KEY. Practice Problems: Applications of Aqueous Equilibria
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1 Practice Problems: Applications of Aqueous Equilibria KEY CHEM 1B 1. Ammonia (NH3) is a weak base with a Kb = 1.8 x 1 5. a) Write the balanced chemical equation for the reaction of ammonia with water. Using the I.C.E. method, calculate the ph and % ionization of a 1.75 M NH3 solution in 2.5 M NHCl. NH3 (aq) + H2O (l) NH + (aq) + OH (aq) I C X +X +X E 1.75 X X X Kb = [NH+ ] [OH ] [NH3] X = 1.26 x 1 5 M assume X is negligible (2.5+X) (X) 2.5X = = 1.8 x 1 5 (1.75 X) 1.75 assume X is negligible Check (% ionization): X [NH 3] init (1%) = 1.26 x 1 5 M 1.75 M [OH ]eq = X = 1.26 x 1 5 M OH poh = log [OH ] poh = log (1.26 x 1 5 M) poh = ph + POH = 1 ph = 9.1 (1%) =.72 % ionization assumption okay (<5%) b) How would the % ionization and ph have been different without the NHCl? What is this effect called? The % ionization would be larger / smaller (circle one), and the ph would be higher / lower (circle one) without the common ion effect. c) Is this solution a buffer? Explain. Yes, this solution is a buffer, because it contains a weak acid - weak base conjugate pair. d) Other than with the ICE method, how else could you have solved this problem? Method: using the Henderson-Hasselbalch equation
2 2. A buffer is made from acetic acid (HC2H3O2) and sodium acetate (NaC2H3O2). 1.5 moles of acetic acid and 2.5 moles of sodium acetate are added to enough water to make 1.5 liter of solution. The ionization constant (Ka) for acetic acid is 1.8 x 1 5. a) Write the equation for the ionization of acetic acid in water, and using a stress-shiftequilibrium arrow diagram, show how the buffer would minimize the effect of adding a base to the solution. Stress: Shift: Equilibrium: HC2H3O2 (aq) b) Using the Henderson-Hasselbalch equation, find the ph of this buffer solution. ph = pka + log H + (aq) + C2H3O2 (aq) [base] [acid] [C2H3O2 ] ph = pka + log [HC2H3O2] (2.5 mol / 1.5 L) ph = log (1.8 x 1 5 ) + log (1.5 mol / 1.5 L) ph = = ph =.97 Note: Assume no ionization (a good assumption due to the common ion effect). Also note: The volume of the solution cancels out. Dilution does not affect the ph of a buffer! c) Calculate the percent ionization of the acetic acid in this solution. ph = log [H + ] [H + ] = 1 ph [H + ] = = 1.8 x 1 5 M H + % ionization = [H + ]eq [HC2H3O2]init (1%) = 1.8 x 1 5 M H+ 1. M HC2H3O2.11 % ionization (1%) =.18 % ionization d) At what ph would this solution have the largest buffering capacity in both directions? phsoln = pka + log [base] Note: Buffering capacity is [acid] maximized in both directions when [acid] = [base]. phsoln = log (1.8 x 1 5 ) + log (1) = ph =.7
