CHEM 1412 Answers to Practice Problems Chapters 15, 16, & 17

CHEM 1412 Answers to Practice Problems Chapters 15, 16, & 17 1. Definitions can be found in the end-of-chapter reviews and in the glossary at the end of the textbook! 2. Conjugate Base Conjugate Acid Compound (subtract H ) (add H ) H 2 O OH O I None (no H to remove!) HI N NH 2 NH 4 H 2 SO 4 HSO 4 SO 4 HS S 2 H 2 S CH 4 C None (C does not exist) HNO 3 NO 3 H 2 NO 3 OH H 2 O 3. a) [H] = antilog ( ph) = 10 ph = 10 6.3 = 5.01 X 10 7 M b) [H] = 10 13.4 = 3.98 X 10-14 M c) ph = 14 poh = 14 1.5 = 12.5, [H] = 10 12.5 = 3.16 X 10 13 M 4. a) Since HBr is a strong acid, [H] = [HBr] = 0.0250 M. ph = log(0.0250) = 1.60 b) Since KOH is a strong base, [OH ] = [KOH] = 15.0 M. poh = log(15.0) = 1.18, ph = 14 ( 1.18) = 15.18 c) Since Ca(OH) 2 is a strong base, [OH ] = 2[Ca(OH) 2 ] = (2)(0.0175 M) = 0.0350 M. [H ][OH ] = 1.0 X 10 14, [H ] = 1.0 X 10 14 / [OH ] = 1.0 X 10 14 / 0.0350 = 2.86 X 10 13 M, ph = log(2.86 X 10 13 ) = 12.54

d) moles of HNO 3 = Volume in Liters X Molarity = (0.050 L) (0.150 mole/l) = 0.0075 mole moles of NaOH = (0.200 L) (0.0750 mole/l) = 0.0150 mole HNO 3 (aq) NaOH (aq) NaNO 3 (aq) H 2 O (l) initially 0.0075 mole 0.0150 mole 0 - after reaction 0 0.0075 mole 0.0075 mole - HNO3 is the limiting reactant, NaOH is in excess. The amount of NaOH wich remains after all of the HNO3 reacts is 0.0150 mole 0.0075 mole = 0.0075 mole = moles of OH in the solution. [OH ] = 0.0075 mole / (0.050 L 0.200 L) = 0.030 M. poh = log(0.030) = 1.52, ph = 14 1.52 = 12.48 Note that the solution also contains NaNO 3, but Na and NO 3 ions will not affect the ph (see problem 5). e) moles of Ca(OH) 2 = (2.50 g) / (74.093 g/mole) = 0.03374 mole moles of HCl = (0.075 L) (3.00 mole/l) = 0.225 mole 2 HCl (aq) Ca(OH) 2 (aq) CaCl 2 (aq) 2 H 2 O (l) In this reaction, HCl is in excess. The moles of HCl that react with Ca(OH) 2 = (2)(0.03374 mole) = 0.06748 mole. The moles of HCl that remain = 0.225 mole 0.06748 mole = 0.15752 mole = moles of H. [H] = 0.15752 mole / 0.0750 L = 2.10 M. ph = log(2.10) = 0.322 5. a) K is neutral because it is a group IA metal cation. NO3 is the conjugate base of the strong acid HNO 3 (nitric acid) and is therefore also neutral. Therefore, the solution will be neutral. b) Ba(OH) 2 is a strong base (thought not very water-soluble). Basic solution. c) Na is neutral, C 2 is the conjugate base of the weak acid acetic acid and is therefore a weak base. Basic solution. d) Na is neutral, Cl is the conjugate base of the strong acid HCl and is therefore neutral. Neutral solution. e) Na is neutral, F is the conjugate base of the weak acid HF and is therefore a weak base. Basic solution. f) Cl is neutral, Zn 2 is a transition metal ion and will increase [H ] according to the reaction [Zn(H 2 O) 4 ] 2 (aq) [Zn(H 2 O) 3 (OH)] (aq) H (aq). Acidic solution.

