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Applications of Aqueous Equilibria

Chapter 8. Applications of Aqueous Equilibria. Chapter 8: Applications of Aqueous Equilibria. 8.1 Solutions of Acids or Bases Containing a Common Ion 8.2 Buffered Solutions 8.3 Exact Treatment of Buffered Solutions 8.4 Buffer Capacity 8.5 Titrations and pH Curves

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Applications of Aqueous Equilibria

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  1. Chapter 8 Applications of Aqueous Equilibria

  2. Chapter 8: Applications of Aqueous Equilibria 8.1 Solutions of Acids or Bases Containing a Common Ion 8.2 Buffered Solutions 8.3 Exact Treatment of Buffered Solutions 8.4 Buffer Capacity 8.5 Titrations and pH Curves 8.6 Acid-Base Indicators 8.7 Titration of Polyprotic Acids 8.8 Solubility Equilibria and The Solubility Product 8.9 Precipitation and Qualitative Analysis 8.10 Complex Ion Equilibria

  3. A base swirling in a solution containing phenolphthalein

  4. Like Example 8.1 (P 274-5) - I [H+] [NO2-] [HNO2] Ka = = 4.0 x 10-4 Nitrous acid, a very weak acid, is only 2.0% ionized in a 0.12 M solution. Calculate the [H+], the pH, and the percent dissociation of HNO2 in a 1.0 M solution that is also 1.0 M in NaNO2! HNO2(aq) H+(aq) + NO2-(aq) Initial Concentration (mol/L) Equilibrium Concentration (mol/L) [HNO2]0 = 1.0 M [HNO2] = 1.0 – x (from dissolved HNO2) [NO2-]0 = 1.0 M [NO2-] = 1.0 + x (from dissolved NaNO2) [H+]0 = 0 [H+] = x (neglect the contribution from water)

  5. Like Example 8.1 (P 274-5) - II 4.0 x 10-4 x 100 = ________ % 1.0 [H+] [NO2-] [HNO2] ( x ) ( 1.0 + x ) (1.0 – x ) Ka = = = 4.0 x 10-4 Assume x is small as compared to 1.0: X (1.0) (1.0) = [H+] = 4.0 x 10-4 or x = 4.0 x 10-4 Therefore pH = - log [H+] = - log ( 4.0 x 10-4 ) = 3.40 The percent dissociation is: Nitrous acid Nitrous acid alone + NaNO2 [H+] 2.0 x 10-2 4.0 x 10-4 pH 1.70 3.40 % Diss 2.0 0.040

  6. Example 8.2 (P276-9) - I A buffered solution contains 0.50 M acetic acid (HC2H3O2, Ka = 1.8 x 10-5) and o.50 M sodium acetate (NaC2H3O2). Calculate the pH of this solution, and the pH when 0.010 M of solid NaOH is added to 1.0 L of this buffer and to pure water. HC2H3O2 (aq) H+(aq) + C2H3O2 (aq) [H+] [C2H3O2-] [HC2H3O2] Ka = 1.8 x 10-5 = Initial Concentration (mol/L) Equilibrium Concentration (mol/L) X mol/L of HC2H3O2 dissociates to reach equilibrium [HC2H3O2]0 = 0.50 [HC2H3O2] = 0.50 – x [C2H3O2-]0 = 0.50 [C2H3O2-] = 0.50 +x [H+]0 = 0 [H+] = x ~

  7. Example 8.2 (P276-9) - II [H+][C2H3O2-] [HC2H3O2] ( x ) ( 0.50 + x) 0.50 - x (x) (0.50) 0.50 ~ Ka = 1.8 x 10-5 = = = x = 1.8 x 10-5 The approximation by the 5% rule is fine: [H+] = x = 1.8 x 10-5 M and pH = 4.74 To calculate the pH and concentrations after adding the base: OH- + HC2H3O2 H2O + C2H3O2- Before reaction: 0.010 mol 0.50 mol - 0.50 mol After reaction: 0.010 – 0.010 0.50 – 0.10 - 0.50 + 0.10 = 0 mol = 0.49 mol = 0.51 mol Note that 0.01 mol of acetic acid has been converted to acetate ion by the addition of the base.

