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Acid Base Equilibria maybe do packet 2 before packet 3? Combine concepts from 1 and 3?

Acid Base Equilibria maybe do packet 2 before packet 3? Combine concepts from 1 and 3?. Arrhenius Acids & Bases (1 st year chem definition!). An Arrhenius acid is a substance that, when dissolved in water, increases the concentration of H + (needs H + in it)

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Acid Base Equilibria maybe do packet 2 before packet 3? Combine concepts from 1 and 3?

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  1. Acid Base Equilibriamaybe do packet 2 before packet 3? Combine concepts from 1 and 3?

  2. Arrhenius Acids & Bases(1st year chem definition!) An Arrhenius acid is a substance that, when dissolved in water, increases the concentration of H+ (needs H+ in it) Example: HCl (monoprotic) H2SO4 (diprotic) An Arrhenius base is a substance that, when dissolved in water, increases the concentration of OH– Example: NaOH(monobasic) Ca(OH)2 (dibasic)

  3. Brønsted-Lowry Acids and Bases Brønsted-Lowry acid is a species that donates H+ (proton) Brønsted-Lowry base is a species that accepts H+(proton) Brønsted-Lowry definition of a base does not mention OH–and the reaction does not need to be aqueous.

  4. The H+ Ion in Water • The H+(aq) ion is simply a proton with no surrounding valence electrons. • In water, clusters of hydrated H+(aq) ions form.

  5. The simplest cluster is H3O+(aq) We call this a hydronium ion. • Larger clusters are also possible (such as H5O2+ and H9O4+). • Generally we use H+(aq) and H3O+(aq) interchangeably.

  6. Proton-Transfer Reactions • Consider NH3(aq) + H2O(l) NH4+(aq) + OH–(aq) • H2O donates a proton to ammonia. • Therefore, water is acting as an acid. • NH3 accepts a proton from water. • Therefore, ammonia is acting as a base.

  7. Amphotericsubstances • Can behave as acids and bases. • Water is an example of an amphoteric species. Water as an acid (proton donor) H2O + NH3NH4+ + OH- Water as a base (proton acceptor) H2O + HNO2H30+ + NO2-

  8. Conjugate Acid-Base Pairs • A conjugate acid is the substance formed by adding a proton to the base. • A conjugate base is the substance left over after the acid donates a proton. Within a pair the acid has more hydrogen!

  9. Strong Acids & Bases All other acids and bases are weak!

  10. Ions Most anions are weak bases Most cations are weak acids Anions of strong acids and cations of strong bases are neutral

  11. Strengths of Acids and Bases Strong acids completely ionize in water. HCl + H2O  H3O+ + Cl- HCl H+ + Cl- • Essentially no un-ionized molecules remain in solution so the equation usually does not contain equilibrium arrows. Keq >>1 • Their conjugate bases have negligible tendencies to become protonated Cl- + H+  HCl. The conjugate base of a strong acid is a neutral anion.

  12. Strengths of Acids and Bases Strong bases completely dissociate in water NaOH(aq)  Na+(aq) + OH-(aq) • Essentially no undissociated compound remains in solution so the equation usually does not contain equilibrium arrows. Keq >>1 • The ions have negligible tendencies to attract OH- in solution. Na+(aq) + OH-(aq)  NaOH(aq) The cation of a strong base is a neutral cation.

  13. All other acids are Weak acids. They only partially dissociate in aqueous solution. HC2H3O2(aq) + H2O(l)  H3O+(aq) + C2H3O2-(aq) • They exist in solution as a mixture of molecules and component ions. (usually mostly molecules in equilibrium) • Their conjugate bases are weak bases. • Besides non-neutral anions, weak bases tend to be nitrogen containing organic compounds • Example: Acetic acid is a weak acid; acetate ion (conjugate base) is a weak base. conjugateacid conjugatebase acid base

  14. Non-acid compounds with hydrogen Not all compounds containing hydrogen are acidic. These are extremely weak acids…so much that we don’t consider them acids at all. Their conjugate bases are strong bases! Negligible acidity: OH- H2 CH4 Strong bases: O2- H- CH3-

  15. The stronger an acid is, the weaker its conjugate base will be. In acid-base reactions, the reaction favors the transfer of a proton from the stronger acid to the stronger baseto form a weaker acid and weaker base. We need a more specific way to determine acid strength!