g) N is a weak base: N (aq) H 2 O (l) NH 4 (aq) OH (aq). Basic solution. h) Br is the conjugate base of the strong acid HBr and is therefore neutral. NH 4 ion is the conjugate acid of the weak base N and is therefore a weak acid. Acidic solution. i) K is a group IA metal cation and is therefore neutral. N (nitrite ion) is the conjugate base of the weak acid HN (nitrous acid) and is therefore a weak base. Basic solution. 6. a) HF (aq) H (aq) F (aq) Ka = [H][F ] / [HF] b) F (aq) H 2 O (l) HF (aq) OH (aq) Kb = [HF][OH ] / [F ] c) N (aq) H 2 O (l) NH 4 (aq) OH (aq) Kb = [NH 4 ][OH ] / [N ] d) NH 4 (aq) H (aq) N (aq) Ka = [H][N ] / [NH 4 ] e) C 7 (aq) H 2 O (l) HC 7 (aq) OH aq) Kb = [HC 7 ][OH ] / [C 7 ] f) [Fe(H 2 O) 6 ] 3 (aq) [Fe(H 2 O) 5 (OH)] 2 (aq) H (aq) Ka = [[Fe(H 2 O) 5 (OH)] 2 ][H] / [[Fe(H 2 O) 6 ] 3 ] 7. C 6 OH (aq) H (aq) C 6 O (aq) initially 0.100 M 0 0 at equilib 0.1 X X X K a = [H ][C 6 O ] / [C 6 OH] = 1.3 X 10 10 K a = 1.3 X 10 10 = X 2 / (0.1 X) which we can approximate = X 2 / 0.1 X = 3.61 X 10 6 M = [H]. ph = log(3.61 X 10 6 ) = 5.44.

8. C 4 H 7 (aq) H 2 O (l) HC 4 H 7 (aq) OH (aq) initially 0.250 M - 0 0 at equilib 0.25 X - X X K b = [HC 4 H 7 ][OH ] / [C 4 H 7 ] K a K b = 1.0 X 10 14, Kb = 1.0 X 10 14 / Ka = 1.0 X 10 14 / 1.5 X 10 5 = 6.67 X 10 10 6.67 X 10 10 = X 2 / (0.25 X) which we can approximate = X 2 / 0.25 X = 1.29 X 10 5 M = [OH ]. poh = log(1.29 X 10 5 ) = 4.89, ph = 14 4.89 = 9.11. 9. 3% hydrogen peroxide will have 3 g of H 2 per 100 g of solution (the solvent is water). The moles of H 2 will be 3 g / (34.015 g/mole) = 0.0882 mole, and 100 g of solution will be 100 ml or 0.100 L. Therefore, the molarity of the solution is 0.0882 mole /0.100 L = 0.882 M. H 2 (aq) H (aq) H (aq) initially 0.882 M 0 0 at equilib 0.882 X X X K a = 2.4 X 10 12 = X 2 / (0.882 X) which we can approximate = X 2 / 0.882 X = 1.45 X 10 6 M = [H ], ph = log(1.45 X 10 6 ) = 5.84. 10. Part I. moles of acetic acid = (0.100L)(0.250 mole/l) = 0.0250 mole moles of NaOH = (0.400 L)(0.05 mole/l) = 0.020 mole. HC 2 (aq) NaOH (aq) NaC 2 (aq) H 2 O (l) initially 0.0250 mole 0.020 mole 0 - after reaction 0.0050 mole 0 0.020 mole - Part II. We have a solution of 0.0050 mole of acetic acid and 0.020 mole of sodium acetate in a volume of 100 ml 400 ml = 500 ml = 0.500 L. [HC 2 ] = 0.0050 mole / 0.500 L = 0.010 M. [C 2 ] = 0.020 mole / 0.500 L = 0.040 M. Since this is a solution of a weak acid and its conjugate base, or a buffer solution, we can use the Henderson-Hasselbalch equation to calculate the ph. ph = pka log{[c 2 ] / [HC 2 ]} = log(1.8 X 10 5 ) log(0.040 M / 0.010 M) = 4.74 0.60 = 5.34.

