Chemistry 202 Exam 4 KEY March 21, 2005 (Print Clearly) 1. (16 pts) Fill in the blanks with the best answer. 2 points each. (a) Consider the reaction 3A(g) + B(g) 3 C(s) + 3 D(g).!H reaction = -150.0 kj/mol. How will each of the following changes influence the position of equilibrium. Will it shift the equilibrium to the left (toward reactants), to the right (toward products), or have no effect. Answer left, right, or no effect. B is added to the system: right ; C is removed from the system: no effect ; Catalyst X (which lowers E a by a factor of 2) is added: no effect ; The temperature is reduced: right ; An inert gas is added to the system in such a way that total pressure increases, but the volume remains constant: no effect ; The volume of the container is doubled (e.g. by raising a piston in a cylinder: _left. (b) Adding P 4 O 10 (s) into an aqueous solution will cause the ph to decrease (increase/decrease/stay the same). (c) Adding barium oxide to an aqueous solution will cause the ph to increase (increase/decrease/stay the same). o 2. (9 pts) Fill in the blanks, and indicate if the aqueous solution is acid, base, neutral NAME FORMULA Acidic/Basic/Neutral Lithium nitrate LiNO3 Neutral Potassium phosphate K3PO4 Basic Ammonium perchlorate NH 4 ClO 4 Acidic Sodium acetate NaC 2 H 3 O 2 Basic Pyridinium bromide Acidic H N Br Chemistry 202, Exam 4, Page 2 /25
Acid Formula K a1 K a2 K a3 Sulfurous H 2 SO 3 1.54 x 10-2 1.02 x 10-7 Chlorous HClO 2 1.1 x 10-2 Phosphoric H 3 PO 4 7.52 x 10-3 6.23 x 10-8 2.2 x 10-13 Nitrous HNO 2 4.6 x 10-4 Formic HCOOH 1.77 x 10-4 Acetic CH 3 COOH 1.76 x 10-5 3. (4 points) Using the data provided above, rank the following anions in order of increasing basicity: ClO 2 -, PO 4 3-, HPO 4 2-, CH 3 COO -. least basic _ ClO 2 - < CH 3 COO -. < _ HPO 4 2-, _ < _ PO 4 3- _ most basic 4. (4 points) Explain the relative acidities of chlorous and nitrous acids as revealed in the table above. Ka HOClO > Ka HONO because Cl is more electronegative than N, polarizing the H-O more. 5. (1 point each) Designate the Bronsted-Lowrey acid (A) and the Bronsted-Lowrey base (B) on the left side of each of the following equations, and also designate the conjugate acid (CA) and conjugate base (CB) on the right side. + H 2 O NH 3 + OH - B A CA CB NH 2 - [Cr(H 2 O) 6 ] 3+ + H 2 O [Cr(H 2 O) 5 (OH)] 2+ + H 3 O + A B CB CA O 2- + H 2 O OH - + OH - B A CA CB 6. (5 points) The strengths of weak acids are often reported in terms pk a rather than K a, where p has the same mathematical meaning that it has in ph and poh. Calculate pk a values for chlorous acid and acetic acid. What general relationship would you expect between pk a and acid strength? pka HClO2 = -log(.011) = 1.96; pka HC2H3O2 = -log(1.76x10-5) = 4.754 The larger the pka value, the weaker the acid. /25 Chemistry 202, Exam 4, Page 3
7. (20 pts) The equilibrium constant, Kp, for the dissociation of phosphorous pentachloride is 4.2 x 10-2 at 250 C. How many moles of PCl 5 must initially be added to a 3.0-liter flask to obtain an equilibrium Cl 2 concentration of 0.15 M? PCl 5 (g) PCl 3 (g) + Cl 2 (g) Kc = Kp ( RT) =.042!n ( " lit atm% + *.0820578 # $ mol K & '[ 523.15K] - ), 1 = 9.8 x 10-4 PCl 5 (g) PCl 3 (g) + Cl 2 (g) [Initial} M 0 0 [Equil] M-x x x x = 0.15M Kc = ( M! x) = (.15) 2 ( M -.15) = 9.8x10-4 Assume that M - 0.15 = M; Then, M =.0225 = 22.9 ; check : 22.9-0.15 = 22.75 = 23M -4 9.8 x 10 Thus the assumption is valid;! moles = 22.9 moles $ # & x ( 3.0 liters) = 68.7 = 69 moles " liter % /20
Chemistry 202, Exam 4, Page 4 8. (a) (5 points) The Ka of hydrofluoric acid (HF) is 7.2 x 10-4. Write the reaction for which that is the equilibrium constant. HF(aq) = H + (aq) + F - (aq) or HF(aq) + H 2 O(l) = H 3 O + (aq) + F - (aq) (b) (10 points) Calculate the ph of a solution that is made by diluting 0.50 moles of HF to a final volume of 1.0 liters of aqueous solution. You must prove any assumptions to receive full credit. HF(aq) = H + (aq) + F - (aq) [Initial] 0.50 0 0 [Equil] 0.50-x x x Ka = = 7.2 x 10-4 ; assume 0.50 - x = 0.50 (.50 - x) Then x = (.50) 7.2x10!4 assumption is not valid! Must do quadratic: x 2 +.00072(x) -.00036 = 0 ( ) =.0189;.50 -.019 =.481 x = -.00072 ± (.00072 )2-4(1)(-.00036) 2 x = -.0193 or 0.0186 ; only positive root makes sense ph = - log(.0186) = 1.73 (c) (5 points) Calculate the value of Kb for F -. Kb = Kw Ka = 1.0x10-14.00072 = 1.4 x 10-11
(d) (10 points) Calculate the ph of 2.0 M NaF(aq). F - (aq) + H 2 O(l) = HF (aq) + OH - (aq) [Initial] 2.00 0 0 [Equil] 2.00-x x x Kb = 2.00 - x = 1.4 x10-11 ; assume 2.00 - x = 2.00 Then x = (2.00)(1.4x10-11 ) = 5.3 x 10-6 Check 2.00 -.0000053 = 1.9999947 = 2.00 (OK) poh = - log(5.3 x 10-6 ) = 5.28; ph = 14.00 - poh = 14.00-5.28 = 8.72