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The Chemistry of Acids and Bases

The Chemistry of Acids and Bases. Acid and Bases. Acid and Bases. Acid and Bases. Acids. General Descriptions. Have a sour taste. Vinegar is a solution of acetic acid. Citrus fruits contain citric acid. React with certain metals to produce hydrogen gas.

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The Chemistry of Acids and Bases

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  1. The Chemistry of Acids and Bases

  2. Acid and Bases

  3. Acid and Bases

  4. Acid and Bases

  5. Acids General Descriptions Have a sour taste. Vinegar is a solution of acetic acid. Citrus fruits contain citric acid. React with certain metals to produce hydrogen gas. React with carbonates and bicarbonates to produce carbon dioxide gas Bases Have a bitter taste. Feel slippery. Many soaps contain bases.

  6. Some Properties of Acids • Produce H+ (as H3O+) ions in water (the hydronium ion is a hydrogen ion attached to a water molecule) see next slide • Taste sour • Corrode metals • Electrolytes • React with bases to form a salt and water • pH is less than 7 • Turns blue litmus paper to red “Blue to Red A-CID”

  7. Lone Hydrogen ions do not exist by themselves in solution. H+ is always bound to a water molecule to form a hydronium ion H3O+ (hydronium ion) also can be thought of as H+ (hydrogen ion) for our purposes

  8. Some Properties of Bases • Produce OH- ions in water • Taste bitter, chalky • Are electrolytes • Feel soapy, slippery • React with acids to form salts and water • pH greater than 7 • Turns red litmus paper to blue “Basic Blue”

  9. Some Common Bases NaOH sodium hydroxide lye KOH potassium hydroxide liquid soap Ba(OH)2 barium hydroxide stabilizer for plastics Mg(OH)2 magnesium hydroxide “MOM” Milk of magnesia Al(OH)3 aluminum hydroxide Maalox (antacid)

  10. Acid/Base definitions • Definition #1: Arrhenius (traditional) Acids – produce H+ ions (or hydronium ions H3O+) Bases – produce OH- ions (problem: some bases don’t have hydroxide ions!)

  11. Arrhenius acid is a substance that produces H+ (H3O+) in water Arrhenius base is a substance that produces OH- in water

  12. Acid/Base Definitions • Definition #2: Brønsted – Lowry Acids – proton donor Bases – proton acceptor A “proton” is really just a hydrogen atom that has lost it’s electron!

  13. Brønsted-Lowry Theory of Acids & Bases Conjugate Acid-Base Pairs General Equation

  14. Brønsted-Lowry Theory of Acids & Bases

  15. HA (aq) + H₂O₍ι₎ → H₃O⁺ (aq)+ A⁻ (aq) Conjugate Acid-Base Pairs base Conjugate acid • In this example: HA donates a proton to water which would make HA the acid and water accepts the proton which make it a base. • The conjugate acid is H₃O⁺ the hydronium ion. • The conjugate base is A⁻, which is everything that remains after the acid donates its proton Acidconjugate base

  16. HCl(aq) + H₂O₍ι₎ → H₃O⁺ (aq)+ Cl⁻ (aq) Conjugate Acid-Base Pairs base Conjugate acid • More specific example of acid-base conjugate pairs Acidconjugate base

  17. Identifying Conjugate acid/base pairs • Which of the following represent conjugate acid base pairs • A. HF and F- • B. NH4+ , NH3 • C. HCl, H2O

  18. Identifying Conjugate acid/base pairs • Which of the following represent conjugate acid base pairs • A. HF and F- YES • B. NH4+ , NH3 YES • C.HCl, H2O NO • * you can identify the pairs because the two species differ by one H+

  19. Writing Conjugate Bases • Write the conjugate base for each: • A. HClO4 • B. H3PO4 • C. CH3NH3+

  20. Writing Conjugate Bases • Write the conjugate base for each: • A. HClO4 • B. H3PO4 • C. CH3NH3+ • A. HClO4 H+ + ClO4- • B. H3PO4 H+ + H2PO4- • C. CH3NH3+ H+ + CH3NH2

