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Ionic Equilibria in Aqueous Systems

Ionic Equilibria in Aqueous Systems. Ionic Equilibria in Aqueous Systems. 19.1 Equilibria of Acid-Base Buffers. 19.2 Acid-Base Titration Curves. 19.3 Equilibria of Slightly Soluble Ionic Compounds. 19.4 Equilibria Involving Complex Ions. Acid-Base Buffers.

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Ionic Equilibria in Aqueous Systems

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  1. Ionic Equilibria in Aqueous Systems

  2. Ionic Equilibria in Aqueous Systems 19.1 Equilibria of Acid-Base Buffers 19.2 Acid-Base Titration Curves 19.3 Equilibria of Slightly Soluble Ionic Compounds 19.4 Equilibria Involving Complex Ions

  3. Acid-Base Buffers An acid-base buffer is a solution that lessens the impact of pH from the addition of acid or base. An acid-base buffer usually consists of a conjugate acid-base pair where both species are present in appreciable quantities in solution. An acid-base buffer is therefore a solution of a weak acid and its conjugate base (e.g., HOAC/OAc-), or a weak base and its conjugate acid (e.g., NH3/NH4+).

  4. The effect of adding acid or base to an unbuffered solution. A 100-mL sample of dilute HCl is adjusted to pH 5.00. The addition of 1 mL of strong acid (left) or strong base (right) changes the pH by several units.

  5. The effect of adding acid or base to a buffered solution. A 100-mL sample of an acetate buffer is adjusted to pH 5.00. The addition of 1 mL of strong acid (left) or strong base (right) changes the pH very little. The acetate buffer is made by mixing 1 M CH3COOH ( a weak acid) with 1 M CH3COONa (which provides the conjugate base, CH3COO-).

  6. Buffers and the Common-ion Effect A buffer works through the common-ion effect. Acetic acid in water dissociates slightly to produce some acetate ion: CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) acetic acid acetate ion If NaCH3COO is added, it provides a source of CH3COO- ion, and the equilibrium shifts to the left. CH3COO- is common to both solutions. The addition of CH3COO- reduces the % dissociation of the acid.

  7. The Effect of Added Acetate Ion on the Dissociation of Acetic Acid [CH3COOH]dissoc [CH3COOH]init * % Dissociation = x 100

  8. How a Buffer Works The buffer components (HA and A-) are able to consume small amounts of added OH- or H3O+by a shift in equilibrium position. CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) Added H3O+ reacts with CH3COO-, causing a shift to the left. Added OH- reacts with CH3COOH, causing a shift to the right. The shift in equilibrium position absorbs the change in [H3O+] or [OH-], and the pH changes only slightly.

  9. H3O+ OH- How a buffer works. Buffer has more HA after addition of H3O+. Buffer has equal concentrations of A- and HA. Buffer has more A- after addition of OH-. H2O + CH3COOH ← H3O+ + CH3COO- CH3COOH + OH- → CH3COO- + H2O

  10. Relative Concentrations of Buffer Components CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) Since Ka is constant, the [H3O+] of the solution depends on the ratio of buffer component concentrations. [CH3COO-][H3O+] [CH3COOH] Ka = [H3O+] = Ka x If the ratio decreases, [H3O+] decreases. If the ratio increases, [H3O+] increases. [HA] [A-] [HA] [A-] [CH3COOH] [CH3COO-]

  11. The Henderson-Hasselbalch Equation HA(aq) + H2O(l) A-(aq) + H3O+(aq) [HA] [A-] [H3O+][A-] [HA] -log[H3O+] = -logKa – log Ka = [H3O+] = Ka x [base] [acid] pH = pKa + log [HA] [A-]

  12. Calculating the pH of a Buffer solution (Assume the additions cause a negligible change in volume.) PROBLEM: Calculate the pH: (a) Of a buffer solution consisting of 0.50 M CH3COOH (HOAc) and 0.50 M CH3COONa (NaOAc) pH = 4.74

  13. Calculating the Effect of Added OH- on Buffer pH (pKa = 4.74) PROBLEM: (b) After adding 0.020 mol of solid NaOH to 1.0 L of the buffer solution Consider chemical reaction first. Which, NaOAc or HOAc, is the Reagent? How much excess reagent remains after reaction? THE ICE table: pH = 4.77

  14. Calculating the Effect of Added H3O+ on Buffer pH (c) After adding 0.020 mol of HCl to 1.0 L of the buffer solution in (a). Ka of CH3COOH = 1.8 x 10-5. PROBLEM: Consider which, NaOAc or HOAc, is the Reagent? How much excess reagent remains after reaction? The ICE table again (group practice) pH = 4.70

  15. Buffer Capacity The buffer capacity is a measure of the “strength” of the buffer, its ability to maintain the pH following addition of strong acid or base. The greater the concentrations of the buffer components, the greater its capacity to resist pH changes. The closer the component concentrations are to each other, the greater the buffer capacity.

