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Acid-Base Equilibria

Acid-Base Equilibria.

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Acid-Base Equilibria

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  1. Acid-Base Equilibria

  2. In this chapter, we study equilibrium problems as related to acids and bases as well as their salts. The main point is to know how to calculate the hydrogen ion concentration, [H+]. However, let us start with definitions of acids and bases and then look at their equilibria.

  3. Acid-Base Theories Four main attempts to define acids and bases are common in the literature of chemistry. Development of these attempts or theories usually followed a desire to explain the behavior of substances and account for their properties as related to having acidic or basic characteristics. Theories of acidity or basicity can be outlined below from oldest to most recent:

  4. 1. Arrhenius theory: This theory is limited to water as a solvent where an acid is defined as a substance which ionizes in water and donates a proton. A base is a substance that ionizes in water to give hydroxide ions. The hydrogen ion reacts with water to give a hydronium ion while the base reacts with water to yield a hydroxide ion. HA + H2O D H3O+ + A- B + H2O D BH+ + OH-

  5. 2. Franklin Theory: This theory introduced the solvent concept where an acid was defined as a substance that reacts with the solvent to produce the cation of the solvent . The base is a substance that reacts with the solvent to yield the anion of the solvent. HA + EtOH D EtOH2+ + A- B + EtOH D BH+ + EtO-

  6. 3. Bronsted-Lowry Theory: Some solvents like hexane or benzene are non ionizable and the Franklin theory can not be used to explain acidic or basic properties of substances. In Bronsted-Lowry theory, an acid is defined as a substance that can donate a proton while a base is a substance that can accept a proton. Also, an acid is composed of two components; a proton and a conjugate base. For example HOAc D H+ + OAc- Acetic acid is an acid which donates a proton and its proton is associated with a base that can accept the proton; this base is the acetate.

  7. 4. Lewis Theory: Lewis introduced the electronic theory for acids and bases where in Lewis theory an acid is defined as a substance that accepts electrons while a base is a substance that donates electrons. Therefore, ammonia is a base because it donates electrons as in the reaction H+ + :NH3D H:NH3+ AlCl3 is an acid because it accepts electrons from a base such as :OR2 AlCl3 + : OR2 = Cl3Al:OR2

  8. Lecture 21 Acid-Base Equilibria, Cont…. Strong Acids/Bases and Their Salts

  9. Acid-Base Equilibria in Water Fortunately, we will only deal with aqueous solutions which means that water will always be our solvent. Water itself undergoes self ionization as follows 2 H2O D H3O+ + OH- K = [H3O+][OH-]/[H2O]2 However, only a very small amount of water does ionize and the overall water concentration will be constant.

  10. Therefore, one can write Kw = [H3O+][OH-] Kw is called autoprotolysis constant of water or ion product of water, we will also refer to [H3O+] as simply [H+] although this is not strictly correct due to the very reactive nature of H+ Kw = [H+][OH-] = 10-14 at 25 oC.

  11. The pH Scale In most cases, the hydrogen ion concentration is very small which makes it difficult to practically express a meaningful concept for such a small value. Currently, the pH scale is used to better have an appreciation of the value of the hydrogen ion concentration where: pH = - log [H+] We also know that kw = [H+][OH-] = 10-14 or pH + pOH = 14 Therefore, calculation of either pH or pOH can be used to find the other.

  12. We are faced with different types of solutions that we should know how to calculate the pH or pOH for. These include calculation of pH for 1. Strong acids and strong bases 2. Weak acids (monoprotic) and weak bases (monobasic) 3. Salts of weak acids and salts of weak bases 4. Mixtures of weak acids and their salts (buffer solutions) 5. Polyprotic acids and their salts and polybasic bases and their salts We shall also look at pH calculations for mixtures of acids and bases as well as pH calculations for very dilute solutions of the abovementioned systems.

  13. pH calculations 1. Strong Acids and Strong Bases Strong acids and strong bases are those substances which are completely dissociated in water and dissociation is represented by one arrow pointing to right. Examples of strong acids include HCl, HBr, HI, HNO3, HClO4, and H2SO4 (only first proton). Examples of strong bases include NaOH, KOH, Ca(OH)2, as well as other metal hydroxides.

