentrations are equal: A buffer whose p. H is equal to or near the p. Ka of its acid component has the highest buffer capacity 9/29/2020 14

Ionic Equilibria in Aqueous Solutions n Buffer Range Ø p. H range over which the buffer acts effectively is related to the relative component concentrations Ø The further the component concentration ratio is from 1, the less effective the buffering action Ø If the [A-]/[HA] ratio is greater than 10 or less than 0. 1 (or one component concentration is more than 10 times the other) the buffering action is poor Ø Buffers have a usable range within: ± 1 p. H unit or p. Ka value of the acid components 9/29/2020 15

Ionic Equilibria in Aqueous Solutions n Equilibria of Acid-Base Systems Ø Preparing a Buffer l Choose the Conjugate Acid-Base pair Driven by p. H Ratio of component concentrations close to 1 and p. H ≈ p. Ka Con’t 9/29/2020 16

Ionic Equilibria in Aqueous Solutions Ex. Assume you need a biochemical buffer whose p. H is 3. 9 1. p. Ka of acid component should be close to 3. 9 Ka = 10 -3. 9 = 1. 3 x 10 -4 2. From a table of p. Ka values select buffer possibilities Lactic acid (p. Ka = 3. 86) Glycolic acid (p. Ka = 3. 83) Formic Acid (p. Ka = 3. 74) 3. To avoid common biological species, select Formic Acid Buffer components of Formic Acid 9/29/2020 Formic Acid – HCOOH (Acid) Formate Ion – HCOO(Conjuagte Base) Obtain soluble Formate salt – HCOONa Con’t 17

Ionic Equilibria in Aqueous Solutions 4. Calculate Ratio of Buffer Component Concentrations ([A-]/[HA]) that gives desired p. H Con’t 9/29/2020 18

Ionic Equilibria in Aqueous Solutions Ø Preparing a Buffer (Con’t) l Determine the Buffer Concentrations The higher the concentration of the components, the higher the buffer capacity Assume 1 Liter of buffer is required and you have a stock of 0. 40 M Formic Acid (HCOOH) Compute moles and then grams of Sodium Formate (CHOONa) needed to produce 1. 4/1. 0 ratio Con’t 9/29/2020 19

Ionic Equilibria in Aqueous Solutions Ø Preparing a Buffer (Con’t) l Mix the solution and adjust the p. H The prepared solution may not be an ideal solution (see Chapter 13, Section 6) The desired p. H (3. 9) may not exactly match the actual value of the buffer solution 9/29/2020 The p. H of the buffer solution can be adjusted by a few tenths of a p. H unit by adding strong acid or strong base. 20

Practice Problem Example of preparing a buffer solution without using the Henderson-Hasselbalch equation How many grams of Sodium Carbonate (Na 2 CO 3) must be added to 1. 5 L of a 0. 20 M Sodium Bicarbonate (Na. HCO 3) solution to make a buffer with p. H = 10. 0? Ka (HCO 3 -) = 4. 7 x 10 -11 (From Appendix C) Conjugate Pair - HCO 3 - (acid) and CO 32 - (base) Convert p. H to [H 3 O+] Con’t 9/29/2020 21

Ionic Equilibria in Aqueous Solutions n 9/29/2020 Acid-Base Titration Curves Ø Acid-Base Indicators l Weak organic acid (HIn) that has a different color than its conjugate base (In-) l The color change occurs over a relatively narrow p. H range l Only small amounts of the indicator are needed; too little to affect the p. H of the solution l The color range of typical indicators reflects a 100 - fold range in the [HIn]/In-] ratio l This corresponds to a p. H range of 2 p. H units 22

Mixing Acids & Bases Acids and bases react through neutralization reactions n The change in p. H of an acid mixed with a base is tracked with an Acid-Base Titration Curve n Titration: Ø Titration Curve: Plot of p. H vs. the volume of the “Strong” acid or “Strong” base being added via buret Ø Solution in buret is called the “titrant” n Equivalence Point: Point in a titration curve where stoichiometric amounts of acid and base have been mixed (point of complete reaction) n 3 Important Cases: Strong Acid + Strong Base (& vice versa) Weak Acid + Strong Base Weak Base + Strong Acid n 9/29/2020 23

