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ADVANCED PLACEMENT CHEMISTRY ACIDS, BASES, AND AQUEOUS EQUILIBRIA

ADVANCED PLACEMENT CHEMISTRY ACIDS, BASES, AND AQUEOUS EQUILIBRIA. Acids - taste sour Bases (alkali)- tastes bitter and feels slippery. Arrhenius concept - acids produce hydrogen ions in aqueous solution while bases produce hydroxide ions.

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ADVANCED PLACEMENT CHEMISTRY ACIDS, BASES, AND AQUEOUS EQUILIBRIA

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  1. ADVANCED PLACEMENT CHEMISTRYACIDS, BASES, AND AQUEOUS EQUILIBRIA

  2. Acids- taste sourBases(alkali)- tastes bitter and feels slippery

  3. Arrhenius concept- acids produce hydrogen ions in aqueous solution while bases produce hydroxide ions

  4. Bronsted-Lowry model- acids are proton (H+) donors and bases are proton acceptors

  5. Lewis model- acids are electron pair acceptors while bases are electron pair donors

  6. hydronium ion (H3O+)- formed on reaction of a proton with a water molecule. H+ and H3O+ are used interchangeably in most situations.

  7. HA(aq) + H2O(l) H3O+(aq) + A-(aq)Acid Base Conjugate Conjugate Acid Baseconjugate base- everything that remains of the acid molecule after a proton is lostconjugate acid- base plus a proton

  8. Acid dissociation constant (Ka)Ka = [H3O+][A-] or Ka = [H+][A-] [HA] [HA]

  9. Strong acid - mostly dissociated - equilibrium lies far to the right - a strong acid yields a weak conjugate base (much weaker than H2O)Weak acid- mostly undissociated - equilibrium lies far to the left - has a strong conjugate base (stronger than water)

  10. Common strong acids -all aqueous solutions (Know these!)H2SO4 (sulfuric)HCl (hydrochloric)HNO3 (nitric)HClO3 (chloric)HClO4 (perchloric)HI (hydroiodic)H2CrO4 (chromic)HMnO4 (permanganic)HBr (hydrobromic)

  11. Sulfuric acid is a diprotic acid which means that it has two acidic protons. The first (H2SO4) is strong and the second (HSO4-) is weak.

  12. Oxyacids- most acids are oxyacids - acidic proton is attached to OWeak oxyacids: H3PO4 (phosphoric) HNO2 (nitrous)HOCl (hypochlorous)

  13. Within a series, acid strength increases with increasing numbers of oxygen atoms. For example: HClO4 > HClO3 > HClO2 > HClO and H2SO4 > H2SO3(Electronegative O draws electrons away from O-H bond)

  14. Acid strength increases with increasing electronegativity of oxyacids. For example: HOCl>HOBr>HOI>HOCH3

  15. Organic acids- O have carboxyl group -C-OH - usually weak acidsCH3COOH acetic C6H5COOH benzoic

  16. Hydrohalic acids- H is attached to a halogen (HCl, HI, etc.)HF is the only weak hydrohalic acid. Although the H-F bond is very polar, the bond is so strong (due to the small F atom) that the acid does not completely dissociate.

  17. Weak acid strength is compared by the Ka values of the acids. The smaller the Ka, the weaker the acid. Strong acids do not have Ka values because the [HA] is so small and can not be measured accurately.

  18. Amphoteric substance- Substance that can act as an acid or as a base. Ex. H2O, NH3, HSO4-(anything that can both accept and donate a proton)

  19. autoionization of water H2O + H2O  H3O+ + OH- base acid conjugate conjugate acid base

  20. Ion product constant for water (Kw) Kw = [H3O+][OH-] Kw = [H+][OH-]At 25oC, Kw = 1 x 10-14 mol2/L2 because [H+] = [OH-] = 1 x 10-7 M

  21. No matter what an aqueous solution contains, at 25oC, [H+][OH-] = 1 x 10-14 Neutral solution [H+] = [OH-] Acidic solution [H+] > [OH-] Basic solution [H+] < [OH-]Kw varies with temperature

  22. pH = -log [H+]If [H+] = 1.0 x 10-7 M, pH = 7.00

  23. Significant figures in pH and other log values: The number of decimal places in the log value should equal the number of significant digits in the original number (concentration).

  24. pOH = -log [OH-]pK = -log KpH and pOH are logarithmic functions. The pH changes by 1 for every power of 10 change in [H+]. pH decreases as [H+] increases.

  25. pH + pOH = 14[H+] = antilog(-pH) [OH-] = antilog(-pOH)

  26. Calculating pH of Strong Acid Solutions Calculating pH of strong acid solutions is generally very simple. The pH is simply calculated by taking the negative logarithm of concentration of a monoprotic strong acid. For example, the pH of 0.1 M HCl is 1.0. However, if the acid concentration is less than 1.0 x 10-7, the water becomes the important source of [H+] and the pH is 7.00. The pH of an acidic solution can not be greater than 7 at 25oC!!!!!

  27. Another exception is calculating the pH of a H2SO4 solution that is more dilute than 1.0 M. At this concentration, the [H+] of the HSO4- must also be calculated.