  16. Where does that acid/base ranking come from? Keq!!! Since weak acids and bases are in equilibrium…we can write equilibrium constant expressions! When looking at the reaction of a weak acid with water we label the equilibrium constant Ka: HF(aq) + H2O(l)  H3O+(aq) + F-(aq) When looking at the reaction of a weak base with water we label the equilibrium constant Kb: NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)

  17. Ka and Kb • The value of Ka or Kb indication the extent to which the weak acid or base ionizes/dissociates. • Larger Kaor Kb means more products! • Weak acids with larger Ka values are stronger. • Weak bases with larger Kb values are stronger. (Values in Appendix D)

  18. Percent Ionization • Percent ionization is another method to assess acid strength. • For the reaction: HA(aq)  H+(aq) + A–(aq) • The higher the percent ionization, the stronger the acid.

  19. Polyprotic Acids Polyprotic acids have more than one ionizable proton. • The protons are removed in successive steps. Consider the weak acid, H2SO3 (sulfurous acid): H2SO3(aq)  H+(aq) + HSO3–(aq) Ka1= 1.7 x 10–2 HSO3–(aq)  H+(aq) + SO32–(aq) Ka2= 6.4 x 10–8 • It is always easier to remove the first proton in a polyprotic acid than the second. Ka1 > Ka2 > Ka3, etc

  20. The Autoionization of Water In pure water the following equilibrium is established: H2O(l)+ H2O(l)H3O+(aq) + OH–(aq) acid base acid base • This process is called the autoionizationof water. We can write an equilibrium constant expression for the autoionization of water: Kw = [H3O+] [OH–] = 1.0*10-14 @25°C Kwis called the “ion-product constant” Kw = Ka*Kb for conjugate pair

  21. The Ion Product Constant • This applies to pure water as well as to aqueous solutions. (for our purposes…) • A solution is neutral if [OH–] = [H3O+]. • If the [H3O+] > [OH–], the solution is acidic. • If the [H3O+] < [OH–], the solution is basic. In a neutral solution at 25oC, [H+] = [OH-] = 1.0 x 10-7 M

  22. Give the conjugate base of the following Bronsted-Lowry acids: (a) HIO3 (b) NH4+1 (c) H2PO4-1 (d) HC7H5O2 remove H+ (a) IO3-1 (b) NH3 (c) HPO4-2 (d) C7H5O2-

  23. Designate the Bronsted-Lowry acid and the Bronsted-Lowry base on the left side of each of the following equations, and also designate the conjucate acid and base on the right side: (a) NH4+1(aq) + CN-1(aq)  HCN(aq) + NH3(aq) acid base acid base (b) (CH3)3N(aq) + H2O  OH-1(aq) + (CH3)3NH+1(aq) base acid base acid (c) HCHO2(aq) + PO4-3(aq)  HPO4-2(aq) + CHO2-1(aq) acid base acid base

  24. (a) The hydrogen oxalate ion (HC2O4-1) is amphoteric. Write a balanced chemical equation showing how it acts as an acid toward water and another equation showing how it acts as a base towards water. (b) What is the conjugate acid and base of HC2O4-1? (a) Acid: HC2O4-1(aq) + H2O(l)  C2O4-2(aq) + H3O+1(aq) acid base conj base conj acid Base: HC2O4-1(aq) + H2O(l)  H2C2O4(aq) + OH-1(aq) base acid conj acid conj base (b)

  25. Label each of the following as being a strong acid, a weak acid, or a species with negligible acidity. In each case write the formula of its conjugate base, and indicate whether the conjugate base is a strong base, a weak base or a species with negligible basicity: (a) HNO2 (b) H2SO4 (c) HPO4-2 (d) CH4 (e) CH3NH3+1 (a) HNO2, (b) H2SO4, (c) HPO4-2 , (d) CH4 , (e) CH3NH3+1 , weak acid NO2-1 , weak base strong acid HSO4-1 , negligible base weak base weak acid PO4-3 , negligible acid strong base CH3-1 , weak base weak acid CH3NH2 , ACID BASE strong negligible weak weak negligible strong IMPORTANT

  26. Which of the following is the stronger acid, HBrO or HBr? • Which is the stronger base, F-1 or Cl-1? Briefly explain your choices. (a) HBr - It is one of the seven strong acids • F-1 HF is a weak acid so F- is a weak base HCl is a strong acid so Cl- is neutral

  27. Predict the products of the following acid-base reactions & determine whether equilibrium lies to the right or left: (a) O-2(aq) + H2O(l)  (b) CH3COOH(aq) + HS-1(aq)  (c) NO3-1(aq) + H2O(l)  * You will need a chart like this or Ka/Kb values to determine equilibrium!