  14. Find the pH of a 0.1 M HCl solution. HCl is a strong acid that completely dissociates in water, therefore we have HCl  H+ + Cl- H2O D H+ + OH- [H+]Solution = [H+]from HCl + [H+]from water However, [H+]from water = 10-7 in absence of a common ion, therefore it will be much less in presence of HCl and can thus be neglected as compared to 0.1 ( 0.1>>[H+]from water) [H+]solution = [H+]HCl = 0.1M pH = -log 0.1 = 1

  15. We can always look at the equilibria present in water to solve such questions, we have only water that we can write an equilibrium constant for and we can write:

  16. Kw = (0.1 + x)(x) However, x is very small as compared to 0.1 (0.1>>x) 10-14 = 0.1 x x = 10-13 M Therefore, the [OH-] = 10-13 M = [H+]from water Relative error = (10-13/0.1) x 100 = 10-10% [H+] = 0.1 + x = 0.1 + 10-13~ 0.1 M pH = 1

  17. Find the pH of a 1x10-5 M HNO3 solution. Nitric acid is a strong acid which means that it dissociates completely in water. Therefore, [H+]from Nitric acid = 1x10-5 M We can now set the equilibria in water as above

  18. Kw = (10-5 + x)(x) However, x is very small as compared to 10-5 (10-5>>x) 10-14 = 10-5 x x = 10-9 M Therefore, the [OH-] = 10-9 M = [H+]from water Relative error = (10-9/10-5) x 100 = 0.01 % [H+] = 10-5 + 10-9~ 10-5 M pH = - log 10-5 = 5

  19. Find the pH of a 10-7 M HCl solution. Solution HCl is a strong acid, therefore the [H+]from HCl =10-7 M

  20. Kw = (10-7 + x)(x) Let us assume that x is very small as compared to 10-7 (10-7>>x) 10-14 = 10-7 x x = 10-7 M Therefore, the [OH-] = 10-7 M = [H+]from water Relative error = (10-7/10-7) x 100 = 100 % Therefore, the assumption is invalid and we have to solve the quadratic equation. Solving the quadratic equation gives: [H+] = 7.62x10-7 M pH = 6.79

  21. Find the pH of a 10-10 M HCl solution. HCl  H+ + Cl- H2O D H+ + OH- [H+]Solution = [H+]from HCl + [H+]from water

  22. Assume 10-10>>x X = 10-4 , it is clear thatthe assumption is invalid. In fact x is much larger than 10-10. Therefore reverse assumption and assume x>>10-4 10-14 = (10-10 + x)(x) X = 10-7 RE = (10-7/10-10)*100% = 0.1% The assumption is valid and [H+] = [OH-] = 10-7 M pH = 7

  23. Calculate the pH of the solution resulting from mixing 50 mL of 0.1 M HCl with 50 mL of 0.2 M NaOH. When HCl is mixed with NaOH neutralization takes place where they react in a 1:1 mole ratio. Therefore, find mmol of each reagent to see if there is an excess of either reagent mmol HCl = 0.1 x 50 = 5 mmol mmol NaOH = 0.2 x 50 = 10 mmol mmol NaOH excess = 10 – 5 = 5 mmol [OH-] = 5/100 = 0.05 M

  24. Find [H+] which is equal to [OH-]water [H+] = 10-14/0.05 = 2*10-13M [OH-]water = 2*10-13M The hydroxide ion concentration from the base is high enough to neglect the contribution from water. pOH = 1.3 and pH = 14 – pOH = 14 – 1.3 = 12.7

  25. It is better to use the equilibrium 10-14 = (0.05 + x)(x) Assume 0.05>>x X= 2*10-13, RE = very small [OH-] = 0.05 M, and pOH = 1.3 pH = 14 – 1.3 = 12.7

  26. Find the pH of a solution prepared by mixing 2.0 mL of a strong acid at pH 3.0 and 3.0 mL of a strong base at pH 10. Solution First, find the concentration of the acid and the base pH = 3.0 means [H+] = 10-3.0 pH = 10 means [H+] = 10-10 M or [OH-] = 10-4 M Now find the number of mmol of each mmol H+ = 10-3 x 2.0 = 2.0x10-3 mmol mmol OH- = 10-4 x 3.0 = 3.0 x 10-4 mmol mmol H+ excess = 2.0x10-3 – 3.0x10-4 = 1.7x10-3 [H+] = 1.7x10-3/5 = 3.4x10-4 M pH = - log 3.4x10-4 = 3.5

