Chemistry 201: General Chemistry II - Lecture Dr. Namphol Sinkaset Chapter 18 Study Guide Concepts 1. A buffer is a solution that resists changes in ph by neutralizing added acid or base. 2. Buffers are created using conjugate acid/base pairs. 3. The ph of a buffer can be calculated by approaching the system as an equilibrium problem. Since a conjugate acid/base pair are both in solution, the problem can be solved from a K a or K b point of view. 4. The Henderson-Hasselbalch equation can be used to calculate the ph of a buffer or the amounts/concentrations necessary to create a buffer. It works provided the x is small approximation is valid. 5. Adding strong acid or strong base will result in small changes in the ph of a buffer. New concentrations of the conjugate acid/base pair must calculated based on the stoichiometric reaction and the new total volume. 6. Note that in the Henderson-Hasselbalch equation, the ratio of base to acid is what s needed. As such, moles can be used in place of concentrations since the volume of the entire buffer is the same for both base and acid. 7. Buffers can be made with a weak base and its conjugate acid as well. In these problems, the pk a of the conjugate acid must be used in the Henderson-Hasselbalch equation. 8. Effective buffers have relative concentrations of conjugate acid/base that do not differ by more than a factor of 10 and have high actual concentrations of conjugate acid and base. The effective range for a buffer system is ±1 ph unit on either side of pk a. 9. Buffer capacity is the amount of acid or base that can be added to a buffer without destroying its effectiveness. 10. A buffer is destroyed when either of the conjugate acid/base pairs has been used up. Thus, higher concentrations of buffer components lead to higher buffer capacity. 11. In an acid-base titration, a basic (or acidic) solution of unknown concentration is reacted with an acidic (or basic) solution of known concentration. 12. An indicator is a substance used to visualize the endpoint of a titration. 13. At the equivalence point, the number of moles of acid equals the number of moles of base. 1
14. Theoretical titration curves (also known as ph curves) can be plotted by calculating the ph along several points of the titration. 15. The two keys to titration curve calculations are writing the titration equation and finding the equivalence point of the titration. 16. Regardless of the type of titration (strong/strong or weak/strong), the titration curve can be broken up into four regions: (1) before the titration begins; (2) pre-equivalence; (3) at equivalence; and (4) post-equivalence. The type of calculation performed depends on the region! 17. For weak/strong titrations, the ph does not equal 7.00 at the equivalence point. Additionally, at half equivalence, ph = pk a. 18. Polyprotic acids will display multiple equivalence points in their titration curves if the K a s are different enough. The volumes between equivalence points will always be equal. 19. An indicator is itself a weak acid that has a different color than its conjugate base. 20. If ph > pk a of HIn, then the color of the solution will be that of In. If ph = pk a of HIn, then the color of the solution will be in between that of HIn and In. If ph < pk a of HIn, then the color of the solution will be that of HIn. 21. Solubility is not as clear cut as it seems; there are different degrees of solubility. 22. The solubility product constant, K sp, is an equilibrium constant that describes the relationship between the concentrations of the constituent ions and their corresponding precipitate. Just like any other equilibrium constant, it tells you how far the reaction goes towards products. 23. Generally, solubility is defined as the amount of grams of a solid that will dissolve in 100 g of water. Molar solubility is an alternative measure of solubility; obviously, it is the number of moles of a solid that can dissolve per L of solution. 24. K sp s are used to calculate molar solubilities, and these problems are treated like any other equilibrium problem. 25. common ion effect: the lowering of the solubility of an ionic compound by the addition of a common ion (i.e., an ion involved in the K sp expression). 26. ph will affect the solubility of an ionic compound if either H 3 O + or OH can react with the cation or anion. 27. Q calculations can be performed to see whether or not a precipitate will form. 28. If Q > K sp, a precipitate will form. 29. If Q = K sp, no precipitate will form, but the solution is saturated. 30. If Q < K sp, no precipitate will form. 2