  28. O-2(aq) + H2O(l) OH-1(aq) + OH-1(aq) base acid acid base *OH- is the weaker acid so equilibrium lies to the right • CH3COOH(aq) = HC2H3O2 (aq) • HC2H3O2 (aq) + HS-1(aq)  C2H3O2-1(aq) + H2S(aq) acid base base acid *H2S is the weaker acid so equilibrium lies to the right • NO3-1(aq) + H2O(l) HNO3(aq) + OH-1(aq) • base acid acid base • *H2O is the weaker acid so equilibrium lies (far!) to the left

  29. The “p” Function • p is short for “– log10” pH = -log10[H+] = -log[H+] pOH = -log10[OH-] = -log [OH-] • Note that this is a logarithmic scale. • Thus a change in [H+] by a factor of 10 causes the pH to change by 1 unit. • Most pH values fall between 0 and 14.

  30. In neutral solutions at 25oC, [H+] = 1.0 x 10-7 pH = -log[1.0 x 10-7] = 7.00 Lower pH = more acidic Higher pH = more basic

  31. Another one: pK • We can use a similar system to describe the equilibrium constant pK= - log[K] • The value of Kw at 25oC is 1.0 x 10–14 pKw= - log (1.0 x 10–14)= 14.0 Kw=[H+][OH-] pKw= pH + pOH = 14.0

  32. The pH Loop [H+]=10-pH pH [H+] pH = -log[H+] [H+] [OH-]=10-14 pH + pOH=14.0 [OH-]=10-pOH [OH-] pOH pOH = -log[OH-]

  33. Measuring pH The most accurate method to measure pH is to use a pH meter. Acid Base Indicators: • certain dyes change color as pH changes. • Indicators are less precise than pH meters. • Many indicators do not have a sharp color change as a function of pH. • Most acid-base indicators can exist as either an acid or a base. • These two forms have different colors. • The relative concentration of the two different forms is sensitive to the pH of the solution. • Thus, if we know the pH at which the indicator turns color, we can use this color change to determine whether a solution has a higher or lower pH than this value.

  34. Example 1: Calculate [H+1] for each of the following solutions and indicate whether the solution is acidic, basic or neutral. (a) [OH-1] = 0.0007 M (b) a solution where [OH-1] is 100 times greater than [H+1]

  35. Example 2: (a) If NaOH is added to water, how does the [H+1] change? How does pH change? (b) If [H+1] = 0.005 M, what is the pH of the solution? Is the solution acidic or basic? (c) If pH = 6.3, what are the molar concentrations of H+1(aq) and OH-1(aq) in the solution? • Kw= [H+] [OH-]. the [OH-] will increase and the [H+] will decrease. [H+] decreases, pH increases. • pH = - log [H+] = - log (0.005) = 2.3 acidic • pH = 6.3 [H+] = 10-pH = 10-6.3 = 5 x 10-7 M pOH= 14 – pH = 14 – 6.3 = 7.7 [OH-] = 10-pOH = 10-7.7 = 2 x 10-8 M

  36. Strong Acid Calculations • In solution the strong acid is usually the only source of H+ • The pH of a solution of a monoprotic acid may usually be calculated directly from the initial molarity of the acid. Caution: If the molarity of the acid is less than 10–6M then the autoionization of water needs to be taken into account.

  37. Example 3: Calculate the pH of each of the following strong acid solutions: (a) 1.8  10-4 M HBr (b) 1.02 g HNO3 in 250 mL of solution (c) 2.00 mL of 0.500 M HClO4 diluted to 50.0 mL (d) a solution formed by mixing 10.0 mL of 0.0100 M HBr with 20.0 mL of 2.5  10-3 M HCl • 1.8 x 10-4M HBr = 1.8 x 10-4M H+ pH = -log(1.8 x 10-4) = 3.74 (b) 0.0647 M HNO3 = 0.0647 M H+ pH = -log (.0647) = 1.19 • M1V1 = M2V2 (0.500 M)(.00200 L) = (x M)(0.0500 L) x = .0200 M HCl (d) pH = -log(.0050) = 2.30 [H+] = .0200 M pH = -log(.0200) = 1.70

  38. Strong Base Calculations • Strong bases are strong electrolytes and dissociate completely in solution. • For example: NaOH(aq)  Na+(aq) + OH–(aq) The pOH (and thus the pH) of a strong base may be calculated using the initial molarity of the base.