  27. 2. Salts of Strong Acids and Bases The pH of the solution of salts of strong acids or bases will remain constant (pH = 7) where the following arguments apply: NaCl dissolves in water to give Cl- and Na+. For the pH to change, either or both ions should react with water. However, is it possible for these ions to react with water? Let us see:

  28. Cl- + H2O D HCl + OH- (wrong equation) Now, the question is whether it is possible for HCl to form as a product in water!! Of course this will not happen as HCl is a strong acid which is 100% dissociated in water. Therefore, Cl- will not react with water but will stay in solution as a spectator ion. The same applies for any metal ion like K+ where if we assume that it reacts with water we will get:

  29. K+ + H2O D KOH + H+ (wrong equation) Now, the question is whether it is possible for KOH to form as a product in water!! Of course this will not happen as KOH is a strong base which is 100% dissociated in water. Therefore, K+ will not react with water but will stay in solution as a spectator ion. It is clear now that neither K+ nor Cl- react with water and the hydrogen ion concentrationof solutions of salts of strong acids and bases comes from water dissociation only and will be 10-7 M (pH =7).

  30. Lecture 22 Acid-Base Equilibria, Cont… Weak acids/Bases and their Salts

  31. 3. Weak Acids and Bases A weak acid or base is an acid or base that partially (less than 100%) dissociated in water. The equilibrium constant is usually small and, in most cases, one can use the concepts mentioned in the equilibrium calculations section discussed previously with application of the assumption that the amount dissociated is negligible as compared to original concentration. This assumption is valid if the equilibrium constant is very small and the concentration of the acid or base is high enough.

  32. Calculate the pH and pOH for a 0.10 M acetic acid solution. Ka= 1.75x10-5 Solution [H+]solution = [H+]HOAc + [H+]water However, in absence of an acid the dissociation of water is extremely small and in presence of an acid dissociation of water becomes negligible due to the common ion effect. Therefore, we can neglect the [H+]water (which is equal to [OH-]) in presence of an acid since the hydroxide ion concentration is insignificant in an acid solution, therefore we can write

  33. [H+]solution = [H+]HOAc The first point is to write the equilibrium where HOAc D H+ + OAc-

  34. Ka = [H+][OAc-]/[HOAc] Ka = x * x / (0.10 – x) Ka is very small. Assume 0.10 >> x 1.75*10-5 = x2/0.10 x = 1.3x10-3 Relative error = (1.3x10-3/0.10) x 100 = 1.3% The assumption is valid and the [H+] = 1.3x10-3 M pH = 2.88 pOH = 14 – 2.88 = 11.12 Now look at the value of [OH-] = 10-11.12 M = 7.6x10-12 M =[H+]from water. Therefore, the amount of H+ from water is negligible

  35. Calculate the pH and pOH for a 1.00x10-3 M acetic acid solution.Ka = 1.75x10-5 Solution The first point is to write the equilibrium where HOAc D H+ + OAc-

  36. Ka = [H+][OAc-]/[HOAc] Ka = x * x / (1.00*10-3 – x) Ka is very small. Assume 1.00*10-3 >> x 1.75*10-5 = x2/1.00*10-3 x = 1.32x10-4 Relative error = (1.32x10-4/1.00x10-3) x 100 = 13.2% The relative error is more than 5% therefore, the assumption is invalid and we have to use the quadratic equation to solve the problem.

  37. Find the pH and pOH of a 0.20 M ammonia solution. Kb = 1.8x10-5. Solution The same treatment above can be used to solve this problem where: [OH-]solution = [OH-]ammonia + [OH-]water [H+] = [OH-]water However, in absence of a base the dissociation of water is extremely small and in presence of the base the dissociation of water becomes negligible due to the common ion effect.

