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Acid-Base Equilibria

Acid-Base Equilibria. Aqueous solutions can be classified as acidic, basic, or neutral. This classification scheme is based on the quantities of 2 ions, hydronium ion, H 3 0 1+ and hydroxide ion, OH 1- . Where do these ions come from in solutions of pure water?

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Acid-Base Equilibria

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  1. Acid-Base Equilibria Aqueous solutions can be classified as acidic, basic, or neutral. This classification scheme is based on the quantities of 2 ions, hydronium ion, H301+ and hydroxide ion, OH1-. Where do these ions come from in solutions of pure water? Water molecules in motion will randomly collide with one another. When this happens occasionally a hydrogen nucleus from one molecule will be transferred from one molecule to the other. This can be illustrated.

  2. Notice the nucleus of one hydrogen atom, a proton, was transferred, but the electron pair was left behind. This produces the H301+ ion (hydronium) and the OH1- ion (hydroxide)

  3. water ionizing Hydronium ion (H3O1+) H2O + H20 H3O1+(aq) + OH1-(aq) H20 H1+(aq) + OH1-(aq) Hydroxide ion (OH1-) the equilibrium expression is Which is usually shortened to:

  4. H20 H1+(aq) + OH1-(aq) The equilibrium constant for is [H1+][OH1-] Ke = [H2O] [H2O] in mol/L is mass of 1 L of water is 1000 g n = (1000g)/(18g/mol) =55.6 molL-1

  5. [H1+][OH1-] [H2O] [H2O] Ke = [H2O] To simplify this equation Here is a new constant [H2O]Ke = Kw the ion product constant of water Kw = [H1+][OH1-] = 1.0 x 10-14 @ 25oC at higher temperatures Kw increases

  6. Kw = [H1+][OH1-] = 1.0 x 10-14 @ 25oC if [H1+]=[OH1-] then [H1+]=[OH1-]= 1.0 x 10-7 which is the situation for neutral water. A scale which simplifies this very small number is the pH (potency of H1+) scale. It is based on powers of ten. Simply take the exponent from 10-7 and then multiply it by -1. This produces the number 7.

  7. In mathematical terms pOH = -log[OH1-] so if in an aqueous solution the [OH1-] = 2.4 x 10-4, the pOH is 3.62 If the pOH = 4.56 then the [OH1-] = 2.8 x 10-5 This is calculated using the equation [OH1-] = 10-pOH , similarly [H1+]= 10-pH.

  8. In mathematical terms pH = -log[H1+] so if in an aqueous solution the [H1+] = 2.4 x 10-8, the pH is 7.62 Remember the whole number portion of a pH doesn’t count as a significant digit (SD), just like in the number 2.4 x 10-8 the exponent -8 doesn’t count as a SD.

  9. If the [H1+][OH1-] = 1.0 x 10-14 then taking the log of each side pH + pOH = 14 so if the pH of a solution is 2.3 the pOH is 11.7 This means for these 4 values [H1+], [OH1-], pH, pOH if one of them is given the other 3 can be calculated. Here is an example

  10. If the pH of a solution is 1.45 find the [H1+], [OH1-], and the pOH. Remember here are the equations needed. If pH = 1.45 6.pOH = 14 - pH pOH = 12.55 3.[OH1-] = 10-12.55 [OH1-] = 2.8 x 10-13M 4.[H1+] = 10-1.45 [H1+] = 0.035 M 1.pH = -log[H1+] 2.pOH = -log[OH1-] 3.[OH1-] = 10-pOH 4.[H1+] = 10-pH. 5.[H1+][OH1-] = 1.0 x 10-14 6.pH + pOH = 14

  11. Take note the equations can be applied in varying orders and there are choices as to exactly which equations are used. These decisions are yours to make.