  39. Example 4: Calculate [OH-1] and pH for (a) 3.5  10-4 M Sr(OH)2 (b) 1.50 g LiOH in 250 mL of solution (c) 1.00 mL of 0.095 M NaOH diluted to 2.00 L (d) a solution formed by adding 5.00 mL of 0.0105 M KOH to 15.0 mL of 3.5  10-3 M Ca(OH)2 • [OH-] = 2[Sr(OH)2] = 2(0.00035 M) = .00070 M OH- pOH = -log(.00070) = 3.15 pH = 14 – 3.15 = 10.85 (b) pOH = -log (.251) = 0.601 pH = 14 - .601 = 13.399 (c) M1V1 = M2V2 (0.095 M)(.00100 L) = (x M)(2.00 L) x = .000048 M NaOH = [OH-] pOH = -log(.000048) = 4.32 pH = 14 – 4.32 = 9.68 (d) pOH = -log(.0079) = 2.1 pH = 14 – 2.1 = 11.9

  40. Weak Acid Calculations • Weak acids are only partially ionized in aqueous solution. • Therefore, weak acids are in equilibrium: HA(aq) + H2O(l)  H3O+(aq) + A–(aq) OR HA(aq)  H+(aq) + A–(aq) • We can write Ka for this dissociation:

  41. Calculating Ka from pH for weak acids • In order to find the value of Ka, we need to know all of the equilibrium concentrations. (ICE Chart) • The pH gives the equilibrium concentration of H+. • We then substitute these equilibrium concentrations into the equilibrium constant expression and solve for Ka.

  42. Example 4: A 0.20 M solution of niacin (a monoprotic weak acid) has a pH of 3.26. What is the Ka for niacin?

  43. Using Ka to Calculate pH for weak acids • Write the balanced chemical equation clearly showing the equilibrium. • Write the equilibrium expression. Look up the value for Ka (in a table). • Write down the initial and equilibrium concentrations for everything except pure water. (ICE table) • We usually assume that the equilibrium concentration of H+ is x. • Substitute into the equilibrium constant expression and solve. • Remember to convert x to pH if necessary.

  44. Polyprotic Acids • Polyprotic acids have more than one ionizable proton. H2SO3(aq)  H+(aq) + HSO3–(aq) Ka1 = 1.7 x 10–2 HSO3–(aq)  H+(aq) + SO32–(aq) Ka2 = 6.4 x 10–8 • The majority of the H+(aq) at equilibrium usually comes from the first ionization • If the successive Ka values differ by a factor of 103, we can usually get a good approximation of the pH of a solution of a polyprotic acid by considering the first ionization only. • If not, then we have to account for the successive ionizations

  45. Example 5: The acid dissociation constant for benzoic acid, HC7H5O2 is 6.5 x 10-5. Calculate the equilibrium concentrations of H3O+, C7H5O2-, and HC7H5O2 . The initial concentration of HC7H5O2 is 0.050M. HC7H5O2 (aq) H+(aq) + C7H5O2- (aq) I 0.050 0 0 C -x x x E (.050-x) x x x2 + (6.5*10-5)x – (3.25*10-6) = 0 x= .0018M = [H+]=[C7H5O2- ] [HC7H5O2] = .050 - .0018 = .048 M

  46. What if I don’t have a quadratic equation program? Algebra Shortcut: Assume x is much smaller (less than 5% of .050) To simplify: .050-x = .050 Now you don’t need the quadratic equation! x= .0018M = [H+]=[C7H5O2- ] [HC7H5O2] = .050 - .0018 = .048 M If you made this assumption you need to check and make sure it’s valid – if this answer isn’t less than 5% use the quadratic equation!

  47. Example 6: Calculate the pH of the following solution (Kaand Kb values are in Appendix D). 0.175 M hydrazoic acid, HN3 • HN3(aq)  H+(aq) + N3-(aq) I 0.175M 0 0 C -x x x E 0.175-x x x x2 + (1.9*10-5)x – (3.325*10-6) = 0 x=.0018M H+ pH = -log(0.0018) = 2.74

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