  38. Therefore, we can neglect the [H+] in presence of ammonia since the hydrogen ion concentration is insignificant in basic solution, therefore we can write OH-]solution = [OH-]ammonia NH3 + H2O D NH4++ OH-

  39. Kb = [NH4+][OH-]/[NH3] 1.8*10-5 = x * x / (0.20 – x) kb is very small that we can assume that 0.20>>x. We then have: 1.8*10-5 = x2 / 0.2 x = 1.9x10-3 M Relative error = (1.9x10-3 /0.2) x 100 = 0.95% The assumption is valid, therefore: [OH-] = 1.9x10-3 M, [H+] = 5.3x10-12 M = [OH-]water which is very small. pOH = 2.72 pH = 11.28

  40. 4. Salts of Weak Acids and Bases a. Salts derived from weak acids/bases and strong bases/acids Imagine that an acid is formed from two species a hydrogen ion and a conjugate base attached to it. The acid is said to be strong if its conjugate base is weak while a weak acid has a strong conjugate base. Therefore, we can fairly recognize conjugate bases like Cl-, NO3-, and ClO4- as weak bases that do not react with water and thus will not change the pH of water (pH = 7). On the other hand, conjugate bases derived from weak acids are strong bases which react with water and alter its pH.

  41. Examples of strong conjugate bases include OAc-, NO2-, CN-, etc.. The same applies for bases where weak bases are weak because their conjugate acids are strong which means they react with water and thus alter its pH. One important piece of information with regards to salts of weak acids and bases is that we have to find their ka or kb as the equilibrium constants given in problems are for the parent acid or base. Let us look at the following argument for acetic acid: HOAc D H+ + OAc- Ka = [H+][OAc-]/[HOAc]

  42. For the conjugate base of acetic acid (acetate) we have OAc- + H2O D HOAc + OH- Kb = [HOAc][OH-]/[OAc-] Let us multiply ka times kb we get Ka kb = [H+][OH-] = kw , or Ka kb = kw Therefore, if we know the ka for the acid we can get the equilibrium constant for its conjugate base since we know kw. We can find ka for the conjugate acid by the knowledge of the equilibrium constant of the parent base.

  43. Find the pH of a 0.10 M solution of sodium acetate. Ka = 1.75x10-5 Solution [OH-]solution = [OH-]acetate + [OH-]water [OH-]water = [H+] Since the hydrogen ion concentration is very small in a solution of a base, we can neglect [OH-]water and we then have [OH-]solution = [OH-]acetate

  44. OAc- + H2O D HOAc + OH- Kb = kw/ka Kb = 10-14/1.75x10-5 = 5.7x10-10

  45. Kb = [HOAc][OH-]/[OAc-] Kb = x * x/(0.10 – x) Kb is very small and we can fairly assume that 0.10>>x 5.7x10-10 = x2/0.1 x = 7.6 x 10-6 Relative error = (7.6x10-6/0.10) x100 = 7.6x10-3% The assumption is valid. [OH-] = 7.6x10-6 M [H+] = 1.3x10-9 M = [OH-]water

  46. The relative error in neglecting OH- from water = (1.3x10-9/7.6x10-6) x 100 = 0.017% This validate our assumption at the beginning of the solution that [OH-]acetate >> [OH-]water pOH = 5.12 pH = 14 – 5.12 = 8.88

  47. Calculate the pH of a 0.25 M ammonium chloride solution. Kb = 1.75x10-5 Solution [H+]solution = [H+]ammonium + [H+]water However, in absence of an acid the dissociation of water is extremely small and in presence of an acid dissociation of water becomes negligible due to the common ion effect. Therefore, we can neglect the [H+]water in presence of an acid since the hydroxide ion concentration is insignificant in an acid solution

  48. therefore we can write [H+]solution = [H+]ammonium The first point is to write the equilibrium where NH4+D H+ + NH3

  49. Ka = 10-14/1.75x10-5 = 5.7x10-10 Ka = [H+][NH3]/[NH4+] Ka = x * x / (0.25 – x) Ka is very small. Assume 0.25 >> x 5.7*10-10 = x2/0.25 x = 1.2x10-5 Relative error = (1.2x10-5/0.25) x 100 = 4.8x10-3 % The assumption is valid and the [H+] = 1.2x10-5 M

  50. Now look at the value of [OH-] = 10-14/1.2x10-5 = 8.3x10-10 M = [H+]from water. Therefore, the amount of H+ from water is negligible when compared to that from the acid. The relative error for neglecting the H+ from water = (8.3x10-10/1.2x10-5) x 100 = 6.9x10-3% pH = 4.92 pOH = 14 – 4.92 = 9.08 We should remember that dissociation of water is negligible in presence of an acid or base.

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