Applications of Aqueous Equilibria Buffers AcidBase Titrations and

Applications of Aqueous Equilibria Buffers, Acid-Base Titrations and Precipitation Reactions

Buffers are solutions of a weak acid and its conjugate base. Buffers resist changes in p. H when small amounts of strong acid or base are added. This is because there is both an acid and its conjugate base present initially. Buffers typically have the acid and its conjugate base (or salt) present in roughly equal concentrations.

Buffers are extremely important in regulating biological processes, as many biochemical reactions are p. H dependent. Blood is a buffer composed of dissolved carbon dioxide (which forms carbonic acid, H 3 CO 3), bicarbonate ion (HCO 3–), and carbonate ion (CO 32 -). The buffer maintains the p. H of blood at a value of between 7. 35 -7. 45.

Buffers Consider a buffer of weak acid HA and Na. A in equimolar concentrations. The predominant reaction is: HA(aq) + H 2 O(l) ↔ H 3 O+(aq) + A-(aq) If a small amount of strong acid is added, there is enough A- available to neutralize the acid and become protonated. The equilibrium shifts to the left, and the [H 3 O+] and p. H do not change much.

Buffers HA(aq) + H 2 O(l) ↔ H 3 O+(aq) + A-(aq) Likewise, if a small amount of strong base is added, some of the hydronium will react with it. H 3 O+(aq) + OH-(aq) 2 H 2 O(l) As hydronium ion reacts, more HA will dissociate so as to replenish the [H 3 O+]. As a result, the p. H remains fairly constant.

Buffers Because buffers contain both a weak acid and its conjugate base, they can react with strong acids or bases and maintain their p. H.

Buffers

Problem: Buffers n Determine the p. H of a solution that contains 0. 50 M HCl. O and 0. 60 M Na. Cl. O. (Ka for HCl. O = 3. 5 x 10 -8) - Write the major reactions: Na. Cl. O(aq) Na+(aq) + Cl. O-(aq) The Na. Cl. O serves as a source of Cl. Oions. The sodium ion is a spectator.

Problem: Buffers n Determine the p. H of a solution that contains 0. 50 M HCl. O and 0. 60 M Na. Cl. O. Even though both HCl. O and Cl. O- are present initially, it is important to realize that they are in equilibrium with each other, and go on opposite sides of the equation. HCl. O(aq) + H 2 O(l) ↔ H 3 O+(aq) + Cl. O(aq)

Problem: Buffers n Determine the p. H of a solution that contains 0. 50 M HCl. O and 0. 60 M Na. Cl. O. (Ka for HCl. O = 3. 5 x 10 -8) HCl. O(aq) + H 2 O(l) ↔ H 3 O+(aq) + Cl. O(aq) Ka = [H 3 O+][Cl. O-] = 3. 5 x 10 -8 [HCl. O]

Problem: Buffers n Determine the p. H of a solution that contains 0. 50 M HCl. O and 0. 60 M Na. Cl. O. (Ka for HCl. O = 3. 5 x 10 -8) HCl. O(aq) + H 2 O(l) ↔ H 3 O+(aq) + Cl. O( aq ) + HCl. O H O Cl. O 3 initial 0. 50 chang -x e equil. . 50 -x 0 0. 60 +x +x x . 60+x

The Henderson-Hasselbalch Equation For buffers, regardless of the value of Ka, you can usually assume that x will be small compared to the concentrations of the acid or its conjugate base. This is because the reaction will proceed only slightly to the right due to the presence of the conjugate base initially. HA(aq) + H 2 O(l) ↔ H 3 O+(aq) + A-(aq)

The Henderson-Hasselbalch Equation As a result, the Ka expression can be written in logarithmic form. Ka = [H 3 O+][A-] [HA] [H 3 O+] = Ka [HA] [A-] p. H = p. Ka + log [A-] [HA]

The Henderson-Hasselbalch Equation In more general terms, the equation is written as: [base] p. H = p. Ka + log [acid] This equation is for buffers only, and is a quick way to calculate p. H.

