Chapter 16 Equilibria in Aqueous Systems
Buffers! buffers are solutions that resist changes in ph when an acid or base is added! they act by neutralizing the added acid or base! but just like everything else, there is a limit to what they can do, eventually the ph changes! many buffers are made by mixing a solution of a weak acid with a solution of soluble salt containing its conjugate base anion
Making a Buffer Solution from a Weak Acid
How Acid Buffers Work HA (aq) + H 2 O (l)! A! (aq) + H 3 O+ (aq)! buffers work by applying Le Châtelier s Principle to weak acid equilibrium! buffer solutions contain significant amounts of the weak acid molecules, HA these molecules react with added base to neutralize it!!you can also think of the H 3 O + combining with the OH! to make H 2 O; the H 3 O + is then replaced by the shifting equilibrium! the buffer solutions also contain significant amounts of the conjugate base anion, A! - these ions combine with added acid to make more HA and keep the H 3 O + constant
Common Ion Effect HA (aq) + H 2 O (l)! A! (aq) + H 3 O+ (aq)! adding a salt containing the anion, NaA, that is the conjugate base of the acid (the common ion) shifts the position of equilibrium to the left! this causes the ph to be higher than the ph of the acid solution!!lowering the H 3 O + ion concentration
The Effect of Added Acetate Ion on the Dissociation of Acetic Acid [CH 3 COOH] initial [CH 3 COO - ] added % Dissociation* ph 0.10 0.00 0.10 0.050 0.10 0.10 1.3 0.036 0.018 2.89 4.44 4.74 0.10 0.15 0.012 4.92 * % Dissociation = [CH 3 COOH] dissoc [CH 3 COOH] initial x 100
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 -
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]
Buffering Effectiveness! a good buffer should be able to neutralize moderate amounts of added acid or base! however, there is a limit to how much can be added before the ph changes significantly! the buffering capacity is the amount of acid or base a buffer can neutralize! the buffering range is the ph range the buffer can be effective! the effectiveness of a buffer depends on two factors (1) the relative amounts of acid and base, and (2) the absolute concentrations of acid and base
Effectiveness of Buffers! a buffer will be most effective when the [base]: [acid] = 1!!equal concentrations of acid and base! effective when 0.1 < [base]:[acid] < 10! a buffer will be most effective when the [acid] and the [base] are large
! we have said that a buffer will be effective when 0.1 < [base]:[acid] < 10! substituting into the Henderson-Hasselbalch we can calculate the maximum and minimum ph at which the buffer will be effective & - # $ [A ] ph = pk a + log! %[HA]" ph ph = Lowest ph pk + log ( 0.10 ) = pk a a! 1 Buffering Range ph ph Highest ph = pk + log10 ( ) = pk a a + 1 therefore, the effective ph range of a buffer is pk a ± 1 when choosing an acid to make a buffer, choose one whose is pk a is closest to the ph of the buffer
Buffering Capacity! buffering capacity is the amount of acid or base that can be added to a buffer without destroying its effectiveness! the buffering capacity increases with increasing absolute concentration of the buffer components! as the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves! buffers that need to work mainly with added acid generally have [base] > [acid]! buffers that need to work mainly with added base generally have [acid] > [base]
How Much Does the ph of a Buffer Change When an Acid or Base Is Added?! though buffers do resist change in ph when acid or base are added to them, their ph does change! calculating the new ph after adding acid or base requires breaking the problem into 2 parts 1.! a stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other!! added acid reacts with the A! to make more HA!! added base reacts with the HA to make more A! 2.! an equilibrium calculation of [H 3 O + ] using the new initial values of [HA] and [A! ]
Basic Buffers B: (aq) + H 2 O (l)! H:B + (aq) + OH! (aq)! buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B + Cl! H 2 O (l) + NH 3 (aq)! NH 4 + (aq) + OH! (aq)
The Common Ion Effect
Titration! in an acid-base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete!!when the reaction is complete we have reached the endpoint of the titration! an indicator may be added to determine the endpoint!!an indicator is a chemical that changes color when the ph changes! when the moles of H 3 O + = moles of OH!, the titration has reached its equivalence point
Titration
Color Changes for Various Acid-Base Indicators
Acid-Base Indicators
Titration Curve! a plot of ph vs. amount of added titrant! the inflection point of the curve is the equivalence point of the titration! prior to the equivalence point, the known solution in the flask is in excess, so the ph is closest to its ph! the ph of the equivalence point depends on the ph of the salt solution!!equivalence point of neutral salt, ph = 7!!equivalence point of acidic salt, ph < 7!!equivalence point of basic salt, ph > 7! beyond the equivalence point, the unknown solution in the burette is in excess, so the ph approaches its ph
Curve for a strong acid-strong base titration
14 Titration Curve of Weak Acid with NaOH ph 12 10 8 ph 6 4 2 Show First Derivative 0 0 5 10 15 20 25 30 35 40 45 50 Volume NaOH Added, ml I
Curve for a weak acidstrong base titration Titration of 40.00mL of 0.1000M HPr with 0.1000M NaOH pk a of HPr = 4.89 ph = 8.80 at equivalence point [HPr] = [Pr - ] methyl red!