Titration Curves n Titration Curve: Plot of measured p. H versus Volume of acid or base added during a “Neutralization” experiment n All Titration Curves have a characteristic “Sigmoid (S-shaped) profile n 9/29/2020 Ø Beginning of Curve: p. H changes slowly Ø Middle of Curve: p. H changes very rapidly Ø End of Curve: p. H changes very slowly again p. H changes very rapidly in the titration as the equivalence point (point of complete reaction) is reached and right after the equivalence point 24

Neutralization Reactions n Acids and Bases react with each other to form salts (not always) and water (not always) through Neutralization Reactions n Neutralization reaction between a strong acid and strong base HNO 3(aq) + KOH(aq) KNO 3(aq) + H 2 O(l) acid n base salt water Neutralization of a strong acid and strong base reaction lies very far to the right (Kn is very large); reaction goes to completion; net ionic equation is H 3 O+(aq) + OH-(aq) 2 H 2 O(l) n 9/29/2020 H 3 O+ and OH- efficiently react with each other to form water 25

Titration Curves Curve for Titration of a Strong Acid by a Strong Base 9/29/2020 26

Neutralization Reactions n Neutralization reactions between weak acids and strong bases (net ionic equation shown below) C 6 H 5 COOH(aq) benzoic acid n + Na. OH C 6 H 5 COONa(aq) base benzoate “salt” + H 2 O(l) water The above equilibrium lies very far to the right The two equilibria below can be added together to provide the overall neutralization reaction shown above (1) C 6 H 5 COOH(aq) + H 2 O(l) C 6 H 5 COO-(aq) + H 3 O+(aq) Ka = 1. 7 x 10 -5 (from Appendix C) (2) H 3 O+(aq) + OH-(aq) 2 H 2 O(l) 9/29/2020 1/Kw = (1 x 10 -14)-1 = 1 x 1014 27

Neutralization Reactions n The overall neutralization equilibrium constant (Kn) is the product of the two intermediate equilibrium constants (Ka & 1/Kw) KN = Ka x 1/Kw = (1. 7 x 10 -5)(1 x 1014) = 1. 9 x 109 (very large!) n 9/29/2020 Weak acids react completely with strong bases 28

Weak Acid-Strong Base Titration Curve n Consider the reaction between Propanoic Acid (weak acid) and Sodium Hydroxide (Na. OH) (strong base) n Ka for CH 3 CH 2 COOH (HPr) - 1. 3 x 10 -5 n The Titration Curve (see previous slide) Ø The bottom dotted curve corresponds to the strong acid-strong base titration Ø The Curve consists of 4 regions, the first 3 of which differ from the strong acid case l The initial p. H is “Higher” ¬ l 9/29/2020 The weak acid dissociates only slightly producing less [H 3 O+] than with a strong acid The gradually arising portion of the curve before the steep rise to the equivalence point is called the “buffer region” 29

Weak Acid-Strong Base Titration Curve As HPr reacts with the strong base, more and more of the conjugate base (Pr-) forms creating an (HPr/Pr-) buffer At the midpoint of the “Buffer” region, half the original HPr has reacted p. H at midpoint of “Buffer” region is common method of estimating p. Ka of an “unknown” acid 9/29/2020 30

Weak Acid-Strong Base Titration Curve l The p. H at the “equivalence point” is greater than 7. 00 l 9/29/2020 The weak-acid anion (Pr-) acts as a weak base to accept a proton from H 2 O forming OHThe additional [OH-] raises the p. H Beyond the equivalence point, the ph increases slowly as excess OH- is added 31

Weak Base-Strong Acid Titration n Neutralization reactions between weak bases and strong acids (net ionic equation shown below) NH 3(aq) + H 3 O+(aq) NH 4+(aq) + H 2 O(l) KN=? base acid salt water n Above equilibrium also lies very far to the right; the two equilibria below can be added together to provide the overall neutralization reaction shown above NH 3(aq) + H 2 O(aq) NH 4+ + OH-(aq) H 3 O+(aq) + OH-(aq) 2 H 2 O(l) Kb = 1. 8 x 10 -5 1/Kw = 1. 0 x 1014 KN = Kb X 1/Kw = (1. 8 x 10 -5)(1 x 1014) = 1. 8 x 109 (very large!) n Weak bases react completely with strong acids 9/29/2020 32

Weak Base-Strong Acid Titration Curve for Titration of a Weak Base by a Strong Acid 9/29/2020 33