We can also use our "old" approach to calculate the ph, using the equation for the weak acid dissociation: HC 2 (aq) H (aq) C 2 (aq) Ka = [H ][C 2 ] / [HC 2 ] initially 0.010 M 0 0.040 M at equilib 0.01 X X 0.04 X 1.8 X 10 5 = (X)(0.04 X) / (0.01 X) which we can approximate as (X)(0.04) / (0.01). 1.8 X 10 5 = 4X, X = 4.5 X 10 6 = [H ]. ph = log (4.5 X 10 6 ) = 5.35. 11. We can use the Henderson-Hasselbalch equation to calculate the ph since this is a buffer solution. ph = pka log([n ] / [NH 4 ] = log(5.6 X 10 10 ) log(0.15 M / 0.25 M) = 9.25 ( 0.22) = 9.03. Or once again, we could use the equation for the weak acid dissociation to calculate the the ph: NH 4 (aq) H (aq) NH3 (aq) K a = [H ][N ] / [NH 4 ] initially 0.25 M 0 0.15 M at equilib 0.25 X X 0.15 X 5.6 X 10 10 = (X)(0.15 X) / (0.25 X) which we can approximate as (X)(0.15) / (0.25). 5.6 X 10 10 = 0.6X, X = 9.33 X 10 10 M = [H ]. ph = log (9.33 X 10 10 ) = 9.03. 12. For a 0.01 M solution of acetic acid, we have 1.8 X 10 5 is approximately = X 2 / 0.01 M, or X = [H] = 4.24 X 10 4 M. The percent dissociation will be 4.24 X 10-4 M / 0.01 M X 100 = 4.24 % (approximately; the quadratic formula would give a more exact value). For a 1.0 M solution, we can similarly calculate X, which in this case is 4.24 X 10 3 M, but the percent dissociation will only be 4.24 X 10 3 M / 1.0 M X 100 = 0.424 %. In other words, the more concentrated a solution of a weak acid is, the greater the H ion concentration in the solution will be, but the percent dissociation will actually be less. The percent dissociation depends on the concentration of the weak acid. 13. If we "plug in" 1.0 X 10 9 into the ph formula, we get ph = log(1.0 X 10 9 ) = 9.0 which indicates a basic solution, but how can even a dilute solution of a strong acid like HCl give a basic solution? The answer is, it will not. The H ions contributed by the very small amount of HCl is less than the H ion concentration in pure water, which is 1.0 X 10 7 M. Therefore, here, the H concentration from the water solvent itself is significant. The total H concentration would be 1.0 X 10 7 M 1.0 X 10 9 M = 1.01 X 10 7 M, and the ph will be log(1.01 X 10 7 ) = 6.996, which is barely acidic.

14. Fe(OH) 3 (s) Fe 3 (aq) 3 OH (aq) Ksp = [Fe 3 (aq)][oh (aq)] 3 initially - 0 0 at equilib - X 3X Ksp = 4.0 X 10 38 = (X)(3X) 3 = 27X 4, X = the molar solubility of Fe(OH) 3 = 1.96 X 10 10 M. 15. a) Ca(OH) 2 (s) Ca 2 (aq) 2 OH (aq) Ksp = [Ca 2 (aq)][oh (aq)] 2 initially - 0 0 at equilib - X 2X Ksp = 5.5 X 10 6 = (X)(2X) 2 = 4X 3, X = the molar solubility of Ca(OH) 2 = 0.011 M. [OH ] = 2X = 2(0.011 M) = 0.022 M. poh = log(0.022) = 1.66, ph = 14 1.66 = 12.34. b) At ph = 13, [H] = 1.0 X 10 13 M and [OH ] will be 1.0 X 10 14 / 1.0 X 10 13 = 0.10 M. All we have to do is to plug in 0.10 M for [OH ] into the Ksp expression and solve for [Ca 2 ], which will be the molar solubility of Ca(OH) 2 since there is one Ca 2 ion per Ca(OH) 2. K sp = 5.5 X 10 6 = [Ca 2 ](0.10 M) 2, [Ca 2 ] = 5.5 X 10 6 / (0.10) 2 = 5.5 X 10 4 M. Note that this is much less than the molar solubility in pure water (0.011 M) calculated above. This is an example of Le Chatelier's principle and the common ion effect. 16. The relevant solubility process is BaF 2 (s) Ba 2 (aq) 2 F (aq) Ksp = [Ba 2 (aq)][f (aq)] 2 First calculate the concentrations of Ba 2 and F- ions. Na and Cl will be spectators since they are not forming any precipitate. moles of Ba 2 = (0.100 L) (0.020 mole/l) = 0.0020 mole. moles of F = (0.100 L) (0.010 mole/l) = 0.0010 mole. The total volume is 200 ml. [Ba 2 ] = 0.0020 mole / 0.200 L = 0.010 M. [F ] = 0.0010 mole / 0.200 L = 0.0050 M.