  12. If the [OH1-] of a solution is 9.4 x 10-9 M find the [H1+], pH, and the pOH. If [OH1-] = 9.4 x 10-9 2.pOH = -log 9.4 x 10-9 pOH = 8.03 6.pH = 14 - 8.03 =5.97 4.[H1+] = 10-5.97 = 1.1 x 10-6 M 1.pH = -log[H1+] 2.pOH = -log[OH1-] 3.[OH1-] = 10-pOH 4.[H1+] = 10-pH. 5.[H1+][OH1-] = 1.0 x 10-14 6.pH + pOH = 14

  13. If the [H1+] of a solution is 6.2 x 10-2 M find the [OH1-], pH, and the pOH. If [H1+] = 6.2 x 10-2M 1. pH = -log 6.2 x 10-2 pH = 1.21 6. pOH = 14 - 1.21 pOH = 12.79 3.[OH1-] = 10-12.79 [OH1-] = 1.6 x 10-13M 1.pH = -log[H1+] 2.pOH = -log[OH1-] 3.[OH1-] = 10-pOH 4.[H1+] = 10-pH. 5.[H1+][OH1-] = 1.0 x 10-14 6.pH + pOH = 14

  14. pH Scale Increasing basicity Increasing acidity -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 N e u t r a l How much more acidic is pH 1 than pH 5? 10 000 x’s What pH is 1000x’s more acidic than pH 2? Since this is a logarithmic scale pH 9 is 10x’s more basic than pH 8, pH 12 is 1000 x’s more basic than pH 9 pH of -1

  15. Conceptual Definitions of Acids and Bases Arrhenius’s Concept - acids are substances which react in water and produce hydronium ions. HCl(g) + H20 -------> H301+(aq) + Cl1-(aq) Bases are substances which react with water and produce hydroxide ions. NH3(g) + H20 ------> NH41+(aq)+ OH1-(aq)

  16. This concept has its limitations however. Can’t substances be classified as acids or bases without the involvement of water?

  17. Bronstead’s Definition of Acids and Bases Acids are substances which donateprotons and bases are substances which acceptprotons. In the examples above HCl(g) is an acid because it donates protons to H2O molecules and NH3 is a base because it accepts protons from H2O molecules.

  18. Strength of Acids and Bases is determined by the degree to which a substance produces ions in solution. A strong acid or base is a substance which completely ionizes. In other words if 100 molecules of a strong acid like HCl are placed in water all 100 of them will react with H2O producing 100 H3O1+ ions and 100 Cl1- ions. Weak acids and bases only partially ionize. Strong Acid - the reaction below goes to completion. HCl(g) + H20 --------> H301+(aq) + Cl1-(aq)

  19. Weak Acid - the reaction occurs to a limited extent. In the example below if 100 acetic acid molecules are placed in water only a few of them will successfully react with water molecules producing hydronium ions. Most CH3COOH molecules remain intact. CH3COOH + H20 H301+(aq)+ CH3COO1-(aq)

  20. Strong Acids in order of decreasing strength are HClO4, HI, HBr, H2SO4, HCl, HNO3 A table with the remaining moderate and weak acids can be found on page 803. Acid strength has to do with the ease with which an acid can lose a proton. If the binary acid strengths (HI, HBr, HCl) are compared it can be seen that HI is the strongest acid of this group because its iodide ion is the largest of the group so the force between the hydrogen ion and the iodide ion is the weakest so it loses its proton most easily.

  21. Force is strongest since the ions are closest Br1- H1+ H1+ H1+ Cl1- Force is weakest since the ions are furthest Remember the weaker the force the stronger the acid I1-

  22. Strong Bases include hydroxides of group 1A and Ca2+, Ba2+, and Sr2+. A table with the remaining moderate and weak bases can be found on page 803. As with acids the weaker the bonds, the stronger the base since liberation of OH1- ions is easiest when the bonds are weakest.

  23. Polyprotic Acids donate protons in steps. For instance carbonic acid, H2CO3 has two protons to donate and it does this in two steps: step 1 H2CO3 + H20 HCO31- + H301+ step 2 HCO31- + H20 CO32- + H301+ note: The arrows are constructed in this manner to show the reverse reaction has a greater tendency than the forward reaction.

  24. conjugate pair conjugate pair Conjugate Acid - Base Pairs - When using the Bronsted concept for acids and bases it is convenient to consider all acid - base reactions as reversible equilibria. For instance when sulfurous acid, H2SO3 reacts with water the following equilibrium is established: H2SO3 + H2O H301+ + HSO31- acid base acid base

  25. conjugate pair conjugate pair acid baseacid base H2SO3 + H2OH301+ + HSO31- In the forward direction the H2SO3 is the proton donor so it’s the acid and the H2O is the proton acceptor so it’s the base. In the reverse direction the H301+ is the proton donor so it’s the acid and the HSO31- is the proton acceptor so it’s a base.