31. A complex ion is comprised of a transition metal Lewis acid that is bound to one or more Lewis bases. In a complex ion, the Lewis bases attached to the central metal are called ligands. 32. Formation of complex ions can also be written as an equilibrium, and the equilibrium constant in this case is called a formation constant, K f. 33. Formation constants are usually very large. 34. Calculations with K f s are a little different as we assume that the equilibrium lies all the way to the right. 35. Formation of complex ions will enhance the solubility of otherwise insoluble ionic compounds. Thus, the addition of Lewis bases that can bind to metals will increase solubility. 36. Some metal hydroxides are amphoteric. Thus, their solubility is very ph dependent. 37. Only aluminum, chromium(iii), zinc(ii), lead(ii), and tin(ii) hydroxides are amphoteric. Equations 1. ph = pk a + log [base] i [acid] i (Henderson-Hasselbalch equation) 2. pk a + pk b = 14.00 (Relationship between pk a and pk b ) 3. K sp = [A + ][B ] (solubility product constant for AB (s) A + (aq) + B (aq) ) 4. K f = [MLx+ y ] [M x+ ][L] y (formation constant for M x+ (aq) + yl (aq) ML x+ y(aq) ) Representative Problems R7. How many grams of sodium formate, NaCHO 2, would have to be dissolved in 1.0 L of 0.12 M formic acid (pk a 3.74) to make the solution a buffer for ph 3.80? Since we know it s a buffer problem, we know we will need to use the Henderson- Hasselbalch equation. ph = pk a + log [A ] [HA] ph = pk a + log [CHO 2 ] [CHO 2 H] 3.80 = 3.74 + log [CHO 2 ] [0.12] 3
0.06 = log [CHO 2 ] [0.12] 1.14 8 = [CHO 2 ] [0.12] [CHO 2 ] = 0.13 776 We need to do a stoichiometry calculation to get the grams of sodium formate needed. 1.0 L 0.13 776 mole CHO 2 1 L 1 mole NaCHO 2 1 mole CHO 2 68.008 g NaCHO 2 1 mole NaCHO 2 = 9.3 7 g NaCHO 2 R22. Consider the titration of 100.0 ml of 0.100 M NaOH with 1.00 M HBr. Find the ph at the following volumes of acid added: V a = 5.0, V eq, and 12.0 ml. As in all titrations, there are two things you should always do first. One is write the titration reaction, and the other is find the equivalence point. The titration is between a strong base and a strong acid. Therefore, the reaction is: OH (aq) + H 3O + (aq) 2H 2O (l) We find the equivalence point by equating the moles of base with the moles of acid. moles base = moles acid (0.1000 L)(0.100 moles OH /L) = (1.00 moles H + /L)x x = 0.0100 0 L We see that the equivalence point is reached at 10.0 ml. Since this is a strong/strong titration, the ph = 7.00 at the equivalence point. That means that 5.0 is pre-equivalence, and 12.0 ml is post-equivalence. In the pre-equivalence region, the added H 3 O + is neutralized by the OH in solution. However, not all of the OH has reacted, so the ph is determined by how much OH is left. In the post-equivalence region, all the OH has reacted, and now we re just adding excess H 3 O + ; the ph is determined by how much extra H 3 O + we ve added. We go through the calculation below, but first we calculate the total moles of base before the titration begins. 100.0 ml 0.100 mole NaOH 1 mole OH 1 mole NaOH = 0.0100 0 mole OH At 5.0 ml, we need to calculate how many moles of H 3 O + have been added and then subtract that from the total moles of hydroxide. 5.0 ml 1.00 mole HBr 1 mole H 3O + 1 mole HBr = 0.0050 0 mole H 3 O + Because of the 1:1 stoichiometry of the titration reaction, the moles of OH remaining is 0.0100 0 0.0050 0 mole = 0.0050 0 mole. To get [OH ], we divide by the new total volume which is 105.0 ml, or 0.1050 L. Thus, [OH ] = 0.047 6 M, poh = 1.32 2, and ph = 12.68. 4