  16. The relation between buffer capacity and pH change. When strong base is added, the pH increases least for the most concentrated buffer. This graph shows the final pH values for four different buffer solutions after the addition of strong base.

  17. Buffer Range The buffer range is the pH range over which the buffer is effective. Buffer range is related to the ratio of buffer component concentrations. [HA] [A-] The closer is to 1, the more effective the buffer. If one component is more than 10 times the other, buffering action is poor. Since log10 = 1, buffers have a usable range within ± 1 pH unit of the pKa of the acid component.

  18. Using Molecular Scenes to Examine Buffers PROBLEM: The molecular scenes below represent samples of four HA/A- buffers. (HA is blue and green, A- is green, and other ions and water are not shown.) (a) Which buffer has the highest pH? (b) Which buffer has the greatest capacity? (c) Should we add a small amount of concentrated strong acid or strong base to convert sample 1 to sample 2 (assuming no volume changes)?

  19. Preparing a Buffer • Choose the conjugate acid-base pair. • The pKa of the weak acid component should be close to the desired pH. • Calculate the ratio of buffer component concentrations. • Determine the buffer concentration, and calculate the required volume of stock solutions and/or masses of components. • Mix the solution and correct the pH. [base] [acid] pH = pKa + log

  20. 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 Na2CO3 must she add to 1.5 L of freshly prepared 0.20 M NaHCO3 to make the buffer? Ka of HCO3- is 4.7x10-11. Preparing a Buffer 15 g Na2CO3

  21. Acid-Base Indicators An acid-base indicator is a weak organic acid (HIn) whose color differs from that of its conjugate base (In-). The ratio [HIn]/[In-] is governed by the [H3O+] of the solution. Indicators can therefore be used to monitor the pH change during an acid-base reaction. The color of an indicator changes over a specific, narrow pH range, a range of about 2 pH units.

  22. pH Colors and approximate pH range of some common acid-base indicators.

  23. The color change of the indicator bromthymol blue. pH < 6.0 pH = 6.0-7.5 pH > 7.5

  24. Acid-Base Titrations In an acid-base titration, the concentration of an acid (or a base) is determined by neutralizing the acid (or base) with a solution of base (or acid) of known concentration. The equivalence point of the reaction occurs when the number of moles of OH- added equals the number of moles of H3O+ originally present, or vice versa. The end point occurs when the indicator changes color. - The indicator should be selected so that its color change occurs at a pH close to that of the equivalence point.

  25. How to calculate pH during an Acid-Base Titrations: HA + OH- Setting up an ICE table for the neutralization reaction: H+ + OH- H2O or HA + OH- A- + H2O When an acid and a base is mixed, the neutralization reaction proceeds quickly to reach completion. The limiting reagent is consumed instantaneously (zero molarity in the ICE table). Molarity of each species in the solution is affected by the increased volume (VHA + VOH-). After reaching the equivalence point, pH of the titration mixture depends on the molarity of excessreagent.

  26. Curve for a strong acid–strong base titration. The pH increases gradually when excess base has been added. The pH rises very rapidly at the equivalence point, which occurs at pH = 7.00. The initial pH is low.

  27. Calculating the pH during a strong acid–strong base titration Initial pH [H3O+] = [HA]init pH = -log[H3O+] mol H3O+remaining Vacid + Vbase pH before equivalence point initial mol H3O+ = Vacid x Macid mol OH- added = Vbase x Mbase mol H3O+remaining = (mol H3O+init) – (mol OH-added) [H3O+] = pH = -log[H3O+]

  28. Calculating the pH during a strong acid–strong base titration (contd.) pH at the equivalence point pH = 7.00 for a strong acid-strong base titration. mol OH-excess Vacid + Vbase pH beyond the equivalence point initial mol H3O+ = Vacid x Macid mol OH- added = Vbase x Mbase mol OH-excess = (mol OH-added) – (mol H3O+init) [OH-] = pOH = -log[OH-] and pH = 14.00 - pOH

