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Acid-Base Equilibria

Acid-Base Equilibria. Arrhenius Definition. Acids are substances that produce hydrogen ions when dissolved in water. HCl → H + + Cl - Bases are substances that produce hydroxide ions when dissolved in water. NaOH → Na + + OH -

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Acid-Base Equilibria

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  1. Acid-Base Equilibria

  2. Arrhenius Definition • Acids are substances that produce hydrogen ions when dissolved in water. HCl → H+ + Cl- • Bases are substances that produce hydroxide ions when dissolved in water. NaOH → Na+ + OH- • Problem: NH3 (ammonia) when dissolved in water forms NH4OH, a weakbase, but NH3 could not be an Arrhenius base based on traditional definition because NH3does not have a hydroxide to donate.

  3. Brønsted-Lowry Definitions A Brønsted–Lowry acid… …must have a removable (acidic) proton. A Brønsted–Lowry base… …must have a pair of nonbonding electrons.

  4. Brønsted-Lowry Definitions According to this theory, an acid is a proton (hydrogen ion, H+) donor and a base is a proton (hydrogen ion, H+) acceptor. …Consider introducing HCl(g) into water EOS

  5. What Happens When an Acid Dissolves in Water? • Water acts as a Brønsted–Lowry base and extracts a proton (H+) from the acid, becoming a proton acceptor. • As a result, the conjugate base of the acid and a hydronium ion (H3O+) are formed.

  6. NH3 works under this definition as a base. Let’s see how. • Notice that water now behaves as an acid

  7. If it can be either an acid or base… ...it is amphiprotic (amphoteric). HCO3− HSO4− H2O

  8. Conjugate Acids and Bases: • From the Latin word conjugare, meaning “to join together.” • Reactions between acids and bases always yield their conjugate bases and acids. Conjugate acid and base pairs are related by a hydrogen ion on either side of the equation If the reaction proceeds in the forward direction, then HNO2 acts as an acid by donating a hydrogen ion (proton). However, if the reaction were to then go backwards, then NO2-1 would act as a base by accepting a hydrogen ion (proton)

  9. The conjugate acidof a base is the base PLUS the attached proton and the conjugate base of an acid is the acid MINUS the proton

  10. Identify the acid, base, conjugate acid, conjugate base and pairs • HCN + NH3 CN- +NH4+ • Acid – HCN • Base – NH3 • Conjugate Base - CN- • Conjugate Acid - NH4+

  11. Acid and Base Strength • A strong acid/base undergoes complete ionization • Reaction goes to completion • Full dissociation. Equilibrium lies completely to the product side • LARGE value of Kc • Ex: HCl(aq)  H3O+(aq) + Cl-(aq) • Ex: NaOH(aq)  Na+(aq) + OH-(aq) • The conjugate base/acid therefore is extremely weak • Ex: Cl- has VERY poor ability to attract protons to itself

  12. Acid and Base Strength • Weak acids/bases have very little ionization. • Very little (partial) dissociation in water • Equilibria lies mostly to the LEFT. SMALL value of Kc • Their conjugate bases/acidss are weak to exceedingly strong • As acid/base strength decreases, the conjugate base/acid strength increases

  13. Acid and Base Strength • Substances with negligible acidity do not dissociate in water. • Their conjugate bases are exceedingly strong. CH4 + H2O   CH3- + H3O+ CH3- is a VERY strong base due to its extreme attraction for H+

  14. Acid and Base Strength In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. HCl(aq) + H2O(l) H3O+(aq) + Cl−(aq) H2O is a much stronger base than Cl−, so the equilibrium lies so far to the right K is not measured (K>>1). ASSUMPTION: 100% dissociation. [H+] or [H3O+] = [ACID given] ** important when calculating pH

  15. HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq) Acid and Base Strength Acetate ion is a stronger base than H2O,Predict K>>>1 K>1 K<1 K<<<1 equilibrium favors the left side (K<<<1).

  16. Consider the four diagrams where in each case, HA is an acid, H+ is a hydrogen ion, A- is an anion, and water molecules HA H+ A- H2O Which diagram represents a relatively concentrated weak acid? Explain Which diagram represents a relatively concentrated strong acid? Explain Assign relative strengths and concentrations to the two remaining diagrams. Explain.

  17. Carefully note that the concentration of H+ and OH- are dependent upon TWO, separate factors. • Strength of acid or base, i.e, degree of dissociation/ionization • Amount of water present, i.e. concentration of sample solution • It is possible to have a dilute, strong acid and a weak, concentrated acid with the SAMEhydronium ion concentration • Therefore, the same pH value

  18. Examples of Acids and Bases

  19. H2O(l) + H2O(l) H3O+(aq) + OH−(aq) Autoionization of Water • As we have seen, water is amphoteric. • In pure water, a few molecules act as bases and a few act as acids. • This is referred to as autoionization.