Polyprotic Acid Titrations n n n 9/29/2020 Polyprotic Acids have more than one ionizable proton Except for Sulfuric Acid (H 2 SO 4), the common polyprotic acids are all weak acids Successive Ka values for a polyprotic acid differ by several orders of magnitude The first H+ is lost much more easily than subsequent ones All “ 1 st” protons (H+) are removed before any of the “ 2 nd” protons 34

Polyprotic Acid Titrations Each mole of H+ is titrated separately n In a diprotic acid two OH- ions are required to completely remove both H+ ions n All H 2 SO 3 molecules lose one H+ before any HSO 3 - ions lose a H+ to form SO 3 n Titration curves looks like two weak acid-strong base curves joined end-to-end HSO 3 - is the conjugate base of H 2 SO 3 (Kb = 7. 1 x 10 -13) HSO 3 - is also an acid (Ka=6. 5 x 10 -8) dissociating to form its conjugate base SO 32 - 9/29/2020 35

Slightly Soluble Compounds n Most solutes, even those called “soluble, ” have a limited solubility in a particular solvent n In a saturated solution, at a particular temperature, equilibrium exists between the dissolved and undissolved solute n Slightly soluble ionic compounds, normally called “insoluble, ” reach equilibrium with very little of the solute dissolved n Slightly soluble compounds can produce complex mixtures of species n Discussion here is to assume that the small amount of a slightly soluble solute that does dissolve, dissociates completely 9/29/2020 36

Equilibria - Slightly Soluble Ionic Compounds 9/29/2020 37

Equilibria - Slightly Soluble Ionic Compounds n Equilibrium exists between solid solute and the aqueous ions Pb. SO 4(s) ⇄ Pb 2+(aq) + SO 42 -(aq) n Set up “Reaction Quotient” Note: concentration of a solid - [Pb. SO 4(s)] = 1 n n 9/29/2020 Define Solubility Product When solid Pb. SO 4 reaches equilibrium with Pb 2+ and SO 4 at saturation, the numerical value of Qsp attains a constant value called the solubility-product constant (K sp) 38

Equilibria - Slightly Soluble Ionic Compounds n 9/29/2020 The Solubility Product Constant (Ksp) Ø Value of Ksp depends only on temperature, not individual ion concentrations Ø Saturated solution of a slightly soluble ionic compound, Mp. Xq, composed of ions Mn+ and Xz-, the equilibrium condition is: 39

Equilibria - Slightly Soluble Ionic Compounds 9/29/2020 Ø Single-Step Process Ø Multi-Step Process 40

Equilibria - Slightly Soluble Ionic Compounds 9/29/2020 41

Equilibria - Slightly Soluble Ionic Compounds n 9/29/2020 The Effect of the Common Ion Ø The presence of a “Common Ion” decreases the solubility of a slightly soluble ionic compound Ø Add some Na 2 Cr. O 4, a soluble salt, to the saturated Pb. Cr. O 4 solution l Concentration of Cr. O 42 - increases l Some of excess Cr. O 42 - combines with Pb 2+ to form Pb. Cr. O 4(s) l Equilibrium shifts to the “Left” l This shift “reduces” the solubility of Pb. Cr. O 4 42

Equilibria - Slightly Soluble Ionic Compounds n The Effect of p. H on Solubility Ø The Hydronium Ion (H 3 O+) of a strong acid increases the solubility of a solution containing the anion of a weak acid (HA-) Ø Adding some strong acid to a saturated solution of Calcium Carbonate (Ca. CO 3) introduces large amount of H 3 O+ ion, which reacts immediately with the Ca. CO 3 to form the weak acid HCO 3 - Ø 9/29/2020 Additional H 3 O+ reacts with the HCO 3 - to form carbonic acid, H 2 CO 3, which immediately decomposes to H 2 O and CO 2 43

Equilibria - Slightly Soluble Ionic Compounds n The equilibrium shifts to the “Right” and more Ca. CO 3 dissolves – increased solubility n The overall reaction is: n Adding H 3 O+ to a saturated solution of a compound with a strong acid anion – Ag. Cl 9/29/2020 Ø Chloride ion, Cl-, is the conjugate base of a strong acid (HCl) Ø It coexists with water, i. e. does not react with water Ø There is “No” effect on the equilibrium 44

