Chemistry 201: General Chemistry II - Lecture

Transcription

1 Chemistry 201: General Chemistry II - Lecture Dr. Namphol Sinkaset Chapter 17 Study Guide Concepts 1. There are multiple definitions for acids and bases. 2. An Arrhenius acid is a substance that produces H + (H 3 O + ) ions in aqueous solution. An Arrhenius base is a substance that produces OH ions in aqueous solution. 3. A Brønsted-Lowry acid is a proton donor whereas a Brønsted-Lowry base is a proton acceptor. 4. Under the Brønsted-Lowry definitions, acids and bases always occur in pairs. 5. Substances that can act as either an acid or base are called amphoteric. 6. Conjugate acid-base pairs are two substances related to each other by a transfer of a proton. 7. A Lewis acid is an electron pair acceptor. A Lewis base is an electron pair donor. 8. A strong acid completely ionizes in solution. 9. A weak acid partially ionizes in solution. 10. Strong acids include: hydrochloric, hydrobromic, hydroiodic, nitric, chloric, perchloric, bromic, perbromic, iodic, periodic, and sulfuric acids. 11. Weak acids establish an equilibrium between their ionized and nonionized forms. 12. The stronger the acid, the weaker its conjugate base. 13. An acid dissociation constant, K a, is an equilibrium constant written for the equilibrium set up by a weak acid. 14. Water reacts with itself in an autoionization reaction. 15. The ion product constant for water, K w, is the equilibrium constant written for the autoionization reaction of water. Note that K w = When [H 3 O + ] = [OH ], a solution is neutral. 17. Acidic solutions have additional H 3 O + ions. Basic solutions have additional OH ions. Regardless, the product of hydronium and hydroxide ions always equals K w. 18. ph is another way to specify the acidity of a solution. ph < 7 is acidic, ph = 7 is neutral, and ph > 7 is basic. 1

2 19. The log function has its own significant figure rule. 20. The p function is the mathematical operation of taking the negative log. 21. Weak acid equilibrium problems can be solved using techniques applied to ordinary equilibria. In these problems, we ignore the insignificant contribution of H 3 O + from the autoionization of water. 22. The strength of a weak acid can also be characterized by its percent ionization. 23. A mixture of a strong acid with one or more weak acids can be treated as if the strong acid is the only component. 24. A mixture of weak acids may need to be solved as a double equilibrium problem depending on their relative K a s. 25. Everything discussed about acids applies to bases. 26. Strong bases ionize completely. Most hydroxides of Group 1 and Group 2 metals are strong bases. 27. Weak bases produce OH by pulling a proton off water. 28. The strength of a weak base depends on its base ionization constant, K b. It is the equilibrium constant for a weak base. 29. Some ions can act as either weak acids or weak bases. 30. Anions that are conjugate bases of weak acids are themselves weak bases. 31. Anions that are conjugate bases of strong acids are ph neutral. 32. Conjugate acid/base pairs have related K a s and K b s. 33. Cations that are conjugate acids of weak bases are themselves weak acids. 34. Small, highly-charged metal cations form weakly acidic solutions. 35. Salt solutions can be either neutral, acidic, or basic. If the cation is a weak acid, and the anion is a weak base, then the ph depends on the K a and K b. 36. Acids that have more than one acidic proton are called polyprotic acids. 37. Generally, successive K a s are much smaller than the 1st, so polyprotic acids only need to be solved for the first ionization. Exceptions include sulfuric acid and when K a s are within a few hundred of each other. Exceptions must be solved as a double equilibrium problem. 38. The strength of a binary acid depends on bond polarity and bond strength. When H holds the partial negative charge and the bond is relatively weak, the acid will be strong. 2

3 39. Oxyacid (oxoacid) strength depends on the electronegativity of Y and the number of O atoms attached to Y. Oxyacids that have Y s with higher electronegativities and more O atoms are stronger acids. Equations 1. K a = [H 3O + ][A ] [HA] (General acid dissociation expression) 2. K w = [H 3 O + ][OH ] (Ion product constant for water) 3. ph = log [H 3 O + ] (Definition of ph) 4. ph + poh = (Relationship between ph and poh) 5. % ionization = [ionized acid] [initial acid] 100% = [H 3O + ] equil [HA] init 100% (% ionization) 6. K b = [BH+ ][OH ] [B] (General base association equation) 7. K a K b = K w = (Relationship between K a and K b ) 8. pk a = log K a (pk a expression) 9. pk b = log K b (pk b expression) Representative Problems R43. What are the concentrations of all the substances in a 1.0 M solution of hydrogen peroxide, H 2 O 2? What is the ph of the solution? For H 2 O 2, K a = We determine that H 2 O 2 must be a weak acid because they give us the K a. The acid dissociation equation is then: The acid dissociation constant is then: We set up our chart and fill it in as usual. H 2 O 2 (aq) + H 2 O (l) H 3 O + (aq) + HOO (aq) K a = [H 3O + ][HOO ] [H 2 O 2 ] Reaction: H 2 O 2(aq) + H 2 O (l) H 3 O + (aq) + HOO (aq) Initial Change x +x +x Equil. 1.0 x x x 3