  29. Example: 40.00 mL of 0.1000 M HCl is titrated with 0.1000 M NaOH. Calculate The initial pH [H+] = [HCl] = 0.1000 M, pH = 1.00 The pH after 20.00 mL of NaOH solution has been added: First, consider the combined volume after addition of NaOH. ICE table: x = 0.03334 M, [H+] = 0.03334 M, pH = 1.48

  30. Example: 40.00 mL of 0.1000 M HCl is titrated with 0.1000 M NaOH. Calculate the pH after 50.00 mL of NaOH solution has been added. First, consider the combined volume after addition of NaOH. ICE table: x = 0.04444 M, [OH-] = 0.01111 M, pH = 12.05

  31. Titration Curve for a weak acid–strong base titration The pH increases slowly beyond the equivalence point. The curve rises gradually in the buffer region. The weak acid and its conjugate base are both present in solution. The pH at the equivalent point is > 7.00 due to the reaction of the conjugate base with H2O. The initial pH is higher than for the strong acid solution.

  32. Calculating the pH during a weak acid–strong base titration: HA + OH-  H2O + A- Before titration: Weak acid only Valid approximation? pH = -log[H3O+] pH before equivalence point (Vbase < Veq): conjugate base A- (from rxn of HA and added OH- ) and excess HA yields a Buffer

  33. Calculating the pH during a weak acid–strong base titration (contd.) pH at the equivalence point: Ionization of the conjugate base A- Where Finally pH beyond the equivalence point depends on excess OH- A-(aq) + H2O(l) HA(aq) + OH-(aq)

  34. Example: Calculate the pH during the titration of 40.00 mL of 0.1000 M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of 0.1000 M NaOH: (a) 0.00 mL; Finding the pH During a Weak Acid–Strong Base Titration Solution: Note that the neutralization rxn of HA with NaOH is instantaneous. Find Veq first. • Before titration begins, only weak acid HPr exists • [H+]  = 1.1 10-3M, • Is the approximation valid? • [H+]/[HA]0 = 0.011 < 5%, so yes • pH = -log[H+] = 2.96

  35. Example: Calculate the pH during the titration of 40.00 mL of 0.1000 M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of 0.1000 M NaOH: (b) 30.00 mL Finding the pH During a Weak Acid–Strong Base Titration (contd.) Solution: (b) Note that the outcome of the reaction depends on the limiting reagent. Use of ICE table (solution table), I + C = E Find [HA] and [OH-] after mixing HA with NaOH (before reaction): Volume after mixing = 40.00 mL + 30.00 mL = 70.00 mL [HA] = 40.00 mL HA x 0.1000 M/70.00mL = 0.05714 M [OH-] = 30.00 mL x 0.1000 M/70.00 mL = 0.04286 M

  36. Example: Calculate the pH during the titration of 40.00 mL of 0.1000 M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of 0.1000 M NaOH: (b) 30.00 mL Finding the pH During a Weak Acid–Strong Base Titration (contd.) Solution: (b) Note that reactant OH- is lower than HA, OH- is the limiting reagent and becomes zero after reaction is complete. Solve for x = 0.04286 Find [A-] = 0.04286 M [HA] = 0.01429 M pH = pKa + log([A-]/[HA]) = 5.37

  37. Time to Discuss and Practice : Calculate the pH when the following volume of NaOH is added to 40.00 mL 0.1000 M HPr solution: a) 10.00 mL; b) 20.00 mL; c) 35.00 mL. Ka = 1.3x10-5 • Step by step before eqiv. point • Find total volume after adding base • Find [HA] and [OH-] before rxn: Dilution • ICE table and find limiting reagent (L.R.) • Solve ICE table: find [HA] and [A-] after rxn. I + C = E • Apply Henderson-Hasselbalch equation Answer keys: a) 4.41; b) 4.89; c) 5.73

  38. PROBLEM: Calculate the pH during the titration of 40.00 mL of 0.1000 M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of 0.1000 M NaOH: (c) 40.00 mL, Finding the pH During a Weak Acid–Strong Base Titration (contd., almost!) (c) Note this is the equivalence point!  All HA is now consumed and converted into the conjugate base Total volume = 40.00 mL + 40.00 mL = 80.00 mL Mol A- = mol HA (stoichiometry) [A-] = 40.00 mL HA x 0.1000 mol/L /80.00 mL = 0.05000 M [OH-] = 6.21x10-6 M, pOH = 5.21, pH = 8.79