  8. Example 8.2 (P276-9) - III Initial Concentration (mol/L) Equilibrium concentration (mol/L) X mol/L of HC2H3O2 Dissociates to reach equilibrium [HC2H3O2]0 = 0.49 [HC2H3O2] = 0.49 – x [C2H3O2-]0 = 0.51 [C2H3O2-] = 0.51 + x [H+]0 = 0 [H+] = x [H+][C2H3O2-] [HC2H3O2] (x)(0.51+ x) 0.49 - x (x)(0.51) 0.49 Ka = 1.8 x 10-5 = = = x = 1.7 x 10-5 and pH = 4.76 If the base is added to pure water without the buffer being present we get an entirely different solution: If the 0.01 mol of NaOH is added to 1.0 L of pure water the Concentration of hydroxide ion is 0.01 M. Kw [OH-] 1.0 x 10-14 1.0 x 10-2 [H+] = = = __________ and the pH = ______

  9. Original buffered solution pH Modified pH (H+/OH- added) Step 1: Do stoichiometric calculations to determine new concentrations. assume reaction with H+/OH- goes to completion. Step 2: Do equilibrium calculations.

  10. Le Châtelier’s principle for the dissociation equilibrium for HF HF(aq) H+(aq) + F-(aq)

  11. Molecular model: F-, Na+, HF, H2O

  12. How Does a Buffer Work Lets add a strong base to a weak acid and see what happens: OH- + HA A- + H2O Final pH of buffer close to original- Original buffer pH Added OH- ions Replaced by A- ions [H+] [A-] [HA] [HA] [A-] Ka = [H+] = Ka

  13. The Effect of Added Acetate Ion on the Dissociation of Acetic Acid [CH3COOH] [CH3COO-]added % Dissociation* pH 0.10 0.00 1.3 2.89 0.10 0.050 0.036 4.44 0.10 0.10 0.018 4.74 0.10 0.15 0.012 4.92 [CH3COOH]dissoc * % Dissociation = x 100 [CH3COOH]init

  14. Human blood is a buffered solution Source: Visuals Unlimited

  15. How a Buffer Works–I [CH3COOH] [CH3COO-] A buffer consists of a solution that contains “high” concentrations of the acidic and basic components. This is normally a weak acid and the anion of that weak acid, or a weak base and the corresponding cation of the weak base. When small quantities of H3O+ or OH- are added to the buffer, they cause a small amount of one buffer component to convert into the other. As long as the amounts of H3O+ and OH- are small as compared to the concentrations of the acid and base in the buffer, the added ions will have little effect on the pH since they are consumed by the buffer components. Consider a buffer made from acetic acid and sodium acetate: CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) [CH3COO-] [H3O+] Ka = or [H3O+] = Ka x [CH3COOH]

  16. [CH3COOH] [CH3COO-] 0.012 mol H3O+ = 0.0399 M H3O+ 301.00 mL soln. How a Buffer Works–II Let’s consider a buffer made by placing 0.25 mol of acetic acid and 0.25 mol of sodium acetate per liter of solution. What is the pH of the buffer? And what will be the pH of 100.00 mL of the buffer before and after 1.00 mL of concentrated HCl (12.0 M) is added to the buffer? What will be the pH of 300.00 mL of pure water if the same acid is added? (0.25) [H3O+] = Ka x = 1.8 x 10-5 x = 1.8 x 10-5 (0.25) pH = -log[H3O+] = -log(1.8 x 10-5) = pH = ____ Before acid added! 1.00 mL conc. HCl 1.00 mL x 12.0 mol/L = 0.012 mol H3O+ Added to 300.00 mL of water : pH = -log(0.0399 M) pH = _____ Without buffer!

  17. How a Buffer Works–III [CH3COOH] [CH3COO-] After acid is added: Conc. (M) CH3COOH(aq) + H2O(aq) CH3COO- + H3O+ Initial 0.250 ---- 0.250 0 Change +0.012 ---- -0.012 0.012 Equilibrium 0.262 ---- 0.238 0.012 Solving for the quantity ionized: Conc. (M) CH3COOH(aq) + H2O(aq) CH3COO- + H3O+ Initial 0.262 ---- 0.238 0 Change -x ---- +x +x Equilibrium 0.262 - x ---- 0.238 + x x Assuming: 0.262 - x = 0.262 & 0.238 + x = 0.238 (0.262) [H3O+] = Ka x =1.8 x 10-5 x = 1.982 x 10-5 (0.238) pH = -log(1.982 x 10-5) = 5.000 - 0.297 = ____ After the acid is added!