At 12.0 ml, we re in the post-equivalence region, so we need to calculate the excess H 3 O + that s been added. Since the equivalence point is at 10.0 ml, we are 2.0 ml past. We calculate the moles of H 3 O + in 2.0 ml and divide by the total volume. 2.0 ml 1.00 mole HBr 1 mole H 3O + 1 mole HBr = 0.0020 0 mole H 3 O + The new total volume is now 112.0 ml, or 0.1120 L, so [H 3 O + ] = 0.017 9 M, and the ph = 1.75. R28. Calculate the ph of a solution made by mixing 50.00 ml of 0.100 M NaCN with (a) 4.20 ml of 0.438 M HClO 4, (b) 11.82 of 0.438 M HClO 4, (c) What is the ph at the equivalence point with 0.438 M HClO 4? This is essentially a titration problem, but it s just been worded differently. First, we write the equation that describes the reaction between the HClO 4 and NaCN. When these are placed in solution, they dissociate completely. The only active species that we re concerned with are the H + and CN. CN (aq) + H+ (aq) HCN (aq) The next step is to find the equivalence point. moles base = moles acid (0.05000 L)(0.100 moles CN /L) = (0.438 moles H + /L)x x = 0.0114 2 L Now we can identify (a) as being in the pre-equivalence region. In this region, we have a buffer problem because although some CN has reacted to form HCN, there s still some left. These two species in solution form the components for a buffer. We set up a chart to organize the numbers. Reaction: CN (aq) + H + (aq) HCN (aq) Initial rel. [ ]: 1 0 0 Change: 4.20/11.4 2 4.20/11.4 2 4.20/11.4 2 Final rel. [ ]: 1 4.20/11.4 2 0 4.20/11.4 2 We use these relative concentrations in the Henderson-Hasselbalch equation. We get pk a from the back of the book; it equals 9.20 8 (K a = 6.2 10 10 ). ph = pk a + log [CN ] [HCN] = 9.20 8 + log 1 4.20/11.4 2 4.20/11.4 2 = 9.44 3 5
We identify (b) as being in the post-equivalence region. The ph is determined by the excess strong acid that has been added; we ignore the contribution of HCN to the acidity. 0.4 ml 0.438 mole HClO 4 1 mole H 3O + 1 1 mole HClO 4 0.06182 L = 0.002 834 M H 3 O + Thus, the ph = 2.5. For (c), at the equivalence point, we ve transformed all the CN to HCN. Now it s a weak acid problem. We just need to find the [HCN] and solve it as a weak acid problem. 50.00 ml 0.100 mole NaCN 1 mole CN 1 mole NaCN 1 mole HCN 1 mole CN 1 0.06142 L = 0.0814 1 M HCN This concentration goes in as the initial concentration in our acid equilibrium chart for HCN. Reaction: HCN (aq) + H 2 O (l) H 3 O + (aq) + CN (aq) Initial 0.0814 1 0 0 Change x +x +x Equil. 0.0814 1 x x x K a = [H 3O + ][CN ] [HCN] 6.2 10 10 = x 2 0.0814 3 x x = 7.1 1 10 6 Using the approximation, [H + ] = 7.1 1 10 6 (checking the approximation yields 0.009%) and the ph = 5.15. R32. Calculate the molar solubility of AgCl in 0.050 M AlCl 3. This is a common ion effect problem because of the common ion Cl found in both the AgCl and the AlCl 3. We need to realize that the 0.050 M AlCl 3 dissolves as 0.050 M Al 3+ and 0.15 0 M Cl. We identify the solubility reaction and its K sp expression. AgCl (s) Ag + (aq) + Cl (aq) K sp = [Ag + ][Cl ] Now we simply fill in the equilibrium table. Reaction: AgCl (s) Ag + (aq) + Cl (aq) Initial 0 0.15 0 Change +S +S Equil. S 0.15 0 + S 6
We use the equilibrium row to solve for S in the K sp expression. We look up the K sp value for AgCl (s), and it is 1.8 10 10. K sp = [Ag + ][Cl ] 1.8 10 10 = (S)(0.15 0 + S) 1.8 10 10 = (S)(S) S = 1.3 4 10 5 Given the small value of x, it looks like we re well within the 5% requirement (it comes out to 0.009%). Since S and AgCl ( s) are in a 1:1 ratio, the molar solubility of AgCl is equal to x. Therefore, the molar solubility of AgCl (s) is 1.3 10 5 M in 0.050 M AlCl 3. R36. Would a precipitate of silver acetate form if 22.0 ml of 0.100 M AgNO 3 were added to 45.0 ml of 0.0260 M NaC 2 H 3 O 2? For AgC 2 H 3 O 2, K sp = 2.3 10 3. This is a Q expression problem; we need to calculate Q and compare its value to K sp. The solubility equation we re interested in is: AgC 2 H 3 O 2(s) Ag + (aq) + C 2H 3 O 2(aq) The Q expression is then: Q = [Ag + ][C 2 H 3 O 2 ] Now we calculate the concentrations of each ion to put into the Q expression. 0.0220 L 0.100 mole Ag+ 1 L 1 0.0670 L = 0.0328 4 M Ag + 0.0450 L 0.0260 mole C 2H 3 O 2 1 L 1 0.0670 L = 0.0174 6 M C 2 H 3 O 2 Now we put these concentrations into the Q expression. Q = [Ag + ][C 2 H 3 O 2 ] = (0.0328 4 )(0.0174 6 ) = 5.73 4 10 4 We see that 5.73 4 10 4 < 2.3 10 3, which means that Q< K sp. Therefore, a precipitate will not form. 7