Acids & Bases - PowerPoint PPT Presentation

patty
acids bases n.
Skip this Video
Loading SlideShow in 5 Seconds..
Acids & Bases PowerPoint Presentation
Download Presentation
Acids & Bases

play fullscreen
1 / 41
Download Presentation
Acids & Bases
127 Views
Download Presentation

Acids & Bases

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Acids & Bases Unit 14 – Chapter 19

  2. ACIDS AND BASES • Properties of Acids: • Sour Taste (Vinegar, Lemon) • React with active metals (Zn, Mg) to produce H2(g) • Neutralize bases to produce a salt and water • Change the color of certain organic dyes (Indicators); Litmus → red • Acids are electrolytes in H2O (may be strong or weak) • [H3O+] > [OH-] • pH is less than 7 • Many of these properties are a result of the release of H+ ion in water. This H+ ion combines with water to produce the hydronium ion – H3O+.

  3. ACIDS AND BASES • Properties of Bases: • Bitter taste (soap) • React with Mg2+(aq) solutions to produce a white precipitate • Neutralize acids to produce a salt and water • Has a slippery feel, but very caustic • Change color of certain indicators – Litmus → blue; phenolphthalein → pink • Often form strong or weak electrolytes when aqueous • [OH-] > [H3O+] • pH is greater than 7

  4. Acid – Base Theories Arrhenius Theory: First acid – base theory, developed by Svante Arrhenius in the late 19th century. Acids are substances which release H+ ions in aqueous solution: HCl → H+(aq) + Cl-(aq) Bases are substances which release OH- ions: NaOH → Na+(aq) + OH-(aq) Brønsted – Lowry Theory: Proposed independently in 1923 by Johannes Brønsted and Thomas Lowry, it is a more general theory than the one developed by Arrhenius.

  5. Acid – Base Theories According to Brønsted – Lowry theory: Any substance which donates a proton (H+) is an acid. Any substance which accepts a proton is a base. HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq) HCl donates a proton to water, so it is the B – L acid H2O accepts the H+, so it is the B – L base In the reverse reaction, hydronium donates a proton to chloride to reform HCl. So H3O+ is the conjugate acid. Since Cl- is the proton acceptor in the reverse reaction, it is the conjugate base. H3O+/H2O and HCl/Cl- are both conjugate (A – B) pairs

  6. Acid – Base Theories For the following reaction, identify the acid, base, conjugate acid, and conjugate base: NH3 + H2O → NH4+ + OH- Base = NH3 Acid = H2O CA = NH4+ CB = OH- You may have noticed that the water was the acid in this reaction and the base in the previous reaction. A substance which can act as a proton donor in one reaction and a proton acceptor in another is said to be amphiprotic

  7. Mono- & Polyprotic Acids If an acid only has one hydrogen that it can lose- MONOPROTIC Ex. – HCl, HNO3, CH3COOH If an acid has more than one hydrogen that it can lose - POLYPROTIC Diprotic – 2 hydrogens to lose (H2SO4, H2CO3) Triprotic – 3 hydrogens to lose (H3PO4) Polyprotic acids lose H+ one at a time!

  8. Acid – Base Theories The same year as Brønsted and Lowry developed their acid – base theory, a third, more general theory was developed by Gilbert Lewis (as in the dot structures). In the Lewis theory, an acid is an electron pair acceptor. A base is an electron pair donor. H+ + OH-→ H2O Lewis acid Lewis base F H F B N H F H H :N – H H F F B F → + Lewis acid Lewis base

  9. Acid – Base Theories Complex ion formation is also a Lewis A-B reaction. Transition metal cations can attract ions or polar molecules (ligands) with an at least one available lone pair to form a dative bond. The cation acts as the LA and the ligand acts as the LB You have already learned about one of these: Fe3+ + SCN- → [Fe(SCN)]2+ LA LB complex ion Fe3+←:SCN- Coordinate Covalent (Dative) bond ligand

  10. Acid – Base Theories Many transition metal cations form complex ions in ammonia or in aqueous solution. Ex: Fe3+ + 6 H2O → [Fe(H2O)6]3+ Identify the Lewis acid and Lewis base in the reaction above. Lewis Base (e- pair donor) Lewis Acid (e- pair acceptor)