  18. How a Buffer Works–IV Suppose we add 1.0 mL of a concentrated base instead of an acid. Add 1.0 mL of 12.0 M NaOH to pure water and our buffer, and let’s see what the impact is: 1.00 mL x 12.0 mol OH-/1000mL = 0.012 mol OH- This will reduce the quantity of acid present and force the equilibrium to produce more hydronium ion to replace that neutralized by the addition of the base! Conc. (M) CH3COOH(aq) + H2O(aq) CH3COO- + H3O+ Initial 0.250 ---- 0.250 0 Change - 0.012 ---- +0.012 +0.012 Equilibrium 0.238 ---- 0.262 +0.012 Assuming: Again, using x as the quantity of acid dissociated we get: our normal assumptions: 0.262 + x = 0.262 & 0.238 - x = 0.238 0.238 [H3O+] = 1.8 x 10-5 x = 1.635 x 10-5 0.262 pH = -log(1.635 x 10-5) = 5.000 - 0.214 = ____ After base is added!

  19. How a Buffer Works–V 0.012 mol OH- [OH-] = = 3.99 x 10-5 M OH- 301.00 mL Kw 1 x 10-14 [H3O+] = = = 2.506 x 10-10 [OH-] 3.99 x 10-5M By adding the 1.00mL base to 300.00 mL of pure water we would get a hydroxide ion concentration of: The hydrogen ion concentration is: This calculates out to give a pH of: pH = -log(2.5 6 x 10-10) = 10.000 - 0.408 = 9.59 With 1.0 mL of the base in pure water! In summary: Buffer alone pH = 4.74 Buffer plus 1.0 mL base pH = 4.79 Base alone pH = 9.59 Buffer plus 1.0 mL acid pH = 4.70 Acid alone pH = 1.40

  20. Molecular model: HC2H3O2, C2H3O2-

  21. The Relation Between Buffer Capacity and pH Change

  22. A digital pH meter shows the pH of the buffered solution to be 4.74

  23. NH3 (aq) + H2O(l) NH4+(aq) + OH-(aq) Preparing a Buffer Problem: The ammonia-ammonium ion buffer has a pH of about 9.2 and can be used to keep solutions in the basic pH range. What mass of ammonium chloride must be added to 400.00 mL of a 3.00 M ammonia solution to prepare a buffer ? Plan: The conjugate pair is the ammonia-ammonium ion pair which has an equilibrium constant Kb = 1.8 x 10 -5. The reaction equation with water can be written along with the Kb expression, since we want to add sufficient ammonium ion to equal the aqueous ammonia concentration. Solution: The reaction for the ammonia-ammonium ion buffer is: NH4Cl = 53.49 g/mol [NH4+] [OH-] Therefore mass =NH4Cl = 1.20 mol x 53.49g/mol mass = _________ g NH4Cl Kb = = 1.8 x 10-5 [NH3] [NH4+] = 3.00 mol x 0.400 L = 1.20 mol L

  24. Pure water at pH 7.00

  25. The Henderson-Hasselbalch Equation [HA] [H3O+] = Ka x [A-] ( ) ( ) [HA] [HA] -log[H3O +] = -log Ka - log pH = -log Ka - log [A-] [A-] [base] [acid] Take the equilibrium ionization of a weak acid: HA(aq) + H2O(aq) = H3O+(aq) + A-(aq) [H3O+] [A-] Ka = [HA] Solving for the hydronium ion concentration gives: Taking the negative logarithm of both sides: Generalizing for any conjugate acid-base pair : ( ) Henderson-Hasselbalch equation pH = log Ka + log

