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2 NONE OF THE MULTIPLE CHOICE PROBLEMS REQUIRES EXTENSIVE OR TIME COMSUMING CALCULATIONS. IF YOUR METHOD REQUIRES EXTENSTIVE CALCULATIONS IT IS EITHER WRONG OR HARDER THAN WHAT IS REQUIRED. 1. The ph of a nitric acid, HNO 3, solution is What is the concentration of the nitrate ion? (a) 7.24 x M (b) 1.38 x 10-3 M (c) 2.69 x M (d) 3.72 x 10-2 M (e) none of the above ph 1.43 ph = log H +. So, H + = 10 = 10 = M Ba OH Careful. 2. Calculate the ph of a M aqueous barium hydroxide solution, ( ) 2. (a) 2.30 (b) (c) 2.00 (d) (e) 8.70 ( ) OH = 2x0.005= 0.01 M ; poh = log 0.01 = 2.00; ph = 14 poh = A solution consists of M HCl, M HBr, and M HCN. What is the ph? (a) 0.23 (b) 0.87 (c) 1.10 (d) 1.26 (e) Insufficient information HCN is a weak acid. So, its concentration can be ignored compared to a strong acid. Both HCl and HBr are strong acids. H + = = M; ph = log H + = What is the ph of a solution formed by mixing 70. ml of 0.25 M HClO 4 (perchloric acid) and 140 ml of 0.15 M NaOH? (a) 2.46 (b) (c) (d) (e) mol 70mL mmol H = x = 17.5mmol L mol 140mL mmol OH = x = 21mmol L 1 I ( ) + ( ) ( ) + H aq OH aq H2O l C mmol E 0 3.5mmol OH = = M 210mL poh ( ) = log = 1.78; ph = 14 poh = A M aqueous solution of a weak acid, HB, has a ph of Calculate K a, the acid-dissociation constant. I (a) 7.2x10-3 M (b) 7.0x10-3 M (c) 0.22 M (d) 1.0x10-3 M (e) none of the above + ( ) + ( ) ( ) + ( ) HB aq H2O l H aq B aq 0.142M C -x +x +x E x x x + ph 1.50 x = H = 10 = 10 = M = = 0.139M ( ) H B M Ka = = = 9.0x10 M 0.139M

3 6. A carbonated drink is bottled under a fixed pressure of carbon dioxide. Which of the following stresses increases the concentration of the carbonate ion, CO 3 2-? (Use Le Chatelier s principle) P CO2 + H 2 O (l) CO 2 (aq) CO 2 (aq) + 2H 2 O (l) H 3 O + (aq) + HCO 3 - (aq) HCO 3 - (aq) + H 2 O (l) H 3 O + (aq) + CO 3 2- (aq) (a) The carbon dioxide pressure is decreased. Stress: too little P CO2. So, each reaction goes to left. (b) Phosphoric acid is added to the water before the carbonation step (above reactions). (c) Sodium Iodide is added to the water before the carbonation step. No effect. (d) Sodium hydroxide is added to the water before the carbonation step. (e) Ammonium iodide is added to the water before the carbonation step. 7. Which of the following compounds when dissolved in water is acidic? Transition metal ions (a) KCN (b) NaCl (c) LiCOO (d) CH 3 NH 2 (e) FeI 3 Basic Neutral Basic Basic are acidic 8. Calculate the ph of a solution that is 0.90 M in CH 3 COOH and 0.10 M KCH 3 COO. (CH 3 COOH: K a = 1.76 x 10-5 M) (a) 4.75 (b) 3.80 (c) 4.00 (d) 5.70 (e) (b) and (e) Stress: too much H. So, rxns go to left. pk a 5 ( x ) = log = 4.75 B 0.1 ph = pka + log = log = Stress: too little H. So, rxns go to right 9. Which of the following buffer solutions best resists a change in ph as either an acid or base is added? (a) 100 ml of 1.0 M HCOOH and 1.0 M NaHCOO 100 mmol acid and base (b) 200 ml of 0.70 M HCOOH and 0.70 M NaHCOO (c) 50 ml of 1.90 M HCOOH and 1.90 M NaHCOO (d) 1300 ml of 0.10 M HCOOH and 0.10 M NaHCOO. (e) 150 ml of 0.75 M HCOOH and 0.75 M NaHCOO. 140 mmol acid and base (largest) 95 mmol acid and base 130 mmol acid and base 113 mmol acid and base 10. A buffer solution consists of 0.3M HF and 0.5 M KF. A small amount of HI is added to the buffer. Which ion is responsible for keeping the ph almost constant? (a) K + Neutral ion (b) I Neutral ion (c) F Conjugate base neutalizes acid: + F + H O HF + H O 3 2 (d) HF (e) H + Acid does not neutralize acid This is what must be neutalized for the ph not to change.

