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Chapter 16

Chapter 16. ACID - BASE. 16.1 Arrhenius Theory. Acid H + in solution Base OH - in solution . Acid Proton donor H + donor Base Proton acceptor H + acceptor. Conjugate base What’s left of acid Conjugate acid Base + H + Hydronium ion H 3 O + Amphoteric Acts as acid or base.

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Chapter 16

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  1. Chapter 16 • ACID-BASE

  2. 16.1 Arrhenius Theory • Acid • H+ in solution • Base • OH- in solution

  3. Acid Proton donor H+ donor Base Proton acceptor H+ acceptor Conjugate base What’s left of acid Conjugate acid Base + H+ Hydronium ion H3O+ Amphoteric Acts as acid or base 16.2 Brönsted- Lowry Theory Proton = H+

  4. Acid – Base Equations HCl + H2O  H3O+ + Cl- Acid Base C.A. C.B. NH3 + H2O  NH4+ + OH- Base Acid C.A. C.B.

  5. 16.2 Strengths of Acids and Bases • Stronger the acid, weaker the CB • Stronger the base, weaker the CA • Ionization • SA completely ionizes • WA do not completely ionize • Central Atom • Higher oxidation number of central atom • Higher electro-negativity of central atom • Binary Acid • HI is the strongest

  6. Which Acid is Stronger? • HClO3 or HClO2 • HClO3 or HBrO3 • HClO3 or ClO3- • HCl or HBr • HClO4 –single strongest acid • HF – weak cuz HB

  7. 16.3 Autoionization of water • H2O + H2O  H3O+ + OH- • H2O  H+ + OH- • pH + pOH = 14 .0000001 M @ pH 7

  8. pH = -log[H+] = -log[H3O+] Acid pH < 7 Neutral pH = 7 Base pH > 7 pOH = -log[OH-] 16.4 pH Scale – power of hydrogen

  9. pH scale

  10. Ways to measure pH

  11. Litmus paper • Acid • Change from blue to red • Base • Change from red to blue

  12. pH paper

  13. Indicators • Phenolphthalein∆ clear to pink at 8.2

  14. Universal Indicator

  15. Indicators

  16. RED cabbage

  17. pH meters

  18. Acid and conjugate base are diff colors.

  19. hydrangea

  20. Polyprotic acids • H2SO3 H+ + HSO3- Ka1 = 1.7x10-2 • HSO3-  H+ + SO3-2Ka2 = 6.4 x 10-8 • Ka are on page 1115. • Kb are on page 1116. • HCl is a monoprotic acid. • H2 SO4is a diprotic acid. • H3PO4 is a triprotic acid.

  21. Hydrolysis • Water + salt  acid + base • HOH + NaCl  HCl + NaOH SA SB neutral • HOH + NH4Cl  HCl + NH4OH SA WB acidic • HOH + KF  HF + KOH WA SB basic • HOH + NH4C2H3O2  HC2H3O2 + NH4OH WA WB

  22. If WA w/ WB…..then • HC2H3O2 • Ka= 1.8 x 10-5 • This one is neutral cuz • Ka = Kb. • NH4OH • Kb = 1.8 x 10-5

  23. LEWIS Theory • Lewis acid - electron pair acceptor • Metals and comp w/ only 6e- around central atom. • Lewis base - electron pair donor • Compounds w/ lone pairs.

  24. What is the pH of a 0.010 M HCl solution? (strong acid) • HCl  H+ + Cl- .010 0 0 start 0 .010 .010 end • pH = -log[H+] = -log(.010) = 2.00

  25. The equation can also be written: • HCl + H2O  H3O+ + Cl-

  26. What is the pH of a 0.010 M NaOH solution? (strong base) • NaOH  Na+ + OH- • .010 0 0 start • 0 .010 .010 end • pOH = -log[OH-] • = -log (0.010) • = 2.00 • pH = 12.00 pH + pOH = 14