Problem: Adding Acid to a Buffer n Calculate the change in p. H if 5. 00 m. L of 0. 010 M HNO 3 is added to 50. 0 m. L of the HCl. O/Na. Cl. O buffer. The nitric acid will protonate some of the Cl. O- to form additional HCl. O. HNO 3(aq) + H 2 O(l) H 3 O+(aq) + NO 3 -(aq) H 3 O+(aq) + Cl. O-(aq) HCl. O (aq) + H 2 O(l)

Problem: Adding Acid to a Buffer HNO 3(aq) + H 2 O(l) H 3 O+(aq) + NO 3 -(aq) H 3 O+(aq) + Cl. O-(aq) HCl. O (aq) + H 2 O(l) Before you can use the Henderson. Hasselbalch equation, you need to calculate the new concentration of HCl. O and Cl. O- that results from adding the nitric acid.

Problem: Adding Acid to a Buffer HNO 3(aq) + H 2 O(l) H 3 O+(aq) + NO 3 -(aq) H 3 O+(aq) + Cl. O-(aq) HCl. O (aq) + H 2 O(l) For each mole of nitric acid added, you form a mole of HCl. O and react a mole of Cl. O-.

Problem: Adding Acid to a Buffer 1. You will need to calculate the moles of HNO 3 added (5. 00 m. L of 0. 010 M HNO 3). 2. For each mole of nitric acid added, you will produce a mole of HCl. O and lose a mole of Cl. O-. 3. Since you need concentrations, remember that the new volume of the solution has changed to 55. 0 m. L to 50. 0 m. L.

Problem: Compare to Water n Calculate the change in p. H when 5. 00 m. Ls of 0. 0100 M HNO 3 is added to 50. 0 m. L of water.

Preparation of Buffer Solutions Scientists often need to make a solution that is buffered to a specific p. H. They may need to mimic biological conditions, test the corrosiveness of metal parts at a specific p. H, etc. Conjugate acid/base pairs have a p. H range at which they can effectively serve as buffers.

Choosing the Acid and Salt For optimum buffering, choose an acid with a Ka value close to the desired [H 3 O+] of the buffer, or with a p. Ka value near the desired p. H. You can then use the Henderson. Hasselbalch equation (or Ka expression) to determine the relative concentration of base to acid needed to prepare the buffer.

Preparing a Buffer n Choose an appropriate acid and base to make a buffer with a p. H of 6. 50. Calculate the relative concentration of acid and base needed.

Acid-Base Titrations Titration is a laboratory technique in which the amount and concentration of one reactant is known, and the concentration of the other reactant is determined. Titrations can be applied to redox reactions, and precipitation reactions, but they are most commonly used for analyzing solutions of acids or bases.

Acid-Base Titrations Acid-base titrations are based on neutralization reactions, in which an acid is completely neutralized by a base. The progress of the reaction may be monitored using a p. H meter, or p. H indicators.

Titrations In either method, base is added to the acid (or vice versa) until the equivalence point is reached. The equivalence point has been reached when the moles of acid exactly equals the moles of base. It is sometime called the stoichiometric point.

Titrations Not all titrations reach the equivalence point at a p. H of 7. The nature of the acid and base (weak or strong) will influence the p. H at the equivalence point. As a result, indicators much be carefully chosen to change color at the desired p. H.

Strong Acid & Strong Base When a strong acid is titrated with a strong base, a “neutral” salt and water result. Since the conjugate base of a strong acid has no tendency to accept protons, and the conjugate acid of a strong base has no tendency to donate protons, the p. H at the equivalence point will be 7.

Strong Acid & Strong Base If the p. H is monitored during the titration using a p. H meter, a graphical presentation of the p. H versus volume of titrant can be obtained.

Strong Acid & Strong Base The graph has certain features characteristic of a strong acid-strong base titration.