4 We put the equilibrium concentrations into the K a expression. K a = [H 3O + ][HOO ] [H 2 O 2 ] = (x)(x) 1.0 x = x2 1.0 x = Checking our approximation, we get %, well within the 5% requirement. The concentrations of everything in solution are: [H + ] = M, [HOO ] = M, [H 2 O 2 ] = 1.0 M. (You d have to go out to M to see a difference, but of course, sig figs limit us from going out that far). To find the ph, we just take the -log of the [H + ]. ph = log [H + ] = log( ) = = 5.87 R55. Quinine, an important drug in treating malaria, is a weak Brønsted base that we may represent as Qu. To make it more soluble in water, it is put into a solution as its conjugate acid, which we may represent as H-QuCl. What is the calculated ph of a 0.15 M solution of H-Qu +? Its pk a is 8.52 at 25 C. Another equilibrium problem, and we use the exact same steps we ve been using. First, identify the equilibrium reaction and write its K a expression. H-Qu + (aq) + H 2O (l) H 3 O + (aq) + Qu (aq) K a = [H 3O + ][Qu] [H-Qu + ] We set up the chart, filling it in with the given information. Reaction: H-Qu + (aq) + H 2O (l) H 3 O + (aq) + Qu (aq) Initial Change x +x +x Equil x x x 4

5 Before we fill in everything into the K a expression, we need to get the value of K a from the given pk a. pk a = log K a 8.52 = log K a 8.52 = log K a K a = K a = [H+ ][Qu] [H-Qu + ] = (x)(x) 0.15 x = x x = Checking our approximation, we get 0.014%. Now, we calculate the solution ph. ph = log [H + ] = log( ) = = 4.67 R102. What is [OH ] in an aqueous solution that is 5.0% NaClO by mass and also % NaOH by mass? What is the ph of the solution? (Assume d of solution = 1.0 g/ml.) In water, NaClO dissociates into Na + and ClO ions. Since Na + cannot donate protons and isn t highly-charged, it is not a weak acid and thus is a neutral cation. ClO can accept a proton to form HClO, hypochlorous acid. This means that ClO is a weak base. We write a base equilibrium for the ClO in water. ClO (aq) + H 2O (l) HClO (aq) + OH (aq) Thus, we need to calculate the concentration of ClO. 5.0 g NaClO 1.0 g solution 1 mole NaClO 1000 ml solution = M g solution 1.0 ml solution g NaClO 1 L solution Since NaClO dissociates completely, [ClO ] = M. Additionally, there is NaOH in the solution. Sodium hydroxide is a strong base, so it dissociates completely into Na + 5

6 (which again is a neutral cation) and OH. Since OH appears in the equilibrium, we need to know its concentration g NaOH 1.0 g solution 1 mole NaOH 1000 ml solution = M g solution 1.0 ml solution g NaOH 1 L solution Again, since NaOH dissociates completely, [OH ] = M. Now we set up our equilibrium chart and fill in. Reaction: OCl (aq) + H 2O (l) HClO (aq) + OH (aq) Initial Change x +x +x Equil x x x To calculate, we need the K b. Conjugate bases are rarely found in a K b table, so we need to use the K a of HClO to find the K b of OCl. Consulting an acid dissociation constant table, we find that K a = for HClO. K a K b = K w ( )K b = Now we set up our K b expression and solve. K b = K b = [HClO][OH ] [OCl ] = (x)(x ) x = x x = Before continuing, we verify that our approximation is valid. Comparing x to , we find that it is %, so we re in trouble. We have to go through and use the quadratic equation = (x)(x ) x ( )( x) = x x ( )x = x x Now we solve for x with the quadratic equation. 0 = x x

7 x = b ± b 2 4ac 2a = ± ( ) 2 4(1)( ) 2(1) ± = 2 = or The negative root doesn t make physical sense, so x = (our sig figs are limited by what s in K b ). From our chart, [OH ] = M which equals M reported to the correct number of sig figs. We can calculate ph by going through poh. poh = log [OH ] = log( ) = Of course, ph + poh = 14.00, so ph of the solution is