  26. conjugate pair conjugate pair acid base acid base H2SO3 + H2O H301+ + HSO31- When looking at both forward and reverse reactions it is easy to pick out a pair of molecules which differ by a single proton (H atom without its electron). These pairs are called conjugate acid-base pairs.

  27. Identify the acid base conjugate pairs in the equilibrium below: conjugate pair conjugate pair PO43-(aq) + H2O(aq) HPO42-(aq) + OH1-(aq) base acid acid base Show how HPO4-2 can act as both an acid and a base.

  28. Amphoteric (Amphiprotic) Substances can behave as both acids or bases dependent on the circumstances. Water molecules, for instance, can sometimes accept protons and behave as bases or donate protons and behave as acids. HBr(g) + H2O H301+(aq) + Br1-(aq) base NH3(g) + H2O 0H1-(aq) + NH41+(aq) acid

  29. Any substance which alters the pH of an aqueous solution is regarded as an acid or a base. Acids and Bases are classified into 2 categories, strong and weak. Strong Acids and Bases are completely ionized. This means in solution no molecules are present. All the molecules break apart (dissociate), to form ions. Weak acids and bases only partially dissociate and ionize and so are involved in equilibria.

  30. Strong Acid Weak Acid H1+ H1+ H1+ H1+ H1+ H1+ H1+ H1+ H1+ H1+ H1+ Cl1- H1+ H1+ H1+ H1+ H1+ H1+ H1+ H1+ H1+ H1+ H1+ H1+ H1+ F1- Cl1- F1- Cl1- F1- Cl1- F1- Cl1- F1- Cl1- F1- Cl1- Cl1- F1- Cl1- F1- Cl1- F1- Cl1- F1- F1- F1- F1-

  31. Is this a Strong or Weak Acid?

  32. Is this a Strong or Weak Acid?

  33. Problems Involving Strong Acids and Bases Before any HClO4 has dissolved Find the pH and pOH of a 0.28 molL-1 solution of HClO4. Since this is a strong acid the ionization is complete. This can be represented by … HClO4 H1+(aq) +ClO41-(aq) (molL-1) initial 0.28 0.0 0.0 -0.28 0.28 0.28 shift 0.00 0.28 0.28 final

  34. HClO4 H1+(aq) +ClO41-(aq) (molL-1) initial 0.28 0.0 0.0 -0.28 0.28 0.28 shift 0.00 0.28 0.28 final Remember pH = -log[H1+] so pH = - log 0.28 pH = 0.55 pH + pOH = 14 so pOH = 14-0.55 pOH = 13.45

  35. If the pOH of a solution of HCl is found to be 12.9 what is the concentration of this solution. Since this is a strong acid the ionization is complete. This can be represented by … H1+(aq) +Cl1-(aq) HCl (molL-1) initial 0.08 0.0 0.0 -0.079 0.079 0.079 shift 0 0.079 0.079 final In this case the pOH can be used to determine the final [H1+], pH = 14 - 12.9, [H1+] = 10-1.1

  36. Find the pH and pOH of a 0.12 molL-1 solution of Sr(OH)2. Since this is a strong base the ionization is complete. This can be represented by … Sr2+(aq) + 2OH1-(aq) Sr(OH)2 (molL-1) initial 0.12 0.0 0.0 -0.12 0.12 0.24 shift 0.00 0.12 0.24 final Remember pOH = -log[OH1-] so pOH = - log 0.24 = 0.62 pH = 14 - 0.62 = 13.38

  37. Problems Involving Weak Acids and Bases Before any HCN has dissociated Find the pH and pOH of a 1.68 molL-1 solution of HCN (Ka = 6.2 x 10-10. Since this is a weak acid the ionization is incomplete. This can be represented by … HCN H1+(aq) +CN1-(aq) (molL-1) initial 1.68 0.0 0.0 - x x x shift 1.68 - x x x final

  38. To determine the value of x the Ke for this reaction must be known. It is found in tables called Ka (equilibrium constants for acids). For HCN Ka = 6.2 x 10-10. HCN H1+(aq) +CN1-(aq) Since the amount of acid which ionizes is very small this x is negligible [x]2 = [1.68-x] [H1+][CN1-] Ka = [HCN] 6.2 x 10-10 = x2 1.68 x = 3.2 x 10-5, pH = 4.49, pOH = 9.51