Equilibria - Slightly Soluble Ionic Compounds n Predicting Formation of a Precipitate Ø Qsp = Ksp when solution is “saturated” Ø Qsp > Ksp solution momentarily “supersaturated” l Ø 9/29/2020 Additional solid precipitates until Qsp = Ksp again Qsp < Ksp solution is “unsaturated” l No precipitate forms at that temperature l More solid dissolves until Qsp = Ksp 45

Equilibria - Slightly Soluble Ionic Compounds n 9/29/2020 Selective Precipitation of Ions Ø Separation of one ion in a solution from another Ø Exploit differences in the solubility of their compounds Ø The Ksp of the less soluble compound is much smaller than the Ksp of the more soluble compound Ø Add solution of precipitating ion until the Qsp value of the more soluble compound is almost equal to its Ksp value Ø The less soluble compound continues to precipitate while the more soluble compound remains dissolved, i. e. , the Ksp of the less soluble compound is always being exceeded, i. e. precipitation is occurring Ø At equilibrium, most of the ions of the less soluble compound have been removed as the precipitate 46

Equilibria - Slightly Soluble Ionic Compounds n Equilibria Involving Complex Ions Ø The product of any Lewis acid-base reaction is called an “Adduct”, a single species that contains a new covalent bond Ø A complex ion consists of a central metal ion covalently bonded to two or more anions or molecules, called ligands Ionic ligands – OH- Cl- CNMolecular ligands – H 2 O CO NH 3 Ex Cr(NH 3)63+ Cr 3+ is the central metal ion 9/29/2020 NH 3 molecules are molecular ligands Ø Metal acts as Lewis Acid by accepting electron pair; ligands acts as Lewis base by donating electron pair Ø All complex ions are Lewis adducts 47

Equilibria - Slightly Soluble Ionic Compounds n Complex Ions Ø Acidic Hydrated metal ions are complex ions with water molecules as ligands Ø When the hydrated cation is treated with a solution of another ligand, the bound water molecules exchange for the other ligand Ø At equilibrium Water is constant in aqueous reactions Ø Incorporate in Kc to define Kf (formation constant) Ø 9/29/2020 48

Equilibria - Slightly Soluble Ionic Compounds n Complex Ions (Con’t) Ø The actual process is stepwise, with ammonia molecules replacing water molecules one at a time to give a series of intermediate species, each with its own formation constant (Kf) The sum of the 4 equations gives the overall equation Ø The product of the individual formation constants gives the overall formation constant Ø The Kf for each step is much larger than “ 1” because “ammonia” is a stronger Lewis base than H 2 O Ø Ø 9/29/2020 Adding excess NH 3 replaces all H 2 O and all M 2+ exists as M(NH 3)42+ 49

Complex Ions – Solubility of Precipitates n Increasing [H 3 O+] increases solubility of slightly soluble ionic compounds if the anion of the compound is that of a weak acid n A Ligand increases the solubility of a slightly soluble ionic compound if it forms a complex ion with the cation n When 1. 0 M Na. CN is added to the above solution, the CN - ions act as ligands and react with the small amount of Zn 2+(aq) to form a complex ion 9/29/2020 50

Complex Ions – Solubility of Precipitates n Add the two reactions and compute the overall K n The overall equilibrium constant, Koverall , is more than a factor of 1019 larger than the original Ksp (2. 0 x 10 -22) n This reflects the increased amount of Zn. S in solution as Zn(CN 42 -) 9/29/2020 51

Ionic Equilibria in Aqueous Solutions n Equilibria Involving Complex Ions Ø Amphoteric Oxides & Hydroxides (Recall Chapter 8) l Some metals and many metalloids form oxides or hydroxides that are amphoteric; they can act as acids or bases in water l These compounds generally have very little solubility in water, but they do dissolve more readily in acids or bases Ex. Aluminum Hydroxide Al(OH)3(s) ⇆ Al 3+(aq) + 3 OH-(ag) Ksp = 3 x 10 -34 (very insoluble in water) In acid solution, the OH- reacts with H 3 O+ to form water 3 H 3 O+(ag) + 3 OH-(aq) 6 H 2 O(l) Al(OH)3(s) + H 3 O+(aq) Al 3+(aq) + 6 H 2 O(l) 9/29/2020 52