3 3. A buffer is needed to maintain ph = a) There are four reagents to choose from to make the buffer: pka1 = 2.22 pka2 = 6.96 pka3 = M H3AsO (aq) (Ka1 = 6. x 1 3, Ka2 = 1.1 x 1 7, Ka3 = 3. x 1 12 ), NaH2AsO H2O (s), Na2HAsO 7H2O (s), and Na3AsO 12H2O (s) What conjugates should be chosen and why? H3AsO Conj. Acid H2AsO Conj. Base This conjugate pair was chosen because pka1 ( log of the ionization constant for losing the first H + from H3AsO) is closest to the ph we want to maintain. When ph = pka, the concentrations of the two conjugates are equal, giving the maximum buffering capacity in both directions. b) It is decided that the concentration of the conjugate acid should be.1 M. Use the Henderson-Hasselbalch equation to calculate the concentration of the conjugate base required to achieve the desired ph = log ph = pka1 + log [H3AsO] = 2.75 = log (6. x 1 3 ) + log 2.75 = log 1. ml buffer 1 ml buffer = [H2AsO ] M = =.3 M H2AsO c) Use stoichiometry to calculate the amount of the solid salt of the conjugate base needed to make 1. ml of buffer. 1 3 L mol g buffer mol H2AsO NaH2AsO H2O NaH2AsO H2O 1 L buffer 1 mol H2AsO 1 mol NaH2AsO H2O = g NaH2AsO H2O 6.1 g NaH2AsO H2O d) Use the dilution equation to calculate the amount of stock aqueous solution needed to make 1. ml of buffer that is.1 M conjugate acid. H 3AsO (1. M) V1 = (1. ml) Stock Buffer M1 V1 = M2 V2 V1 = 1. ml of 1. M H3AsO e) Other than the amounts of solid and aqueous solution calculated in Parts c) and d) above, what makes up the rest of the volume of the buffer? D.I. H2O f) Would this buffer have a better buffering capacity against an acid or a base (circle one)? [H2AsO ] > [H3AsO]
4 . Based on the titration curve to the right and the tables of Ka values below, determine the following: a) The most appropriate indicator for the titration. ph (at eq. pt.) = pka ind = 9.3 Ka ind = 1 pka = = 5 x 1 1 Phenolphthalein b) Whether the unknown acid is a strong acid or a weak acid (circle one). How do you know? ph (at eq. pt.) > 7 c) The most likely identity of the unknown acid. ph (at ½ eq. pt.) = pka unk =.8 Ka unk = 1 pka = 1.8 = 2 x 1 5 acetic acid ph d) The original concentration of the 1. ml sample of unknown monoprotic acid. (1) Ma (1. ml) = (1) (.5 M) (12. ml) Ma =.6 M acetic acid 1 Titration of 1. ml of an 13 Unknown Acid ml of.5 M NaOH added Indicator Ka Metacresol Purple 3.1 x1 2 Thymol Blue 2.2 x 1 2 Methyl Orange 3.5 x 1 Bromophenol Blue 7.9 x 1 5 Bromocresol Green 1.3 x 1 5 Methyl Red 1. x 1 5 Chlorophenol Red 5.6 x 1 7 Bromocresol Purple. x 1 7 Bromothymol Blue 5. x 1 8 Cresol Red 3.5 x 1 9 Phenolphthalein 3 x 1 1 Acid Ka chlorous acid 1.2 x 1 2 hydrofluoric acid 7.2 x 1 nitrous acid. x 1 lactic acid 1.38 x 1 benzoic acid 6. x 1 5 acetic acid 1.8 x 1 5 hypochlorous acid 3.5 x 1 8 ammonium ion 5.6 x 1 1
5 1 Titration of 25. ml of an Unknown Diprotic Acid ph 12 F 1 E 8 D 6 2 A B C ml of.1 M NaOH added 5. Based on the titration curve for a diprotic acid above, answer the following: a) Identify in the table to the right which species exist in solution at each point on the curve (using H2A to represent a generic diprotic acid). b) Given the table of ionization constants, identify the unknown acid. Point A B C D E Species from the unknown H2A H2A, HA HA HA, A 2 A 2 Species from the titrant ph (at 1st ½ eq. pt.) = pka1 unk = 1.8 Ka1 unk = 1 pka1 = = 2 x 1 2 F A 2, OH ph (at 2nd ½ eq. pt.) = pka2 unk = 7. Ka2 unk = 1 pka2 = 1 7. = 1 x 1 7 sulfurous acid