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electrolytes  electrolytes

BASES. ACIDS. sour taste. bitter taste. turn litmus red. turn litmus blue. react with metals to form H 2 gas. slippery feel. ammonia, lye, antacid, baking soda. vinegar, soda, apples, citrus fruits. electrolytes  electrolytes. H. H. –. +. O. O. Cl. Cl. H. H. H. H.

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electrolytes  electrolytes

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  1. BASES ACIDS • sour taste • bitter taste • turn litmus red • turn litmus blue • react with metals to form H2 gas • slippery feel • ammonia, lye, antacid, baking soda • vinegar, soda, apples, citrus fruits • electrolytes  electrolytes

  2. H H – + O O Cl Cl H H H H Arhenius Definition • In aqueous solutions Acidsform hydronium ions (H3O+) HCl + H2O  H3O+ + Cl– Produce Hydronium Acid

  3. H H – + N O O N H H H H H H H H Arhenius Definition • In aqueous solutions Basesform hydroxide ions (OH-) NH3 + H2O  NH4+ + OH- Produce Hydroxide Base

  4. Exceptions? Oh what to do? This did not encompass all of the compounds that we knew were basic and acidic.

  5. Bronsted-Lowry Definition • Acids • Bases are proton (H+) donors. • are proton (H+) acceptors. HCl(aq) + H2O(l) Cl–(aq) + H3O+(aq) acid base conjugate base conjugate acid

  6. Example (Acid) H2O + HNO3 H3O+ + NO3– Conjugate Base Acid Conjugate Acid Base  Hydronium Conjugate Base

  7. Example(Base) NH3 + H2O  NH4+ + OH- B A CA CB  Hydroxide Conjugate Acid

  8. H2O…Acid or Base? So water is an acid? BOTH So water is a base. Amphiprotic: A chemical species that can act as EITHER, an acid or a base.

  9. Practice Activity • Give the conjugate base for each of the following: HF HBr HI H3O+

  10. Practice Activity Partner Up! Partner 1, write the following bases on the back of a cue card. (one acid per card (3 cards). Partner 2, write the conjugate acid for the acid on the other side of the card.

  11. Give the conjugate acid for each of the following: HBr(aq) H2SO4(aq) HCO3-(aq) Br –(aq) HSO4-(aq) CO32-(aq)

  12. Practice Activity Partner Up! Partner 1, write the following acids on the back of a cue card. (one acid per card (3 cards). Partner 2, write the conjugate base for the acid on the other side of the card.

  13. Give the conjugate base for each of the following: SO42-(aq) H2SO4(aq) HCl(aq) HCO3-(aq) HSO4-(aq) H2SO4(aq) H2CO32-(aq)

  14. Section 6.2 pH and pOH calculations

  15. H2O + H2O H3O+ + OH- Auto-ionization of Water Most water molecules do not ionize. Only 1 in 556 000 000 water molecules ionize! The other 555 999 999 remain H2O! Square Brackets indicate concentration Kw = [H3O+][OH-] = 1.0  10-14

  16. pouvoirhydrogène (Fr.) “power of hydrogen” pH Scale 14 0 7 INCREASING ACIDITY INCREASING BASICITY NEUTRAL pH = -log[H3O+]

  17. Can go beyond 0 and 14 pH of Common Substances

  18. Super Acids/Super Bases A very concentrated (really hot) strong acid can have a pH below 0! (-0.5, -1) A very concentrated (really hot) strong base can have a pH above 14! (15, 16)

  19. Relationship Between Hydronium Concentration [H3O+(aq)] and pH pH = 1 [H3O+(aq)] = 1.00 x10-1 What’s the relationship? pH = 3 [H3O+(aq)] = 1.00 x10-3 pH = 5 [H3O+(aq)] = 1.00 x10-5 pH = 9 [H3O+(aq)] = 1.00 x10-9 pH = 13 [H3O+(aq)] = 1.00 x10-13 Relationship: [H30+(aq)] is related to pH by powers of 10.

  20. Formula’s (pH “Box”) Kw = [H+][OH-] [H+] [OH-] pH = -log[H+] pOH = -log[OH-] [H+] = 1x10-pH [OH-] = 1x10-pOH pH pOH pH + pOH = 14

  21. Whenever you deal with pH and pOH, all you need is!..... pH + pOH = 14 pH/H3O+ pOH/OH- Perfect Diamond Shape -log[OH-(aq)] pOH= pH = -log[H3O+(aq)] Mr. K’s pH/pOH Diamond of Awesome-ness 1 x 10-pOH [OH-(aq)]= 1 x 10-pH [H3O+(aq)]= =Kw [H3O+(aq)] [OH-(aq)]= 1.00 x 10-14

  22. -log[H3O+(aq)] pH = pH = -log[4.7 x 10-11] + - log -log( 11 4.7 x 10 - ^ pH =

  23. [H3O+(aq)]= 1 x 10-pH 1 x 10-10.33 [H3O+(aq)]= - ^ 10 1 x 10.33 [H3O+(aq)]=

  24. hydroxide pH + pOH = 14 [OH-(aq)]= 1 x 10-pOH pOH = 14 - pH pOH = 14 – 5.3 [OH-(aq)]= 1 x 10-8.7 pOH = 8.7 - ^ 10 1 x 8.7 [OH-(aq)]=

  25. Example: 1 1 HNO3(aq) + H2O(l) ↔ H3O+(aq) + NO3-(aq) 1 1 C = 0.050 mol/L -log[H3O+(aq)] pH = -log[0.050] pH = pH = What is the pH of 0.050mol/L HNO3?