  20. pH (power of Hydrogen) • Many of the concentration measurements in acid-base problems are given to us in terms of pH (power of H+) and pOH (power of OH-). • p (anything) = -log (anything) pH = -log [H+] pOH= -log [OH-] pKa = -log Ka

  21. pH (power of H) • pH scale is used to indicate the strength of an acid or base SAMPLE. • Traditional range 0-14 • Negative pH is possible • 0- <7 Acidic • 7> - 14 Basic • 7 Neutral

  22. Recall: For a strong acid/base, ionization is considered to be 100%. Therefore • H+ or OH- concentration can be determined directly from the stoichiometric ratio in the balanced equation and the concentration of acid or base. • Ex: HNO3 is a strong acid. If 0.10 M HNO3 dissociates, [H+] is ALSO 0.10 M HNO3  H+ + NO3- 0.10 M 0.10 M 0.10M • Calculate pH of above sample • pH = -log[H+] • pH = -log [0.10] • pH = 1.00

  23. pH • pH= -log[H+] • Used because [H+] is usually very small • As pH decreases, [H+] increases exponentially

  24. Practice • Calculate pH of each of the following solutions • 0.0030 M HCl • 0.030 M HCl • 0.30 M HCl • 3.0 M HCl • Carefully note that since pH is based on a log scale, there is a 10 fold change in concentration associated with a pH change of 1 unit (EXPONENTIAL)

  25. pH – cont. • pH of 3 = 10 times more concentrated than pH of ???, and ???times more concentrated than pH of 5 pH of 3 = 10 times more concentrated than pH of 4, and 100 times more concentrated than pH of 5

  26. EOS Determining [conc] from pH/pOH pH is a measure of the strength of an acid sample; low pH = stronger acid pOH is a measure of the strength of a base sample; low pOH = stronger base (also means HIGH pH pH = –log[H3O+] and [H3O+] = 10(–pH) pOH = –log[OH–] and [OH–] = 10(–pOH)

  27. Practice: Predict the following from highest to lowest [H+]. Then determine [H+] if pH is1.58 9.5 14 -0.87 4.2

  28. Additionally, at 298 K, 14 = pH + pOH • Therefore, ALL values (pH, pOH, OH- or H+ ) can be determined based on ANY one known value

  29. Practice • Calculate the pH of a 0.010 M solution of Lithium hydroxide • Calculate the hydrogen ion concentration in a solution with a pH of 4.32 • Calculate the pH of a solution made by dissolving 2.00g KOH in water to a total volume of 250. mL

  30. pH of weak acids • Recall: • What is the definition of a weak acid? • Incomplete dissociation • What does the pH and/or strength of sample depend on? • amount of acid dissociated [H+] AND amount of water present. • Weak acids do not completely dissociate in water. • Equilibrate!! • Problems solved similar to equilibrium problems.

  31. We do NOT know how much acid actually dissociated • Therefore, CANNOT use pH = - log [H+] because CANNOT assume the [H+] = [acid]. • Carefully note that since ionization is minimal, value of x (dissociation) is small. • Considered NEGLIGIBLE in terms of (X – x)

  32. Additionally • [H+] = [A-] • H2O (pure liquid) does not appear in Ka (Acid dissociation constant) expression • Therefore, the following Ka expression can be derived Ka = OR • And pKa = -log Ka Ka = Mathematically equal. MUST NOT use as the EQUATION on apfrq. SHOW the ACTUAL equation. • [H+][A-] • [HA] • [H+]2 • [HA]

  33. Practice Problems • If ethanoic acid has a pKa 0f 4.74, what is its Ka? • If propanoic acid has a Ka of 1.38 X 10-5, what is it’s pKa? • Which is a stronger acid, propanoic or ethanoic? • What is the pH of a 1.00 M solution of ethanoic acid? • Calculate the Ka value of a 0.100 M solution of a weak acid with a [H+] of 1.75 X 10-3 M • Calculate the Ka of a solution of 0.250 M of a weak acid with a pH of 5.11

  34. ORGANIC LESSON • Carboxylic acid (organic acid) • General makeup • R-COOH • Example: ethanoic • Ethane (C2H6): remove methyl (CH3) and add –COOH • Formula: CH3COOH • Example: propanoic • propane: (C3H8): remove methyl (CH3) and add –COOH • Formula: C2H5COOH • Determine formula for methanoic and butanoic acids. Draw Structure.