  39. PROBLEM: Calculate the pH during the titration of 40.00 mL of 0.1000 M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of 0.1000 M NaOH: (d) 50.00 mL Finding the pH During a Weak Acid–Strong Base Titration (contd., finally!) (d) Note Vbase > Veq, so it is beyond equivalence point All HA has been consumed and there is excess NaOH in the solution. The unreacted NaOH determine the final pH: MolNaOH (reacted) = mol HA x 1 molNaOH/1 mol HA = 0.0004000 mol MolNaOH excess = molNaOH – molNaOH(reacted) = 0.0001000 mol Total volume of the reaction = 40.00 mL + 50.00 mL = 90.00 mL = 0.09000 L [OH-] = 0.0001000 mol/0.09000 L = 0.01111 M pOH = 1.95, pH = 12.05

  40. Curve for a weak base–strong acid titration. The pH decreases gradually in the buffer region. The weak base and its conjugate acid are both present in solution. The pH at the equivalence point is < 7.00 due to the reaction of the conjugate acid with H2O.

  41. Calculating the pH during a weak base-strong acid titration: Common Ion Effect Initial pH [OH-] = pH = 14 - pOH [OH-][HB+] [B] Kw Kb Kb = pH before equivalence point: Use ICE table Ka = [B] [HB+] pH = pKa + log

  42. Calculating the pH during a weak base-strong acid titration (continued) pH at the equivalence point [H+] = pH = -log[H3O+] mol H+excess Vacid + Vbase HB+(aq) + H2O(l) B(aq) + H3O+(aq) pH beyond the equivalence point [H+] = pH = -log[H3O+]

  43. PROBLEM: Calculate the pH during the titration of 40.00 mL of 0.1000 M methylamine (MeNH2; Kb = 4.4x10-4) after adding the following volumes of 0.1000 M HCl: (a) 0.00 mL; (b) 30.00 mL Finding the pH During a Weak base-strong acid Titration (a) [OH-] = 6.6x10-3M, pH = 11.82 x = 0.04286, pH = pKa + log([B]/[HB+]) = 10.16

  44. PROBLEM: Calculate the pH during the titration of 40.00 mL of 0.1000 M methylamine (Kb = 4.4 x10-4) after adding the following volumes of 0.1000 M HCl: (c) 40.00 mL, (d) 50.00 mL Finding the pH during a Weak base-strong acid Titration (c) [HB+] = 0.05000 M, [H+] = = 1.1E-6 M, pH = 5.97 [H+] = 0.01112 M, pH = 1.95

  45. Curve for the titration of a weak polyprotic acid. Titration of 40.00 mL of 0.1000 M H2SO3 with 0.1000 M NaOH pKa2 = 7.19 pKa1 = 1.85

  46. Amino Acids as Polyprotic Acids An amino acid contains a weak base (-NH2) and a weak acid (-COOH) in the same molecule. Both groups are protonated at low pH and the amino acid behaves like a polyprotic acid.

  47. Abnormal shape of red blood cells in sickle cell anemia. Several amino acids have charged R groups in addition to the NH2 and COOH group. These are essential to the normal structure of many proteins. In sickle cell anemia, the hemoglobin has two amino acids with neutral R groups instead of charged groups. The abnormal hemoglobin causes the red blood cells to have a sickle shape, as seen here.

  48. Equilibria of Slightly Soluble Ionic Compounds Any “insoluble” ionic compound is actually slightly soluble in aqueous solution: very small amount of such a compound that dissolves will dissociate completely. For a slightly soluble ionic compound in water, equilibrium exists between solid solute and aqueous ions. Qsp = Qc[PbF2] = [Pb2+][F-]2 PbF2(s) Pb2+(aq) + 2F-(aq) [Pb2+][F-]2 [PbF2] Qc =

  49. Qsp and Ksp Qsp is called the ion-product expression for a slightly soluble ionic compound. For any slightly soluble compound MpXq: MpXq(s) pMn+(aq) + qXz-(aq) Qsp = [Mn+]p[Xz-]q When the solution is saturated, the system is at equilibrium, and Qsp = Ksp, the solubility product constant. The Ksp value of a salt indicates how far the dissolution proceeds at equilibrium (saturation).

  50. Metal Sulfides: Sulfide Ion Interacts with Water Metal sulfides behave differently from most other slightly soluble ionic compounds, since the S2- ion is strongly basic. The dissolution of a metal sulfide as a two-step process: S2-(aq) + H2O(l) → HS-(aq) + OH-(aq) MnS(s) Mn2+(aq) + S2-(aq) MnS(s) + H2O(l) Mn2+(aq) + HS-(aq) + OH-(aq) Ksp = [Mn2+][HS-][OH-]

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