  21. Electrolytes

  22. ELECTROLYTES Electrolytes are species which conducts electricity when dissolved in water. Acids, Bases, and Salts are all electrolytes. Salts and strong Acids or Bases form Strong Electrolytes. Salt and strong acids (and bases) are fully dissociated therefore all of the ions present are available to conduct electricity. HCl(s) + H2O  H3O+ + Cl- Weak Acids and Weak Bases for Weak Electrolytes. Weaks electrolytes are partially dissociated therefore not all species in solution are ions, some of the molecular form is present. Weak electrolytes have less ions avalible to conduct electricity. NH3 + H2O  NH4+ + OH-

  23. Acids & Bases STRONG vs WEAK _ completely ionized _ partially ionized _ strong electrolyte _ weak electrolyte _ ionic/very polar bonds _ some covalent bonds Strong Acids:Strong Bases: HClO4LiOH H2SO4NaOH HIKOH HBrCa(OH)2 HClSr(OH)2 HNO3Ba(OH)2

  24. Homework P 477 (8, 10, 12, 14, 17 and 18)

  25. pH scale

  26. The pH scale is a way of expressing the strength of acids and bases. Instead of using very small numbers, we just use the NEGATIVE power of 10 on the Molarity of the H+ (or OH-) ion.Under 7 = acid 7 = neutralOver 7 = base pH Scale

  27. pH of Common Substances

  28. pHA measure of the hydronium ion • The scale for measuring the hydronium ion concentration [H3O+] in any solution must be able to cover a large range. A logarithmic scale covers factors of 10. The “p” in pH stands for log. • A solution with a pH of 1 has [H3O+] of 0.1 mol/L or 10-1 • A solution with a pH of 3 has [H3O+] of 0.001 mol/L or 10-3 • A solution with a pH of 7 has [H3O+] of 0.0000001 mol/L or 10-7 pH = - log [H3O+]

  29. Calculating the pH pH = - log [H+] (Remember that the [ ] mean Molarity) Example: If [H+] = 1 X 10-10pH = - log 1 X 10-10 pH = - (- 10) pH = 10 Example: If [H+] = 1.8 X 10-5pH = - log 1.8 X 10-5 pH = - (- 4.74) pH = 4.74

  30. pH calculations – Solving for H+ If the pH of Coke is 3.12, [H+] = ??? Because pH = - log [H+] then - pH = log [H+] Take antilog (10x) of both sides and get 10-pH =[H+] [H+] = 10-3.12 = 7.6 x 10-4 M *** to find antilog on your calculator, look for “Shift” or “2nd function” and then the log button

  31. pH calculations – Solving for H+ • A solution has a pH of 8.5. What is the Molarity of hydrogen ions in the solution?

  32. pH calculations – Solving for H+ • A solution has a pH of 8.5. What is the Molarity of hydrogen ions in the solution? pH = - log [H+] 8.5 = - log [H+] -8.5 = log [H+] Antilog -8.5 = antilog (log [H+]) 10-8.5 = [H+] 3.16 X 10-9 = [H+]

  33. Try These! Find the pH of these: 1.) A 0.15 M solution of Hydrochloric acid 2) A 3.00 X 10-7 M solution of Nitric acid

  34. pOH • Since acids and bases are opposites, pH and pOH are opposites! • pOH does not really exist, but it is useful for changing bases to pH. • pOH looks at the perspective of a base pOH = - log [OH-] Since pH and pOH are on opposite ends, pH + pOH = 14

  35. Water Equilibrium Kw = [H+] [OH-] = 1.0 x 10-14 Equilibrium constant for water • Water or water solutions in which [H+] = [OH-] = 10-7 M are neutral solutions. • A solution in which [H+] > [OH-] is acidic • A solution in which [H+] < [OH-] is basic

  36. [H3O+], [OH-] and pH What is the pH of the 0.0010 M NaOH solution? [OH-] = 0.0010 (or 1.0 X 10-3 M) pOH = - log 0.0010 pOH = 3 pH = 14 – 3 = 11 OR Kw = [H3O+] [OH-] [H3O+] = 1.0 x 10-11 M pH = - log (1.0 x 10-11) = 11.00

  37. Titrations

  38. Adding acid to base = • Acids can react with bases to form a salt and water. • Salt definition: sometimes 'salt' simply refers to table salt, which is sodium chloride. Usually the term is applied to an ionic compound produced by reacting an acid with a base. • Examples: NaCl, KCl, CuSO4