Now we just need to calculate Q and compare it to Ksp. If Q > Ksp, the reverse reaction will occur spontaneously, precipitating BaF2 (s). If Q < Ksp, no precipitate will form. Q = [Ba 2 ] i [F ] i 2 = (0.010 M)(0.0050 M) 2 = 2.5 X 10 7 < 1.0 X 10 6. Since Q < Ksp, no precipitate of BaF 2 (s) will form when the solutions are mixed. 17. The neutralization reaction for this titration is PO 4 (aq) 3 NaOH (aq) Na 3 PO 4 (aq) 3 H 2 O (l) We can use a general acid-base titration formula to calculate the molarity of the phosphoric acid solution: (ml of acid) (M of acid) (no. of H in the acid) = (ml of base) (M of base) ( no. of OH in the base) (50.0 ml) (M of acid) (3) = (27.5 ml) (0.100 M) (1), M of acid = 0.0183 M. 18. See the definitions at the end of chapter 18! 19. a) one mole of liquid water has a higher entropy, or disorder, than one mole of solid ice. b) one mole of N 2 gas at 0.1 atm will occupy a volume 10 times greater than the volume of one mole of N 2 gas at 1 atm pressure, if the temperature is not changed. A mole of gas in a large volume is more random, or disordered, than the same amount of gas confined in a smaller volume. c) Iron metal at 100 o C will have a higher entropy than at 25 o C since entropy increases with temperature. d) Similarly, solid ice at 0 o C will have a higher entropy than ice at 50 o C. 20. a) S is negative since there are fewer product gas molecules. b) S is negative. c) S is negative. d) S is positive. 21. S o = [(2 moles)(0.256 kj/mole K)] [(2 moles)(0.249 kj/mole K) (1 mole)(0.205 kj/mole)] = 0.512 kj/k 0.703 kj/k = 0.191 kj/k. A negative entropy change is expected since there are fewer gaseous product molecules.

22. H o = [(2 moles)( 435.9 kj/mole) (3 moles)(0)] [(2 moles)( 391.2 kj/mole)] = 871.8 kj ( 782.4 kj) = 89.4 kj. S o = [(2 moles)(0.0827 kj/mole K) (3 moles)(0.205 kj/mole K)] [(2 moles)(0.143 kj/mole K)] = 0.7804 kj/k - 0.286 kj/k = 0.4944 kj/k. A positive S is expected since a solid reactant is forming a gaseous product (in addition to another solid product). Now we can plug into G o = H o T S o (T = 25 o C = 298 K): G o = 89.4 kj (298 K)(0.4944 kj/k) = 236.7 kj. The negative sign of G o indicates that the reaction should occur spontaneously. 23. G o = RT lnkeq = (8.314 X 10 3 kj/mole K)(298 K) ln(1.8 X 10 5 ) = 27.1 kj/mole. The positive sign of G o indicates that, under the following initial conditions: HC 2 (aq) H (aq) C 2 H 2 (aq) initially 1 M 1 M 1 M the reaction will not go in the forward ( ) direction spontaneously to reach equilibrium, but rather, the reverse reaction ( ) will occur until equilibrium is reached. 24. First calculate Q: Qp = [N ] i 2 / [N2 ] i [H 2 ] i 3 = (1.0 atm) 2 / {(2.0 atm)(2.0 atm) 3 } = 0.0625 Now use the equation G = G o RT lnq to calculate G: G = 33.32 kj/mole (8.314 X 10 3 kj/mole K)(298 K) ln(0.0625) = 33.32 kj/mole ( 6.87 kj/mole) = 40.19 kj/mole.

25. G o for the reaction = [(2 moles)(0) (1 mole)(0)] [(2 moles)( 236.81 kj/mole)] = 0 ( 473.62 kj) = 473.62 kj (very nonspontaneous!) Note: Even though they were not given, we know that H f o for H 2 (g) and (g) are both zero since these are elements in their natural states (gases) at 25 o C and 1 atm pressure. Now we will use G o = RT lnkeq to calculate Kp for the reaction: 473.62 kj/mole = (8.314 X 10 3 kj/mole K)(298 K) ln Kp lnkp = 191.16, Kp = e 191.16 = 9.53 X 10 84 (!) = 1 / [H 2 (g)] 2 [ (g)] For an equilibrium involving gases, the value of Keq that we calculate from G o will be Kp (for reactions in solution, the Keq calculated from G o will be Kc). Note that, in this reaction, such a small equilibrium constant indicates that there will be a negligible amounts of products ( H 2 (g) and (g) ) present at "equilibrium". 26. Turn the second reaction around (don't forget to change the sign of G o ) and add it to the first reaction to obtain the desired reaction: Cu2S (s) O2 (g) 2 Cu (s) SO2 (g) G o = 213.9 kj SO2 (g) S (s) O2 (g) G o = 300.1 kj Cu2S (s) 2 Cu (s) S (s) G o = 86.2 kj Note that this is just like Hess's law, which we used to calculate H for reactions back in Chapter 6. Here we are using the same approach to calculate G (it works with S also).