4 11. Which of the following weak acids or bases is best suited to prepare a buffer whose ph is 8.50? (a) C 6 H 5 OH: pk a = 10 (b) ClCH 2 COOH: pk a = 2.87 (c) HNO 2 : pk a = 3.35 (d) NH 3 : pk b =4.76; pk a =? pk a = = 9.24 (e) HCN: K a = 4.79 x M; pk a =? pka ( x ) = log = A buffer with a ph = 3.35 is prepared using formic acid, HCOOH, and its conjugate base, HCOO - (pk a = 3.75). If the conjugate base concentration is 0.20 M, what is the approximate acid concentration? (a) 0.5 M (b) 0.4 M (c) 0.08 M (d) 0.8 M (e) 1 M 0.4 log ; ; 0.50 B 0.2M ph = pka + log = 3.35 = log = 3.75 log 0.2 M = = = = = 0.2M 0.2M M 13. What volume of 0.08 M Ba(OH) 2 completely neutralizes 35 ml 0.30 M HNO 3? Answers are for 1 sig. fig. (a) 4x10 ml (b) 5x10 ml ( ) ( ) + 2 ( ) 2 ( ) + ( ) ( ) Ba OH aq HNO aq H O l Ba NO aq (c) 7x10 ml (d) 9x10 ml (e) 1x10 2 ml ( ) 0.30mol HNO L mol Ba OH 2 L x x x = 0.066L= 66mL L 1 2mol HNO 0.08molBa OH ( ) 3 2

5 14. For the reaction As ( ) + Bg ( ) C( g) + Dg ( ) K= ( concentrationunitsomitted) 2 2, seconds into the reaction, there is some A, 0.4 moles B, 0.7 moles C, and 3 moles D in a 2.0 L container. What is the magnitude of Q and in which direction does the reaction go to reach equilibrium? (a) 3.7 and to the left (b) 3.7 and to the right 0.40mol 0.70mol 3mol B = = 0.20 M, C = = 0.35 M, D = = 1.5M 2L 2L 2L [ ] [ ] [ ] (c) 2.6 and to the right (d) 0.92 and to the left 2 [ C] [ D] [ B] ( 0.35M) 2 ( 1.5M) Q = = = 0.92M 0.2M 2 (e) 0.92 and to the right 15. What is the ph of a 10-9 M NaOH? (a) 5 The ph of pure water is 7. When you have a very dilute base, the solution (b) 9 must be slightly basic, that is, the ph must be slightly greater than 7. (c) slightly less than 7 The correct calculation includes the OH from water because the base (d) exactly 7 concentraion is so dilute. (e) slightly greater than For reasonable concentrations, which pair of ions would produce a precipitate in a solution containing these ions? (a) (b) Ti Ti Cl 2 +, Cl 4 +, Rule 4: choride, bromide, and iodide compounds are soluble. (c) Ni (d) Ni I 2 +,, S Rule 5b: CHOPS or carbonate, hydoxide, oxide, phosphate, and sulfide compounds are insoluble. (e) NH +, S 2 4 Rule 1: sodium, potassium, and ammonium compounds are soluble.