  11. - (Self) of ionization Water Auto Because of the hydrogen bonding between them, one water molecule can pull an H+ ion from another. 2 H2O (l) ↔ H3O+(aq) + OH-(aq) Since water is a pure liquid, it would not be included in the equilibrium expression. Keq = [H3O+] [OH-] The self – ionization of water can also be represented simply by showing a water molecule losing an H+ ion. H2O (l) ↔ H+(aq) + OH-(aq) Kw Keq = [H+] [OH-] Conductivity studies show that water ionizes only slightly. In pure water at 25 0C, [H3O+] = [OH-] = 1.00 x 10-7M Kw = (1.00 x 10-7 M)(1.00 x 10-7 M) = 1.00 x 10-14 M2 This value is known as the ion – product constant for water.

  12. The pH Scale Knowledge of the ionization constant for water has enabled chemists to develop a simple acidity scale. In pure water, the [H3O+] and the [OH-] are both 1.00 x 10-7 so it is neutral (that is the 2 concentrations are the same). If the [H3O+] > [OH-], then the solution is acidic. If the [OH-] > [H3O+], then the solution is basic. The pH scale is a measure of the [H3O+] ([H+]). pH = - log [H3O+] In water [H3O+] is 10-7 - log 10-7 = 7 , so in pure water pH = 7 - log 10-4 = 4, so pH = 4 If the [H+] = 1.00 x 10-4, Notice that even though 4 is smaller than 7, 10-4 is a larger number than 10-7. That is why pH gets smaller as the solution becomes more acidic.

  13. The pH Scale This value is equal to the – log[OH-]. There is also a value known as the pOH. If the [OH-] = 1.00 x 10-9, then the pOH = ? 9 Since the ion-product constant of water, Kw = 1.00 x 10-14, then in aqueous solution pH + pOH is always = to 14 This means if you know one, you can easily calculate the other. Ex: If the pOH = 9, then the pH = 5 One final helpful thing to remember when you do pH and related calculations: If [H+][OH-] = 1.00 x 10-14 you can determine one by dividing 1.00 x 10-14 by the other. If the [H+] = 1.20 x 10-2, what is [OH-] ? Example: 1.00 x 10-14/1.20 x 10-2 = 8.33 x 10-13 If we take the –log, 14.0-12.1 = 1.9 = pH -log (8.33 x 10-13) = 12.1 = pOH.

  14. The pH Scale Practice Examples: If the [OH-] = 3.90 x 10-8, what are the [H+], pH and pOH? If the pH is 4.67, determine the [H+], [OH-], and pOH. Here’s something you may not have realized. Neutral pH is 7 only at 25 0C. It is different at different temperatures b/c of Le Chatelier’s principle. For example: The Kw at 30 0C is 1.47 x 10-14 mol2 dm-6. what is the pH of pure water at this temperature?

  15. Strong and Weak Acids and Bases Strong acids are completely (more than 99%) ionized in H2O. Example: HBr → H+(aq) + Br-(aq) The strong acids are: HCl, HBr, HI, HNO3, H2SO4, HClO4 Weak acids are only partially ionized in water. Any acid other than the 6 listed as strong acids are weak. A few examples: HF, H2S, H2CO3, H3PO4, CH3COOH Strong bases are completely converted to OH- ions in H2O Weak bases are only partially converted to OH- ion in water Strong bases: Hydroxides of group 1 metals and heavier group 2 hydroxides – NaOH, Ca(OH)2, etc. Others are weak bases – Mg(OH)2, Al(OH)3, NH3, etc. The way to compare relative strengths of acids and bases is K.

  16. Strong and Weak Acids and Bases How could you distinguish between a strong or weak acid experimentally? • First – make sure you are working with equal concentrations (equimolar) • Determine the pH with an indicator or pH probe • Measure conductivity • React the acid with an active metal or something else (ex: NaHCO3) which produces gas or an obviously visible rxn and measure rate