  26. When 0.01 mol NaOH is added to 1.0 L of pure water, the pH jumps to 12.00

  27. Like Example 8.3 (P 282-3) -I Problem: Instructions for making a buffer say to mix 60.0 ml of 0.100 M NH3 with 40.0 ml of 0.100 M NH4Cl. What is the pH of this buffer? The combined volume is 60.0 ml + 40.0 ml = 100.0 ml Moles of Ammonia = VolNH3 x MNH3 = 0.060 L x 0.100 M = 0.0060 mol Moles of Ammonium ion = VolNH4Cl x MNH4Cl = 0.040 L x 0.100 M = = 0.0040 mol [NH3] = = 0.060 M ; [NH4+] = = 0.040 M 0.0060 mol 0.100 L 0.0040 mol 0.100 L Concentration (M) NH3 (aq) + H2O(l) NH4+(aq) + OH-(aq) Starting 0.060 0.040 0 Change -x +x +x Equilibrium 0.060 – x 0.040 – x x

  28. Like Example 8.3 (P 282-3) - II Substituting into the equation for Kb: [NH4+] [OH-] [NH3] (0.040 + x) (x) (0.060 – x) Kb = = 1.8 x 10-5 = ~ ~ Assume : 0.060 – x = 0.060 ; 0.040 + x = 0.040 0.040 (x) 0.060 Kb = 1.8 x 10-5 = x = 2.7 x 10-5 Check assumptions: 0.040 + 0.000027 = 0.040 or 0.068% 0.060 – 0.000027 = 0.060 or 0.045% [OH-] = 2.7 x 10-5 ; pOH = - log[OH-] = - log (2.7 x 10-5) = 5 – 0.43 pOH = 4.57 pH = 14.00 – pOH = 14.00 – 4.57 = __________

  29. pH Box [H3O+] = 10-pH pH[H3O+] pH = -log[H3O+] Kw = 1 x 10-14 @ 25oC [H3O+][OH-]= =1 x 10-14 pH + pOH = = 14 @ 25oC [OH-] = 10-pOH pOH [OH-] pOH = -log[OH-]

  30. When a strong acid or base is added to a buffered solution, it is best to deal with the stoichiometry of the resulting reaction first. After the stoichiometric calculations are completed, then consider the equilibrium calculations. This procedure can be represented as follows:

  31. Summary: Characteristics of Buffered Solutions Buffered solutions contain relatively large concentrations of a weak acid and its corresponding weak base. They can involve a weak acid HA and the conjugate base A- or a weak base B and the conjugate acid BH+. When H+ is added to a buffered solution, it reacts essentially to completion with the weak base present: H+ + A- HA or H+ + B BH+ When OH- is added to a buffered solution, it reacts essentially to completion with the weak acid present. OH- + HA A- + H2O or OH- + BH+ B + H2O The pH of the buffered solution is determined by the ratio of the concentrations of the weak base and weak acid. As long as this ratio remains virtually constant, the pH will remain virtually constant. This will be the case as long as the concentrations of the buffering materials (HA and A- or G and BH+) are large compared with the amounts of H+ or OH- added.

  32. OH- ions are not allowed to accumulate but are replaced by A- ions.

  33. Exact Treatment of Buffer Solutions [A-] = [A-]0 + [H+]2 – Kw [H+] [H+]{ [A-]0 + } [H+][A-] [HA] Ka = = [H+]2 – Kw [H+] [HA]0 - We can use several relationships to calculate the exact solution to buffered solution problems: Charge – balance equation: [Na+] + [H+] = [A-] + [OH-] Material – Balance equation: [A-]0 + [HA]0 = [HA] + [A-] Since [A-]0 = [Na+] and Kw = [OH-][H+] , we can rewrite the charge balance equation, and solve for [A-] : [H+]2 – Kw [H+] From the mass balance equation solved for [HA] we get: [HA] = [A-]0 + [HA]0 – [A-] Substituting the expression for [A-], and substituting into the Ka expression for HA we obtain:

  34. When the OH- is added, the concentrations of HA and A- change, but only by small amounts. Under these conditions the [HA]/[A-] ratio and thus the [H+] stay virtually constant.