  27. WEAK ACID • What is the pH of a 0.010 M H2CO3 solution? Ka = 4.3 x 10-7 H2CO3  H+ + HCO3- • I 0.010 0 0 • C -x +x +x • E 0.010 – x x x

  28. Ka = [H+][HCO3-]= x2 = 4.3 x 10-7 [H2CO3] 0.010 - x • x2 = 4.3 x 10-9 • x = 6.56 x 10-5 • pH = -log [H+] = -log(6.56 x 10-5) • = 4.18 -x 4.1831 or 4.1871 w/ QUAD

  29. WEAK BASE • What is the pH of a 0.010 M NH4OH (NH3)aq solution? Kb = 1.8 x 10-5 NH4OH   NH4+ + OH- I 0.010 0 0 C -x +x +x E 0.010 – x x x

  30. Kb = [NH4+][OH-] [NH4OH] • 1.8 x 10-5 = x2 0.010 – x • x2 = 1.8 x 10-7 • x = 4.24 x 10-4 • pOH = -log[OH-] = -log(4.24 x 10-4) pOH = 3.37 pH = 10.623 10.615 w/ QUAD -x pH + pOH = 14

  31. Sample problems • 1.What is the pH of a 0.0050M HC2H3O2 solution? (CH3COOH) Ka = 1.8 x 10-5 (3.52) • 2. What is the pH of a 0.0050M KOH solution?(11.70) • 3. What is the pH of a 2.0 x 10-3 M NH4OH solution? (NH3) Kb = 1.8 x 10-5(10.28) • 4. What is the pH of a 0.0010 M HNO3 solution? (3.00)

  32. Sample Problems • What is the pH of a 3.69 x 10-3 M solution of NH4OH? Kb= 1.8 x 10-5 2. What is the pH of a 0.045M HClO4 solution? 3. What is the pH of a 0.002M solution of NaOH? 4. What is the pH of a 1.0 x 10-5 M HF solution? Ka=3.53 x 10-4 (10.41) 10.43 w/ Quad (1.35) (11.30) (4.23) or 5.01 w/ Quad

  33. pH of a salt • What is the pH of a 0.500M solution of NH4Cl? Kb = 1.8 x 10-5 • HOH + NH4Cl  NH4OH + HCl • WB SA acid • HOH + NH4+  NH4OH + H+ • 0.500 x x • K = [NH4OH][H+] Is this Ka or Kb? • [NH4+ ]

  34. pH of a salt • Normally….. • NH4OH   NH4+ + OH- • Kb = [NH4+][OH-] [NH4OH] • So the K on the previous page is Ka!!! • Ka x Kb = 1 x 10-14 • Ka = 1 x 10-14 = 5.6 x 10-10 • 1.8 x 10-5

  35. pH of a salt • K a= [NH4OH][H+] Kb = [NH4+][OH-] [NH4+ ] [NH4OH] • [H+][OH-] = Ka x Kb = 1x10-14 • Back to the problem…. • = 5.6 x 10-10 = x2/ 0.500 • x2 = 2.8 x 10-10 • x = 1.7 x10-5 pH = 4.78

  36. So in general…. • x2 = 1x10-14 • [conc] Ka or Kb

  37. What is the pH of a 0.500M NaCl solution? • HOH + NaCl HCl + NaOH • SA SB neutral • pH = 7.00

  38. What is the pH of a 0.250M NaCN solution? (Ka = 4.9 x 10-10) • HOH + NaCN HCN + NaOH • HOH + CN-  HCN + OH- • WA SB basic • K = [HCN][OH-] [CN-]

  39. x2 = 1x10-14 • 0.250M 4.9 x 10-10 • x2 = 5.10 x 10-6 • x = 2.26 x 10-3 • pOH = 2.64 • pH = 11.35

  40. KHP

  41. KHP + NaOH

  42. (#H)M1V1 = M2V2(#OH) • How many ml of 0.100 M H2SO4 are needed to neutralize 45.0ml of 0.200 MNaOH? (2)(0.100M)(x) = (0.200M)(45.0ml)(1) x = 45.0 ml

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