  35. Strength of carboxylic acid depends upon the stability of the anion (called: ___oate) formed when labile (mobile) proton is lost. • All carboxylic acids that dissociates, give the following equilibrium, where R (hydrocarbon chain) can vary • RCOOH  RCOO- + H+ • Anions may form, releasing H+ to form acidic solution • Stability of anion depends on size of R group. • Smaller R group = FEWER electrons “pumped” to COO- anion = HIGHER the stability of anion = STRONGER acid = LARGER Ka = SMALLER pKa • Explain in terms of polarity: • Higher polarity of a smaller particle (methanoate) will cause greater attraction to water, therefore higher chance of H+ ionizing. Lower polarity of a larger non-polar R group (butanoate) will have far less attraction to water, therefore smaller chance of H+ ionizing. • RECALL: carboxylic acids up to 4 carbons are soluble/acidic

  36. Carefully note: as carboxylic acid R group increases in size, acid strength decreases. • Ka decreases, pKa increases.

  37. Other related molecules • Similar effects are observed when comparing relative strenghts of halogen substituted carboxylic acids. • As the number of chlorine atoms present increases, so does the acid strength. • The high electronegativity of Cl atoms attract electron density away from COO-

  38. High Halogen Electronegativity = high Ka = Stronger Acid = Low pKa • Why? • Greater attraction for electrons, so flourine pulls electrons away from the COO-, allowing easier removal of H+

  39. Kb: Equilibrium Constant for Weak Bases • Similar to weak acid equilibria, weak base ionization is incomplete • Cannot directly determine amount of base dissociated based on Molarity of solution • Cannot use pOH = -log [Base] • Must determine and use ACTUAL [OH-] • Same assumptions as weak acid equilibria calculations • Assume x (dissociated ion) is negligible relative to X (base molarity) • Assume [cation] = [OH-]

  40. NH3 + H2O  NH4+ + OH- Therefore: Kb = Often written as Kb = And pKb = -log Kb [OH-][NH4+] [NH3] [OH-]2 [NH3] DO NOT USE ON AP FRQ. Use ACTUAL Kb expression. CLEARLY SPECIFY that you are ASSUMING [OH-] = [Cation] CLEARLY SPECIFY that you are ASSUMING [OH-] to be negligible, therefore using [Base] as given

  41. Practice

  42. Autoionization of Water • Although pure water is essentially covalent, a small amount of self-ionization occurs • Water autoionizes, by donating a proton to another water molecule: H2O + H2O  H3O+ + OH- or H2O  H+ + OH- • No individual ion remains ionized for long • At 298 K, only about 2 out of every 109 molecules are ionized at any given moment

  43. Autoionization of Water • Equilibrium constant, Kw, written as Kw = [H+][OH-] • At 298 K, Kw = 1.0 X 10-14 • Since [H+] = [OH-], therefore 1.0 X 10-14 = [H+][OH-] • and [H+] = [OH-] = = 1.0 X 10-7 M • Therefore, pH = -log [1.0 X 10-7 M] = 7 √1.0 X 10-14

  44. Determine pKw • pKw = -log Kw = -log [1.0 X 10-14] = 14 • Since we know that Kw = [H+][OH-] therefore Kw = Ka · Kb thereforepKw = pKa + pKbtherefore 14 = pH + pOH *** KNOW THIS

  45. Temperature dependent • Carefully note that at temperatures other than 298 K, Kw value may be different, HOWEVER, [H+] = [OH-] • pH may be different from 7 • HOWEVER, still considered NEUTRAL due to [H+] = [OH-] (no excess H+ or OH-)

  46. Autoionization of Water H2O  H+ + OH- • Since this is specifically for the autoionization of water we use Kw as the equilibrium constant. • At 25oC: Kw = [H+][OH-] = 1.0 x 10-14 • [H+] > [OH-]  acid • [H+] < [OH-]  base • [H+] = [OH-]  neutral

  47. Autoionization of Water H2O  H+ + OH- Kw = [H+][OH-] = 1.0 x 10-14 • At this point it is very useful to remember that pH = - log [H+] • Since Kw is a constant, if we are given [OH-] it is a very simple matter to calculate pH • Since pH, pOH, [H+], [OH-], pKa, pKb, pKw and Kw are ALL related, knowing ANY of these values is sufficient to calculate ALL others.

  48. Titrations • Titration: experimental technique to perform a stoichiometrically balanced neutralization reaction. • Accurately graduated glassware (volumetric flasks, graduated glass pipets and burets) is used in quantitative manner • Helps determine unknown concentration of an acid or a base

  49. Titrations • Terms to Know: • Titrant:solution added FROM the buret • Titrate: solution that titrant is added to. • Equivalence point: 100% neutralization. [H+] = [OH-] • End point: Change in color of indicator • Half Equivalence point: 1/2 total amount of titrant added. Buffer solution pH = pKa

  50. Titrations • As acid or base is added, there is • Very little change in pH • pH change of less than 1.5 units expected upto the point where 90% of acid/base is neutralized • At equivalence point, when 100% of acid or base is neutralized, RAPID change in pH is observed • Summarized using titration curve plots

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