Titrating Weak Acid with a Strong Base! the initial ph is that of the weak acid solution!!calculate like a weak acid equilibrium problem "!e.g., 15.5 and 15.6! before the equivalence point, the solution becomes a buffer!!calculate mol HA init and mol A! init using reaction stoichiometry!!calculate ph with Henderson-Hasselbalch using mol HA init and mol A! init! half-neutralization ph = pk a
Titrating Weak Acid with a Strong Base! at the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established!!mol A! = original mole HA "!calculate the volume of added base like Ex 4.8!![A! ] init = mol A! /total liters!!calculate like a weak base equilibrium problem "!e.g., 15.14! beyond equivalence point, the OH is in excess!![oh! ] = mol MOH xs/total liters!![h 3 O + ][OH! ]=1 x 10-14
Comparison of Titrations
Various weak acid-strong base titrations
Titration of 40.00mL of 0.1000M NH 3 with 0.1000M HCl pk a of NH 4 + = 9.25 Curve for a weak basestrong acid titration ph = 5.27 at equivalence point
Weak base-strong acid titration
Curve for the titration of a weak polyprotic acid. pk a = 7.19 pk a = 1.85 Titration of 40.00mL of 0.1000M H 2 SO 3 with 0.1000M NaOH
Titration of a polyprotic acid
Solubility Equilibria! all ionic compounds dissolve in water to some degree!!however, many compounds have such low solubility in water that we classify them as insoluble! we can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water
Solubility Product! the equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, K sp! for an ionic solid M n X m, the dissociation reaction is: M n X m (s)! nm m+ (aq) + mx n! (aq)! the solubility product would be K sp = [M m+ ] n [X n! ] m! for example, the dissociation reaction for PbCl 2 is PbCl 2 (s)! Pb 2+ (aq) + 2 Cl! (aq)! and its equilibrium constant is K sp = [Pb 2+ ][Cl! ] 2
Molar Solubility! solubility is the amount of solute that will dissolve in a given amount of solution!!at a particular temperature! the molar solubility is the number of moles of solute that will dissolve in a liter of solution!!the molarity of the dissolved solute in a saturated solution! for the general reaction M n X m (s)! nm m+ (aq) + mx n! (aq) Molar Solubility =
K sp and Relative Solubility! molar solubility is related to K sp! but you cannot always compare solubilities of compounds by comparing their K sp s! in order to compare K sp s, the compounds must have the same dissociation stoichiometry
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 10-34 1.0 x 10-10 4.1 x 10-15 3.6 x 10-8 1.6 x 10-8 4.7 x 10-29 8 x 10-48 3.9 x 10-6
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 10-8 1.9 x 10-4 2 PbSO 4 1:1 1.6 x 10-8 1.3 x 10-4 2 BaCrO 4 1:1 2.1 x 10-10 1.4 x 10-5 3 Ca(OH) 2 1:2 5.5 x 10-6 1.2 x 10-2 3 BaF 2 1:2 1.5 x 10-6 7.2 x 10-3 3 CaF 2 1:2 3.2 x 10-11 2.0 x 10-4 3 Ag 2 CrO 4 2:1 2.6 x 10-12 8.7 x 10-5
The Effect of Common Ion on Solubility! addition of a soluble salt that contains one of the ions of the insoluble salt, decreases the solubility of the insoluble salt! for example, addition of NaCl to the solubility equilibrium of solid PbCl 2 decreases the solubility of PbCl 2 PbCl 2 (s)! Pb 2+ (aq) + 2 Cl! (aq) addition of Cl! shifts the equilibrium to the left
The Effect of ph on Solubility! for insoluble ionic hydroxides, the higher the ph, the lower the solubility of the ionic hydroxide!!and the lower the ph, the higher the solubility!!higher ph = increased [OH! ] M(OH) n (s)! M n+ (aq) + noh! (aq)! for insoluble ionic compounds that contain anions of weak acids, the lower the ph, the higher the solubility M 2 (CO 3 ) n (s)! 2 M n+ (aq) + nco 3 2! (aq) H 3 O + (aq) + CO 3 2! (aq)! HCO 3! (aq) + H 2 O(l)
Precipitation! precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound! if we compare the reaction quotient, Q, for the current solution concentrations to the value of K sp, we can determine if precipitation will occur!!q = K sp, the solution is saturated, no precipitation!!q < K sp, the solution is unsaturated, no precipitation!!q > K sp, the solution would be above saturation, the salt above saturation will precipitate! some solutions with Q > K sp will not precipitate unless disturbed these are called supersaturated solutions
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)
Cr(NH 3 ) 6 3+, a typical complex ion.
The stepwise exchange of NH 3 for H 2 O in M(H 2 O) 4 2+. NH 3 M(H 2 O) 4 2+ 3NH 3 M(H 2 O) 3 (NH 3 ) 2+ M(NH 3 ) 4 2+
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)
Solubility of amphoteric hydroxides
The general procedure for separating ions in qualitative analysis. Add precipitating ion Add precipitating ion Centrifuge Centrifuge