  28. Ex. Calculate the [H+] and pH in a 1.0 M solution of HCl. HCl is a strong monoprotic acid, therefore its concentration is equal to the hydrogen ion concentration. [H+] = 1.0 M pH = - log (1.0) = 0.00

  29. Ex. Calculate the pH of 1.0 x 10-10 M HCl. Since the [H+] is less than 1.0 x 10-7, the [H+] from the acid is negligible and the pH = 7.00

  30. Calculating pH of Weak Acid Solutions Calculating pH of weak acids involves setting up an equilibrium. Always start by writing the equation, setting up the acid equilibrium expression (Ka), defining initial concentrations, changes, and final concentrations in terms of X, substituting values and variables into the Ka expression and solving for X.

  31. Ex. Calculate the pH of a 1.00 x 10-4 M solution of acetic acid. The Ka of acetic acid is 1.8 x 10-5HC2H3O2 H+ + C2H3O2-Ka = [H+][C2H3O2-] = 1.8 x 10-5 [HC2H3O2]

  32. Reaction HC2H3O2 H+ + C2H3O2- Initial 1.00 x 10-4 0 0 Change -x +x +x Equilibrium 1.00 x 10-4 - x x x Often, the -x in a Ka expression 1.8 x 10-5 = (x)(x) can be treated as negligible. 1.00x10-4 - x 1.8 x 10-5 (x)(x) x = 4.2 x 10-5 1.00 x 10-4

  33. When you assume that x is negligible, you must check the validity of this assumption. To be valid, x must be less than 5% of the number that it was to be subtracted from. In this example 4.2 x 10-5 is greater than 5% of 1.00 x 10-4. This means that the assumption that x was negligible was invalid and x must be solved for using the quadratic equation or the method of successive approximation.

  34. Use of the quadratic equation:x2 + 1.8 x 10-5x - 1.8 x 10-9 = 0x = 3.5 x 10-5 and -5.2 x 10-5Since a concentration can not be negative, x= 3.5 x 10-5 M x = [H+] = 3.5 x 10-5 pH = -log 3.5 x 10-5 = 4.46

  35. Another method which some people prefer is the method of successive approximations. In this method, you start out assuming that x is negligible, solve for x, and repeatedly plug your value of x into the equation again until you get the same value of x two successive times.

  36. Using successive approximation for the previous example would go as follows: x = 4.2 x 10-5 x = 3.2 x 10-5 x = 3.5 x 10-5 x = 3.4 x 10-5 x = 3.4 x 10-5[H+] = 3.4 x 10-5pH = 4.47

  37. or/ with a graphing calculator:(1.8 x 10-5 x 1.00 x 10-4) = 4.2 x 10-5 (not negl)((1.8 x 10-5)( 1.00 x 10-4 -ans)) = 3.2 x 10-5 =3.5 x 10-5 =3.4 x 10-5 =3.4 x 10-5 Use answer key on calculator for this! Press Enter key repeatedly until you get the same answer each time

  38. Calculating pH of polyprotic acidsAll polyprotic acids dissociate stepwise. Each dissociation has its own Ka value. As each H is removed, the remaining acid gets weaker and therefore has a smaller Ka. As the negative charge on the acid increases it becomes more difficult to remove the positively charged proton.

  39. Except for H2SO4, polyprotic acids have Ka2 and Ka3 values so much weaker than their Ka1 value that the 2nd and 3rd (if applicable) dissociation can be ignored. The [H+] obtained from this 2nd and 3rd dissociation is negligible compared to the [H+] from the 1st dissociation. Because H2SO4 is a strong acid in its first dissociation and a weak acid in its second, we need to consider both if the concentration is more dilute than 1.0 M. The quadratic equation is needed to work this type of problem.

  40. Ex. Calculate the pH of a 1.00 x 10-2 M H2SO4 solution. The Ka of HSO4- is 1.2 x 10-2 H2SO4 H + + HSO4- Before 1.00 x 10-2 0 0 Change -1.00 x 10-2 +1.00 x 10-2 +1.00 x 10-2 After 0 1.00 x 10-2 1.00 x 10-2 Reaction HSO4- H+ + SO4- Initial 1 x 10-2 1x 10-2 0 Change -x +x +x Equil. 1 x 10-2 -x 1 x 10-2 +x x

  41. Ka = [H+][SO4-]= 1.2 x 10-2 [HSO4-] 1.2 x 10-2 = (1 x 10-2 + x)(x) (1 x 10-2 -x) Using the quadratic equation, x = 4.52 x 10-3 [H+]= 1 x 10-2 + (4.52 x 10-3) = 1.45 x 10-2 pH = 1.84

  42. Determination of the pH of a Mixture of Weak AcidsOnly the acid with the largest Ka value will contribute an appreciable [H+]. Determine the pH based on this acid and ignore any others.

  43. Determination of the Percent Dissociation of a Weak Acid% dissociation = amt. dissociated (mol/L) x100 initial concentration (mol/L) = final [H+] x 100initial [HA]

  44. For a weak acid, percent dissociation (or ionization) increases as the acid becomes more dilute. Equilibrium shifts to the right.

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