  17. Calculating pH of Strong Acids Ex: Calculate the pH of 0.100 M HNO3

  18. Calculating pH of Strong Bases Calculate the pH for a 0.250 mol dm-3 of potassium hydroxide

  19. Ionization Constant for Weak Acids and Bases The ionization constant, Ka, for a weak acid, is simply the equilibrium constant for the dissociation of the acid in water. HC2H3O2 + H2O → H3O+ + C2H3O2- Ka = [H3O+][C2H3O2-] _____________ [HC2H3O2] Since the ionization may also be represented as HC2H3O2→ H+ + C2H3O2- Ka may also be written: Ka = [H+][C2H3O2-] ___________ [HC2H3O2] For a weak base like NH3, we use Kb Kb = [NH4+][OH-] _________ NH3 + H2O → NH4+ + OH- [NH3]

  20. Calculating pH of Weak Acids List major species in solution. Choose species that produce H+, write balanced equation Use Ka values to determine dominant equilibrium Write the Ka expression for the dominant equilibrium List initial concentrations Define x Write [equilibrium]s in terms of x; (n.b. #s 5 - 7 = ICE) Substitute [equilibrium]s into the expression Solve for x (almost always by assuming that [HA]0-x = [HA]0 Use 5% rule to verify approximation Calculate [H+] and pH

  21. Calculating pH of Weak Acids Example: Calculate the pH of 0.100 M HOCl(aq); Ka = 3.5 x 10 -8

  22. Example (mixed solutions) • Calculate the pH of an aqueous solution of 1.00 M HCN (Ka = 6.2 x 10-10) and 5.00 M HNO2 (Ka = 4.0 x 10-4). Also calculate [CN-]. • Since HCN and HNO2 are both weak, the major species will be HCN, HNO2, and H2O. All three can produce H+ but since the Ka for HNO2 is much larger than the other two, it will dominate the equilibrium.

  23. Solution

  24. Example (weak base) • Calculate pH for a 15.0M solution of NH3. (Kb = 1.8 x 10-5) Major species are NH3 and H2O. Both can produce OH-. • NH3(aq) + H20(l ) NH4+(aq) + OH-(aq) Kb= 1.8 x 10-5 • H20(l )  H+(aq) + OH-(aq) Kw=1.0 x 10-14

  25. Solution • Since Kb>>KwNH3 will dominate equilibrium, we use

  26. SALTS Arrhenius acids react with Arrhenius bases to form a salt and water. This reaction is called a neutralization. Ex: HCl + NaOH  NaCl + H2O Following a neutralization reaction, if the water is evaporated, the anion from the acid will combine with the cation from the base. This ionic compound is called a salt. Although a salt is formed from a neutralization reaction, hydrolysis (dissociation in water) of a salt does not always produce a neutral solution. 1. Salt of a strong acid and base = neutral. Ex: KNO3 2. Salt of a strong acid and weak base = acidic Ex: NH4ClO4 3. Salt of a weak acid and strong base = basic Ex: NaC2H3O2 • Salt of a weak acid and weak base = depends on relative strength of • acid and base. Need to compare Ka of acid and conjugate acid.

  27. Acid-Base Properties of Salts • Example: Calculate the pH of a .10M NH4Cl solution Kb= 1.8 x 10-5 • Major species = NH4+, Cl-, and H2O. • NH4+ NH3 + H+

  28. Buffer Solutions • A buffer solution is used to resist a change in pH when moderate amounts of strong acid or base are added to a solution. • It is produced by using fairly large amounts of a weak acid or base and a conjugate salt. • An example would be acetic acid with sodium acetate, or ammonia with ammonium chloride. • If a strong acid is added the weak conjugate base will react with excess H+. If a strong base is added the weak conjugate acid will protonate (neutralize) the excess OH-.

  29. Buffer Solutions As a result the H+ ions and OH- ions do not accumulate. Example: A buffer with NH3/NH4+ Add HCl → H+ + Cl- H+ + NH3 → NH4+ Add NaOH→ Na+ + OH- OH- + NH4+→ H2O + NH3

  30. Buffer Solutions Example: A buffer with CH3COOH/NaCH3COO Add HCl → H+ + Cl- H+ + CH3COO- → CH3COOH Add NaOH→ Na+ + OH- OH- + CH3COOH → H2O + CH3COO- Since the H+ and OH- ions are neutralized and what remains are simply the weak conjugates, the pH does not change significantly.

  31. Buffer Calculations Example: A buffer solution is made with 0.50 M acetic acid (Ka = 1.8 x 10-5) with 0.50 M sodium acetate. Calculate the pH of the solution.