6 I 100. ml of 0.40 M formic acid, HCOOH, are titrated with 0.60 M NaOH. K a is 1.78 x 10-4 M. (a) The ph of the formic acid solution after 20. ml of the NaOH solution are added will be calculated? (1 PT) Calculate the initial mmol of formic acid. 0.40MHCOOHx100mL= 40. mmol (1 PT) Calculate the mmol of NaOH added. 0.60MOH x20. ml= 12mmol (2 PTS) Write the appropriate reaction and ICE table ( ) ( ) ( ) ( ) Reaction: HCOOH aq OH + aq H 2O l + HCOO aq Initial: 40mmol 12mmol (1 PTS) ph ph B 12mmol = pka + log = log = mmol (b) (3 PTS) At the midpoint, what is the ph? At the midpoint, the acid and base concentrations are equal. B ph = pka + log = log () 1 = 3.75 (c) What is the ph after 90 ml of the NaOH solution are added? (2 PTS) Write the appropriate reaction and ICE table ( ) ( ) ( ) ( ) Reaction: HCOOH aq OH + aq H 2O l + HCOO aq Initial: 40mmol 54mmol (2 PTS) ph 14mmol OH = = M ; poh = log ( ) = mL ph = 14 poh = 12.87

7 II. The K sp of M(OH) 3 is 9.0 x at 25ºC where M 3+ is a hypothetical non-acidic metal ion. (a) 4PTS What is the solubility of M(OH) 3 in pure water at 25ºC in mol/l? Write the reaction and make an ICE Table. 3+ M ( OH) ( s) " M ( aq) + 3OH ( aq) Reaction: 3 Initial: excess 0 0 -s +s +3s excess s 3s (b) 4PTS What is the solubility of M(OH) 3 in 0.60 M MCl 3 at 25ºC in mol/l? Write the reaction and make an ICE Table. Reaction: ()( s 3 ) M OH s = ( ) ( ) ( ) ( ) x 5 sp; / = K s = = x mol L M OH s " M aq + 3OH aq 3 Initial: excess s +s +3s excess 0.6+s 3s ( 0.6+s)( 3 ) M OH s = = ; because is small, s must be small. So, compared x K sp 18 1/3 910 x 7 to 0.6 is can be negected. Therefore = = / s x mol L 0.6x27 (c) 4PTS What is the solubility of M(OH) 3 in a solution buffered to a ph of 9.00 at 25ºC? Write the reaction and make an ICE Table. Reaction: Initial: Change: poh = 14 9= 5; OH = 10 = 1.00x10 poh 5 ( ) ( ) ( ) ( ) 3+ M OH s " M aq + 3OH aq 3-5 excess s +s +0 (buffered solution!) -5 excess s x10 3 s10 ( ) = x ; s = = x mol/ L 15 10

8 III. The following questions are independent. So, you don't have to get the previous question right to answer the next one. (a) 3PTS The graph to the right shows the titration of a weak acid with a strong base. What is the approximate value of K a for this acid? AT the equivalence point: = B. B So, ph 4.7 = pka + log = pk K a = 10 = 2x (1x10 to 6 x 10 for credit) a (b) 4PTS 10 ml of 1M HNO 3 is added to 40 ml of 0.25 M ammonia, NH 3. What is the approximate equilibrium concentration of each of the following ions? For example, x10 0 or3.6629x10 4x10. 10mL 1MNO3x 0.2M NO3 = 50mL + H 0M = NH3 = 0M + NH4 = 0.2M (c) 5PTS Calculate the ph of 0.40 M NH 4 Cl. Include the appropriate reaction and ICE table. K b (for NH 3!!!) is 1.8 x Reaction: Initial: = + + HO ( aq) + NH ( aq) HO( l) + NH ( aq) I 10 mmol 10 mmol mmol K NH 4 ( aq) + H2O ( l) NH3( aq) + H3O ( aq w 10 ) Ka = = = 5.56x10 5 Kb 1.8x M x +x +x 0.4 x x x C E + 10mmol NH4 = = 0.2 M 50mL 2 10 x 10 5 Ka = 5.56x10 = ; x = 5.56x10 ( 0.4 x) = 1.5x10 M 0.4 x 0 10 ph. 5 ( x ) ph = = 4.83