c) Calculate the concentration of the original diprotic acid solution. Acid Ka1 Ka2 oxalic acid 6.5 x x 1 5 sulfurous acid 1.5 x x 1 7 ascorbic acid 7.9 x x 1 12 carbonic acid.3 x x 1 11 Note: The 2 nd eq. pt. is a better choice since it is sharper (making estimating more precise), and has a larger volume (making estimating errors proportionally smaller). From the first equivalence point (1) Ma (25. ml) = (1) (.1 M) (7. ml) Ma =.28 M sulfurous acid From the second equivalence point (2) Ma (25. ml) = (1) (.1 M) (1. ml) Ma =.28 M sulfurous acid 1 H + neutralized both H + neutralized
6 6. Based on the titration curve to the right and the tables of Kb values below, determine the following: a) The most appropriate indicator for the titration. ph (at eq. pt.) = pka ind = 5.2 Ka ind = 1 pka = = 6 x 1 6 Methyl Red b) Whether the unknown base is a strong base or a weak base (circle one). How do you know? ph (at eq. pt.) < 7 c) The most likely identity of the unknown base. ph (at ½ eq. pt.) = pka unk. conj. acid = = pka + pkb pkb unk =.2 Kb unk = 1 pkb = 1.2 = 6 x 1 5 trimethylamine ph d) The original concentration of the 15. ml sample of unknown base. (1) (.25 M) (1. ml) = (1) Mb (15. ml) Mb =.23 M trimethylamine 1 Titration of 15. ml of an 13 Unknown Base ml of.25 M HCl added Indicator Ka Metacresol Purple 3.1 x1 2 Thymol Blue 2.2 x 1 2 Methyl Orange 3.5 x 1 Bromophenol Blue 7.9 x 1 5 Bromocresol Green 1.3 x 1 5 Methyl Red 1. x 1 5 Chlorophenol Red 5.6 x 1 7 Bromocresol Purple. x 1 7 Bromothymol Blue 5. x 1 8 Cresol Red 3.5 x 1 9 Phenolphthalein 3 x 1 1 Base Kb ethylamine 6. x 1 dimethylamine 5. x 1 methylamine 3.7 x 1 trimethylamine 6.5 x 1 5 ammonia 1.8 x 1 5 hydroxylamine 9.1 x 1 9 phenylamine.3 x 1 1
7 7. The Ksp of silver chromate (Ag2CrO) is 1.9 x a) Write the balanced complete ionic equation for the dissociation of Ag2CrO. b) Using the I.C.E. method and the equilibrium expression, find the molar solubility of Ag2CrO. (Hint: you are solving for X) Ag2CrO (s) 2 Ag + (aq) + CrO 2 (aq) Initial Conc. Change in Conc. X +2X +X Equilib. Conc. 2X X Ksp = [Ag + ] 2 [CrO 2 ] = (2X) 2 (X) Ksp = X x 1 12 = X 3 X = 1.9 x 1 12 ( 1 /3) = 7.8 x 1 5 M 7.8 x 1 5 M Ag2CrO 8. The molar solubility of chromium (III) hydroxide in water at 25 C is 1.26 x 1 8 M Cr(OH)3. a) Write the balanced complete ionic equation for the dissociation of Cr(OH)3. b) Using the I.C.E. method and the equilibrium expression, find the Ksp of Cr(OH)3. Cr(OH)3 (s) Cr 3+ (aq) + 3 OH (aq) Initial Conc. Change in Conc. X +X +3X Equilib. Conc. X 3X Ksp = [Cr 3+ ] [OH ] 3 = (X) (3X) 3 Ksp = 27X Ksp = 27(1.26 x 1 8 M) Ksp = x 1 31 = 6.81 x The molar solubilities of two ionic compounds can be directly compared, but not always the Ksp values. Circle the two compounds for which the Ksp values could be compared and indicate which one of the two is more soluble in water. # ions: 1 & 1 1 & 2 3 & 2 2 & 1 1 & 3 PbCrO Mg(OH)2 Cu3(PO)2 Ag2SO Al(OH)3 K sp = 2.8 x 1 13 K sp = 5.6 x 1 12 K sp = 1. x 1 37 K sp = 1.2 x 1 5 K sp = 1.9 x 1 33 more soluble
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