  32. Buffer Calculations The calculation may also be done using the Henderson-Hasselbalch equation: pH = pKa + log ([A-]/[HA]) n.b. This is just a modified version of the Ka expression. Now calculate the pH of the same buffer solution after adding 0.10 mol of NaOH to 1.00L of the solution. Compare that to the pH change that occurs when adding the same amount of NaOH to 1.00 L of water.

  33. Buffer Calculations Example: A buffer solution is made with 0.25 M aqueous ammonia (Kb = 1.8 x 10-5) with 0.40 M ammonium chloride. Calculate the pH of the solution.

  34. pH Titration Curves Remember in an acid – base titration, a solution of known concentration (acid or base) is added to an unknown solution (base or acid) until the equivalence pointis reached. This point may be determined using a pH meter or the appropriate indicator. A pH meter may also be used to collect data to produce a pH titration curve. This can be used to more accurately identify the equivalence point, or to select a suitable indicator. It is often easier for us to use concentration as mmol/mL and keep track of our substance amounts in mmol rather than using moles.

  35. Simple pH curves All the following titration curves are based on both acid and alkali having a concentration of 1 mol dm-3. In each case, you start with 25 cm3 of one of the solutions in the flask, and the other one in a burette. Titration curves for strong acid v strong base We'll take hydrochloric acid and sodium hydroxide as typical of a strong acid and a strong base.

  36. Running alkali into the acid You can see that the pH only increases a very small amount until quite near the equivalence point. Then there is a really steep increase. If you calculate the values, the pH goes from 2.7 when you have added 24.9 cm3 to 11.3 when you have added 25.1 cm3. Running acid into the alkali • This is very similar to the previous curve except, of course, that the pH starts off high and increases as you add more HCl solution.

  37. Titration curves for weak acid v strong base We'll take ethanoic acid and sodium hydroxide as typical of a weak acid and a strong base. Running alkali into the acid The start of the graph shows a relatively rapid rise in pH but this slows down as a buffer solution containing ethanoic acid and sodium ethanoate is produced. Close to the equivalence point there is a steep rise as the acid disappears and OH- starts to accumulate.

  38. Titration curves for strong acid v weak base This time we are going to use hydrochloric acid as the strong acid and ammonia solution as the weak base. Running acid into the alkali Because you have got a weak base, the beginning of the curve is obviously going to be different. However, once you have got an excess of acid, the curve is essentially the same as before. At the very beginning of the curve, the pH starts by falling quite quickly as the acid is added, but the curve very soon gets less steep. This is because a buffer solution is being set up - composed of the excess ammonia and the ammonium chloride being formed. Notice that the equivalence point is now somewhat acidic ( a bit less than pH 5), because pure ammonium chloride isn't neutral. However, the equivalence point still falls on the steepest bit of the curve. That will turn out to be important in choosing a suitable indicator for the titration.

  39. Indicators One fairly simple way of determining the pH of a solution is to use an indicator. An indicator is a weak organic acid or base that has a conjugate with a different color. The transition from one form (color) to the other occurs over a fairly narrow pH range. Here is an imaginary example: OH- H+In-↔ H2O + In- H+ Yellow (acid) Red (conjugate base)

  40. Indicators Here are some actual indicators and their color ranges: yellow green blue Bromthymol blue: 0 – 6.3 6.4 – 7.5 7.6 - 14 pink/red orange yellow Methyl orange: 0 – 2.8 3.0 – 4.2 4.3 - 14 red orange yellow Methyl red: 0 – 4.2 4.4 – 6.2 6.2 - 14 clear pink magenta Phenolphthalein: 0 – 8.5 8.6 - 10 10.1 - 14 These indicators can also be combined into a universal indicator which will show a number of different color changes. This allows for more precise determination of pH over a large range

  41. pH Indicators Indicator pKa Acid Form Base Form ______________________________________________ thymolblue (pKa1) 2.0 R Y methyl orange 3.8 R Or bromcresolgreen 4.6 Y B bromthymol blue 7.1 Y B phenyl red 7.3 Y R thymol blue (pKa2) 8.8 Y B phenolphthlein 9.1 - V alizarine yellow 11.0 Y R indigocarmine 12.2 B Y