# KEY. Practice Problems: Applications of Aqueous Equilibria

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1 Practice Problems: Applications of Aqueous Equilibria KEY CHEM 1B 1. Ammonia (NH 3 ) is a weak base with a K b = 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 NH 3 solution in 2.5 M NH Cl. NH 3 (aq) + H 2 O (l) NH + (aq) + OH (aq) I C X +X +X E 1.75 X X X K b = [NH + ] [OH ] [NH 3 ] 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 ] (1%) 1.26 x 1 = init 1.75 M 5 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 NH Cl? What is this effect called? The % ionization would be larger / smaller (circle one), and the ph would be higher / lower (circle one) common ion effect without the. 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? using the Henderson-Hasselbalch equation Method:

2 2. A buffer is made from acetic acid (HC 2 H 3 O 2 ) and sodium acetate (NaC 2 H 3 O 2 ). 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 (K a ) 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: HC 2 H 3 O 2 (aq) b) Using the Henderson-Hasselbalch equation, find the ph of this buffer solution. ph = pk a + log H + (aq) + C 2 H 3 O 2 (aq) [base] [acid] [C 2 H 3 O 2 ] ph = pk a + log [HC 2 H 3 O 2 ] (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 [HC 2 H 3 O 2 ] init 1.8 x 1 5 M H + (1%) = 1. M HC 2 H 3 O 2 (1%) =.18 % ionization.11 % ionization d) At what ph would this solution have the largest buffering capacity in both directions? [base] ph soln = pk a + log [acid] ph soln = log (1.8 x 1 5 ) + ph =.7 Note: Buffering capacity is maximized in both directions when [acid] = [base]. log (1) =

3 3. Based on the titration curve to the right and the tables of K a values below, determine the following: a) The most appropriate indicator for the titration. ph (at eq. pt.) = pk a ind = 9.3 K a 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.) = pk a unk =.8 K a 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) M a (1. ml) = (1) (.5 M) (12. ml) M a =.6 M acetic acid 1 Titration of 1. ml of an 13 Unknown Acid ml of.5 M NaOH added Indicator K a 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 K a 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

4 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. 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 H 2 A to represent a generic diprotic acid). b) Given the table of ionization constants, identify the unknown acid. ph (at 1st ½ eq. pt.) = pk a1 unk = 1.8 K a1 unk = 1 pka1 = = 2 x 1 2 Point A B C D E F Species from the unknown H 2 A H 2 A, HA HA HA, A 2 A 2 A 2 Species from the titrant, OH ph (at 2nd ½ eq. pt.) = pk a2 unk = 7. K a2 unk = 1 pka2 = 1 7. = 1 x 1 7 sulfurous acid c) Calculate the concentration of the original diprotic acid solution. Acid K a1 K a2 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) M a (25. ml) = (1) (.1 M) (7. ml) M a =.28 M sulfurous acid From the second equivalence point (2) M a (25. ml) = (1) (.1 M) (1. ml) M a =.28 M sulfurous acid 1 H + neutralized both H + neutralized

5 5. Based on the titration curve to the right and the tables of K b values below, determine the following: a) The most appropriate indicator for the titration. ph (at eq. pt.) = pk a ind = 5.2 K a 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.) = pk a unk. conj. acid = = pk a + pk b pk b unk =.2 K b 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) M b (15. ml) M b =.23 M trimethylamine 1 Titration of 15. ml of an 13 Unknown Base ml of.25 M HCl added Indicator K a 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 K b 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

6 6. The K sp of silver chromate (Ag 2 CrO ) is 1.9 x a) Write the balanced complete ionic equation for the dissociation of Ag 2 CrO. b) Using the I.C.E. method and the equilibrium expression, find the molar solubility of Ag 2 CrO. (Hint: you are solving for X) Ag 2 CrO (s) 2 Ag + (aq) + CrO 2 (aq) Initial Conc. Change in Conc. X +2X +X Equilib. Conc. 2X X K sp = [Ag + ] 2 [CrO 2 ] = (2X) 2 (X) K sp = 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 Ag 2 CrO 7. 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 K sp 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 K sp = [Cr 3+ ] [OH ] 3 = (X) (3X) 3 K sp = 27X K sp = 27(1.26 x 1 8 M) K sp = x 1 31 = 6.81 x The molar solubilities of two ionic compounds can be directly compared, but not always the K sp values. Circle the two compounds for which the K sp 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 Cu 3 (PO ) 2 Ag 2 SO 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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