  35. Example 8.4 (P 286-7) - I [H+] = x = = 1.05 x 10-7 M = 1.1 x 10-7 M Calculate the pH of a buffered solution containing 3.0 x 10-4 M HOCl (Ka = 3.5 x 10-8) and 1.0 x 10-4 M NaOCl. Let x = [H+] then: [OCl-] = 1.0 x 10-4 + x [HOCl] = 3.0 x 10-4 - x [H+] [OCl-] [HOCl] Ka = = 3.5 x 10-8 [H+] [OCl-] [HOCl] (x)(1.0 x 10-4 + x) (3.0 x 10-4 – x ) 3.5 x 10-8 = = Assuming x is small compared to 1.0 x 10-4 and solving for x we have: 1.05 x 10-11 1.0 x 10-4 Since this is close to that of water we must use the equation that uses water, and takes it’s ionization into account.

  36. Example 8.4 (P 286-7) - II [H+]2 – 1.0 x 10-14 [HOCl]0 - [H+] [OCl-]0 = 1.0 x 10-4 M >>> [H+]2 – 1.0 x 10-14 [H+] [H+][OCl-] [HOCl] [H+](1.0 x 10-4) (3.0 x 10-4) 3.5 x 10-8 = = [H+]2 – 1.0 x 10-14 [H+]{[OCl-]0 + } [H+] Ka = 3.5 x 10-8 = Where: [OCl-]0 = 1.0 x 10-4 M [HOCl]0 = 3.0 x 10-4 M We expect [H+] to be close to 1.0 x 10-7, so [H+]2 to be about 1.0 x 10-14 [H+]2 – 1.0 x 10-14 [HOCl]0 = 1.0 x 10-4 M >>> [H+] The expression becomes:

  37. Example 8.4 (P 286-7) - III [H+][OCl-] [HOCl] [H+](1.0 x 10-4) (3.0 x 10-4) 3.5 x 10-8 = = [H+] = 1.05 x 10-7 M = 1.1 x 10-7 M Using this result, we can check the magnitude of the neglected term: (1.05 x 10-7)2 – 1.0 x 10-14 [H+]2 – 1.0 x 10-14 = =__________ 1.05 x 10-7 [H+] This result suggests that the approximation was fine!

  38. Molecular model: Cl-, NH4+

  39. pH and Capacity of Buffered Solutions The pH of a buffered solution is determined by the ratio[A-]/[HA]. The capacity of a buffered solution is determined by the magnitudes Of [HA] and [A-]

  40. Original solution and new solution

  41. Original solution and new solution

  42. Solution A and Solution B

  43. Example 8.5 (P 287-9) - I Calculate the change in pH that occurs when 0.010 mol of gaseous HCl is added to 1.0 L of each of the following solutions: Solution A: 5.00 M HC2H3O2 and 5.00 M NaC2H3O2 Solution B: 0.050 M HC2H3O2 and 0.0500 M NaC2H3O2 For Acetic acid, Ka = 1.8 x 10-5 Use the Henderson-Hasselbalch equation for initial pH: [C2H3O2-] [H C2H3O2] Since [C2H3O2-] = [H C2H3O2] The equation becomes: pH = pKa + log{ } pH = pKa + log (1) = pKa = -log(1.8 x 10-5) = 4.74 Adding 0.010 mol of HCl will cause a shift in the equilibrium due to: H+(aq) + C2H3O2-(aq) H C2H3O2 (aq)

  44. Example 8.5 (P 287-9) - II 4.99 5.01 [C2H3O2-] [H C2H3O2] pH = pKa + log ( ) = 4.74 + log ( ) = 4.74 – 0.0017 = _____ For Solution A: H+ + C2H3O2- H C2H3O2 Before reaction 0.010 M 5.00 M 5.00 M After reaction 0 4.99 M 5.01 M Calculate the new pH using the Henderson-Hasselbalch equation: For Solution B: H+ + C2H3O2- H C2H3O2 Before reaction 0.010 M 0.050 M 0.050 M After reaction 0 0.040 M 0.060 M 0.040 0.060 The new pH is: pH = 4.74 + log( ) = 4.74 – 0.18 = ______

  45. A burette valve Source: American Color

  46. Figure 8.1: The pH curve for the titration of 50.0

  47. Vol NaOH added (mL)

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