  39. Homework P 478 (32, 34, 36, 37, 40, 42, 44, 46, 48) complete only part a and b of each question

  40. TITRATION Titration of a strong acid with a strong base ENDPOINT = POINT OF NEUTRALIZATION = EQUIVALENCE POINT At the end point for the titration of a strong acid with a strong base, the moles of acid (H+) equals the moles of base (OH-) to produce the neutral species water (H2O). If the mole ratio in the balanced chemical equation is 1:1 then the following equation can be used. MOLES OF ACID = MOLES OF BASE nacid = nbase Since M=n/V MAVA = MBVB

  41. TITRATION MAVA = MBVB 1. Suppose 75.00 mL of hydrochloric acid was required to neutralize 22.50 mLof 0.52 M NaOH. What is the molarity of the acid? HCl + NaOH  H2O + NaCl Ma Va = Mb Vb rearranges to Ma = Mb Vb / Va so Ma = (0.52 M) (22.50 mL) / (75.00 mL) = 0.16 M Now you try: 2. If 37.12 mL of 0.843 M HNO3 neutralized 40.50 mL of KOH, what is the molarity of the base? Mb = 0.773 mol/L

  42. Molarity and Titration

  43. TITRATION Titration of a strong acid with a strong base ENDPOINT = POINT OF NEUTRALIZATION = EQUIVALENCE POINT At the end point for the titration of a strong acid with a strong base, the moles of acid (H+) equals the moles of base (OH-) to produce the neutral species water (H2O). If the mole ratio in the balanced chemical equation is NOT 1:1 then you must rely on the mole relationship and handle the problem like any other stoichiometry problem. MOLES OF ACID = MOLES OF BASE nacid = nbase

  44. TITRATION 1. If 37.12 mL of 0.543 M LiOH neutralized 40.50 mL of H2SO4, what is the molarity of the acid? 2 LiOH + H2SO4 Li2SO4 + 2 H2O First calculate the moles of base: 0.03712 L LiOH (0.543 mol/1 L) = 0.0202 mol LiOH Next calculate the moles of acid: 0.0202 mol LiOH (1 mol H2SO4 / 2 mol LiOH)= 0.0101 mol H2SO4 Last calculate the Molarity: Ma = n/V = 0.010 mol H2SO4 / 0.4050 L = 0.248 M 2. If 20.42 mL of Ba(OH)2 solution was used to titrate29.26 mL of 0.430 M HCl, what is the molarity of the barium hydroxide solution? Mb = 0.308 mol/L

  45. Molarity and Titration • A student finds that 23.54 mL of a 0.122 M NaOH solution is required to titrate a 30.00-mL sample of hydr acid solution. What is the molarity of the acid? • A student finds that 37.80 mL of a 0.4052 M NaHCO3 solution is required to titrate a 20.00-mL sample of sulfuric acid solution. What is the molarity of the acid? • The reaction equation is: H2SO4 + 2 NaHCO3 → Na2SO4 + 2 H2O + 2 CO2

  46. PRACTICE PROBLEM #25 10.8 mL 1. How many milliliters of 1.25 M LiOH must be added to neutralize 34.7 mL of 0.389 M HNO3? 2. What mass of Sr(OH)2 will be required to neutralize 19.54 mL of 0.00850 M HBr solution? 3. How many mL of 0.998 M H2SO4 must be added to neutralize 47.9 mL of 1.233 M KOH? 4. What is the molar concentration of hydronium ion in a solution of pH 8.25? 5. What is the pH of a solution that has a molar concentration of hydronium ion of 9.15 x 10-5? 6. What is the pOH of a solution that has a molar concentration of hydronium ion of 8.55 x 10-10? 0.0101 g 29.6 mL 5.623 x 10-9 M pH = 4.0 pOH = 4.9

  47. GROUP STUDY PROBLEM #25 ______1. How many milliliters of 0.75 M KOH must be added to neutralize 50.0 mL of 2.50 M HCl? ______2. What mass of Ca(OH)2 will be required to neutralize 100 mL of 0.170 M HCl solution? ______3. How many mL of 0.554 M H2SO4 must be added to neutralize 25.0 mL of 0.9855 M NaOH? ______ 4. What is the molar concentration of hydronium ion in a solution of pH 2.45? ______ 5. What is the pH of a solution that has a molar concentration of hydronium ion of 3.75 x 10-9? ______ 6. What is the pOH of a solution that has a molar concentration of hydronium ion of 4.99 x 10-4?

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