Strong Acid & Strong Base Note the equivalence point of 7, and the nearly vertical rise in p. H from 4 -10.

Strong Acid & Strong Base Because the rise in p. H is vertical over such a broad range, there are several indicators that may be used with accurate results.

Strong Acid & Strong Base Phenolphthalei n, which changes color at a p. H of around 9, and methyl red, which changes color at a p. H of approximately 5 will both give highly accurate

Strong Acid & Strong Base Phenolphthalei n is usually used because it goes from colorless to pink, and our eyes are better at detecting the presence of color than a change in color.

Weak Acid & Strong Base When a weak acid is titrated with a strong base, the salt that is produced will contain the conjugate base of the weak acid. As a result, the p. H will be greater than 7 at the equivalence point.

Weak Acid & Strong Base When a weak acid is titrated with a strong base, the salt that is produced will contain the conjugate base of the weak acid. As a result, the p. H will be greater than 7 at the equivalence point.

Weak Acid & Strong Base The titration curve for acetic acid with Na. OH shows an equivalence point at a p. H of 9. The lack of a steep vertical rise in p. H means that selection of the proper indicator is crucial.

Weak Acid & Strong Base Another feature of the curve is the flattening out or leveling off of p. H near the half-way point. This is not observed in strong acid-strong base titrations.

Weak Acid & Strong Base Another feature of the curve is the flattening out or leveling off of p. H near the half-way point. This is not observed in strong acid-strong base titrations.

Weak Acid & Strong Base The p. H levels off due to the formation of a buffer mid-way through the titration. Half of the acid has been deprotonated to form its conjugate base.

Titration of Benzoic Acid n 25. 0 m. L of 0. 10 M benzoic acid (Ka = 6. 4 x 10 -5) is titrated with 0. 10 M Na. OH. Calculate the p. H: - Initially - After 12. 5 m. L of Na. OH have been added - At the equivalence point Choose an appropriate indicator for the titration.

Titration of Benzoic Acid n 25. 0 m. L of 0. 10 M benzoic acid (Ka = 6. 4 x 10 -5) is titrated with 0. 10 M Na. OH. We will need to know how much Na. OH is needed for complete neutralization. In this case, since the concentrations of acid and base are the same, 25. 0 m. L of 0. 10 M acid will require an equal volume of 0. 10 M base.

Calculating Volume for Neutralization In less obvious cases, the following relationship can be used. We are assuming complete neutralization of a monoprotic acid by a monobasic base. moles acid = moles base Macid. Vacid = Mbase. Vbase

Titration of Benzoic Acid 25. 0 m. L of 0. 10 M benzoic acid (Ka = 6. 4 x 10 -5) is titrated with 0. 10 M Na. OH. Calculate the p. H: - Initially Major Reaction: HA(aq) + H 2 O(l) ↔ H 3 O+(aq) + A-(aq) init: 0. 10 0 0 chg: -x +x +x equ. . 10 -x x x n

Titration of Benzoic Acid 25. 0 m. L of 0. 10 M benzoic acid (Ka = 6. 4 x 10 -5) is titrated with 0. 10 M Na. OH. Calculate the p. H: - After 12. 5 m. L of Na. OH have been added Titration Reaction: HA(aq) + OH-(aq) H 2 O(l) + A-(aq) For each mole of Na. OH added, a mole of HA has been converted to A-. n

Titration of Benzoic Acid n 25. 0 m. L of 0. 10 M benzoic acid (Ka = 6. 4 x 10 -5) is titrated with 0. 10 M Na. OH. Calculate the p. H: - After 12. 5 m. L of Na. OH have been added This problem can be greatly simplified if you realize that this is the mid-point of the titration. Since complete neutralization will require 25. 0 m. Ls of Na. OH, at 12. 5 m. L base, half of the acid has been converted

Titration of Benzoic Acid n 25. 0 m. L of 0. 10 M benzoic acid (Ka = 6. 4 x 10 -5) is titrated with 0. 10 M Na. OH. Calculate the p. H: - After 12. 5 m. L of Na. OH have been added In essence, a buffer has been formed, with [HA] = [A-]. (This is the place in the titration curve where the p. H begins to level off. )