  39. What is the percentage ionization of the 1.68 mol/L HCN solution?

  40. If an unknown 0.34 mol/L monoprotic acid is 1.3 x 10-5% ionized what is its Ka?

  41. If the pH of a solution of HOCl is found to be 4.9 what is the concentration of this solution. Ka = 3.0 x 10-8 Since this is a weak acid the ionization is incomplete. This can be represented by … H1+(aq) +OCl1-(aq) HOCl (molL-1) initial x 0.0 0.0 -1.26 x 10-5 1.26 x 10-5 1.26 x 10-5 shift 1.26 x 10-5 1.26 x 10-5 x - 1.26 x 10-5 final In this case the pH can be used to determine the final [H1+], [H1+] = 10-4.9 = 1.26 x 10-5.

  42. H1+(aq) +OCl1-(aq) HOCl (molL-1) initial x 0.0 0.0 -1.26 x 10-5 1.26 x 10-5 1.26 x 10-5 shift 1.26 x 10-5 1.26 x 10-5 x - 1.26 x 10-5 final This amount is negligible [1.26 x 10-5]2 [x - 1.26 x 10-5 ] [H1+][OCl1-] Ka = = [HOCl] 3.0 x 10-8 = 1.58 x 10-10 x x = 5 x 10-3 molL-1

  43. Negative ions Which Act Like Weak Bases NO31 H 1+ H1+ Strong acids totally ionize because the force of attraction between the two ions is relatively weak and water molecules pulling on them can separate them. NO31 This means NO31- has a weak pull on H1+ ions

  44. Conversely weak acids have negative ions with relatively strong attractions for H1+ ions These ions cannot be easily ripped apart Water molecules can’t do it NO21- H1+ If a salt like NaNO2 is added to water it is easily pulled apart since sodium ions are quite large compared to H1+ ions making the attraction between Na1+ and NO21- fairly weak.

  45. O2- O2- NO21 NO21 NO21- NO21 NO21 NO21- O2- O2- O2- H1+ H1+ Na1+ H1+ H1+ H1+ H1+ Na1+ H1+ H1+ H1+ H1+ NO21- All negative ions except those found in strong acids, Cl1-, Br1-, I1-, NO31-, ClO41-, HSO41-, are capable of creating OH1- ions, this means they act like bases because they raise pH

  46. The equilibrium for NO21- where it alters pH is NO21-(aq) + H20 HNO2(aq)+ 0H1-(aq) Show how K2CO3 alters pH CO32-(aq) + H20 HCO31-(aq)+0H1-(aq) Show how NaF alters pH F1-(aq) + H20 HF(aq)+0H1-(aq) Show how NaCl alters pH it doesn’t, Cl1- has too weak a pull on H1+ remember, Cl1- is part of the strong acid HCl

  47. What is the pH of a 0.24 molL-1 solution of potassium sulfite Kb of sulfite ion is 1.5 x 10-7? When potassium sulfite is placed in water it dissociates into ions. This can be shown by: K2SO3(s) 2K1+(aq) + SO32-(aq) Since SO32- is not part of a strong acid it is capable of pulling apart (hydrolyzing) water molecules. This can be shown by: SO32-(aq) + H20 HSO31-(aq) + 0H1- molL-1 initial 0.24 0.00 0.00 -x +x +x shift 0.24-x +x +x @E

  48. SO32-(aq) + H20 HSO31-(aq) + 0H1- molL-1 initial 0.24 0.00 0.00 -x +x +x shift 0.24-x +x +x @E The Kb of SO32- is 1.5 x 10-7. So to find x x2 1.5 x 10-7 = 0.24 -x X is so small compared to 0.24 it can be regarded as negligible x = 1.9 x 10-4 = [OH1-] pOH = -log (1.9 x 10-4) pOH = 3.72 pH = 14 - 3.72 = 10.28

  49. Positive Ions Which Act Like Acids Positive ions like NH41+, can donate H1+ ions to water molecules. This increases the amount of hydronium (H301+) creating an acidic affect. This can be illustrated by the following:

  50. NH41+ H3O1+ NH3 H2O NH41+ + H2O NH3 + H3O1+

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