  26. Example: 1 1 HBr(aq) 1 + H2O(l) ↔ H3O+(aq) + Br-(aq) 1 C = ? [H3O+(aq)]= 1 x 10-pH [H3O+(aq)]= 1 x 10-4.4 pH + pOH = 14 [H3O+(aq)]= pH = 14 - pOH pH = 14 – 9.6 = 4.4 What is the amount concentration of HBr in a solution that has a pOH of 9.6?

  27. Example: N G 1 1 HBr(aq) 1 + H2O(l) ↔ H3O+(aq) + Br-(aq) 1 C = ? C = C = (N/G) (1/1) [HBr(aq)] = What is the amount concentration of HBr in a solution that has a pOH of 9.6?

  28. Why is the concentration of HBr the same as the hydronium ion concentration? Look at the dissociation/ionization equation HBr(aq)  H +(aq) + Br -(aq) HBr(aq) + H2O(l)  H30+(aq) + Br-(aq) There is a 1:1 relationship between HBr and the ions

  29. Acid – Base Indicators: Substances that change colour due to the acidity of a solution.

  30. They are a weak acid – conjugate base pair that exist in two forms (two different colours) due to presence or lack of a single proton (Hydrogen atom) in the chemical formula.

  31. Because of the complex nature of the chemical formula of each indicator. • Abbreviations are usually used to make using indicators less complex. • Ex: HLt – Lt- are the acid and conjugate base of litumus with Hlt being the red form and Lt- being the blue form

  32. Example Reactions Placing red litmus paper in a base: HLt(aq) + NaOH (aq) H2O (l) + Na+ (aq) + Lt- (aq) Placing blue litmus in an acid: HCl (aq) + Lt- (aq)HLt (aq) + Cl- (aq)

  33. Universal Indicators An indicator substance that changes a variety of different colours to indicate a more precise acidity of the solution being tested. ***most indicators DO NOT DO THIS*** **Usually only do two colours**

  34. Uses of Indicators • Mark the end point of a titration (chp. 8) • to estimate the pH of a solution. ***We can use a series of indicators to get a fairly precise pH instead of using the more expensive pH meter.**** pH = 2.8 – 3.2 2.8-8.0 0-4.8 7 0 14 0-3.2 Indicator Table (Pg. 10)

  35. Example (You Try) 0-8.2 Phenolphthalein 7.6-8.2 7.6-14 Bromothymol Blue pH = 7.6-8.0 8.0-14 Phenol Red

  36. Indicator Practice Activity

  37. Defining Acids • According to the modified Arrhenius theory • Acids are substances that react with water (ionize in water) to produce hydronium ions. • Alternately, according to Bronsted – Lowry theory • Acids are proton donors that become basic (conjugate bases) once they donate their proton.

  38. Example CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO-(aq) This reaction can be explained using either definition and requires the acid react with water. H2O(l) H3O+(aq) CH3COOH(aq) CH3COO-(aq)

  39. Example • How can we explain using either the modified arrhenius theory or the Bronsted-Lowry theory that a solution of NaHSO4 will turn blue litmus paper red? Acidic Arhenius Bronsted-Lowry NaHSO4(aq) Na+(aq) HSO4-(aq) SO42-(aq) HSO4-(aq) + + H2O(l) → H3O+(aq)

  40. Example • How can we explain using either the modified arrhenius theory or the Bronsted-Lowry theory that a solution of NaHSO4 will turn blue litmus paper red? Acidic Bronsted-Lowry Na+(aq) HSO4-(aq) SO42-(aq) + + H2O(l) → H3O+(aq) HSO4-(aq) Conjugate Base Acid

  41. Bases • So far bases have been metal hydroxides that can be explained by simple dissociation to produce hydroxide ions according to the Arrhenius theory. Ex: Ca(OH)2(aq) Ca2+(aq) + 2OH-(aq)

  42. Modified Arhenius Theory • The original Arrhenius theory doesn’t account for the basic nature of ammonia or baking soda. • The modified Arrhenius theory, that bases ionize in water (react with water) to produce hydroxide ions. Helps to explain why such substances are in fact basic.

  43. Example Na2CO3(s) 2Na+(aq) + CO32-(aq) Then, CO32-(aq) + H2O(l)OH-(aq) + HCO3-(aq) • The Bronsted – Lowry definition of a base as a proton acceptor can also help explain this reaction.

  44. How did we know that the carbonate ion was going to produce a basic solution and that the sodium ion was a spectator? • Bases are proton acceptors and usually have a negative charge (water and ammonia are exceptions) and acids are proton donors. Therefore, the sodium ion cannot act as an acid or a base. Na does not have H to give away and is + so it can accept an H.

  45. A Special Case • Nonmetal oxides in water will form acidic solutions. There is a two step process to explain how this occurs.

  46. Ex: • CO2(g) + H2O(l) H2CO3(aq) • H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-(aq)

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