Titration of Benzoic Acid n 25. 0 m. L of 0. 10 M benzoic acid (Ka = 6. 4 x 10 -5) is titrated with 0. 10 M Na. OH. Calculate the p. H: - After 12. 5 m. L of Na. OH have been added We can use the Henderson. Hasselbalch equation, because a buffer is present. Since [HA] = [A-], we need not calculate the actual concentration, as they

Titration of Benzoic Acid n 25. 0 m. L of 0. 10 M benzoic acid (Ka = 6. 4 x 10 -5) is titrated with 0. 10 M Na. OH. Calculate the p. H: - After 12. 5 m. L of Na. OH have been added p. H = p. Ka + log ([A-]/[HA]) p. H = -log (6. 4 x 10 -5) + log (1. 0) p. H = 4. 19

Titration of Benzoic Acid Calculate the p. H at the equivalence point. At the equivalence point, enough Na. OH has been added (25. 0 m. L) to completely neutralize the benzoic acid. There are two reactions to consider. 1. The neutralization reaction. 2. The subsequent reaction of the benzoate ion with water. n

Titration of Benzoic Acid n At the equivalence point, benzoate ion will react with water, since it is the conjugate base of a weak acid. Major reactions: HA(aq) + OH-(aq) H 2 O(l) + A-(aq) + H 2 O(aq) ↔ OH-(aq) + HA(aq)

Titration of Benzoic Acid - At the equivalence point, the solution will be basic, due to the high concentration of benzoate ion. Major Reaction: A-(aq) + H 2 O(aq) ↔ OH-(aq) + HA(aq) The [A-] equals the number of moles of HA initially present divided by the total volume of solution at the equivalence point.

Titration of Benzoic Acid Major Reaction: A-(aq) + H 2 O(aq) ↔ OH-(aq) + HA(aq) The [A-] equals the number of moles of HA initially present divided by the total volume of solution at the equivalence point. [A-] = 25. 0 m. L(0. 10 M)/50. 0 m. L = 0. 050 M

Titration of Benzoic Acid Major Reaction: A-(aq) + H 2 O(aq) ↔ OH-(aq) + HA(aq) init. 0. 050 0 0 chg. -x +x +x equil. 050 -x x x Since the reaction involves a conjugate base reacting with water, we need to write the Kb expression, and calculate the value of Kb.

Titration of Benzoic Acid Major Reaction: A-(aq) + H 2 O(aq) ↔ OH-(aq) + HA(aq) Since the reaction involves a conjugate base reacting with water, we need to write the Kb expression, and calculate the value of Kb. Kb = [OH-][HA] = Kw [A-] Ka Kb = 1. 0 x 10 -14/ 6. 4 x 10 -5 = 1. 6 x 10 -10

Titration of Benzoic Acid Major Reaction: A-(aq) + H 2 O(aq) ↔ OH-(aq) + HA(aq) init. 0. 050 0 0 chg. -x +x +x equil. 050 -x x x 1. 6 x 10 -10 = (x)(x)/(. 050 -x)

Titration of Benzoic Acid The p. H at the equivalence point will be 8. 45. Selection of a indicator must be made carefully, as there isn’t a steep vertical rise in p. H near the end point when titrating a weak

Titration of Weak Acids As the acid being titrated gets weaker, the titration curve flattens out, and the end point becomes less obvious.

Titration of Weak Acids Very weak acids cannot be analyzed using titration.

Titration of a Weak Base NH 3(aq) + HCl(aq) NH 4+(aq) + Cl-(aq) The titration of ammonia with a strong acid will produce a solution of ammonium ion. Since ammonium ion is the conjugate acid of a weak base, the p. H at the equivalence point will be less than 7.

Titration of a Weak Base Again, selection of the proper indicator is crucial.