1) Write the Brønsted-Lowry reaction for weak acid HCN reacting with H 2 O.

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1 1) Write the Brønsted-Lowry reaction for weak acid HCN reacting with H O. HCN + H O º H O + + CN ) Write the Brønsted-Lowry reaction for weak base NH reacting with H O NH + H O º OH + NH + ) Using the information from the Chemical Equilibria Eercises, write the equilibria epression for the acid HCN in 1) above using the substitution K a K K a + - [HO ][CN ] [HCN] ) Write the equilibrium epression for the base NH in ) above using the substitution K. + [OH ][NH ]

2 5) Calculate the ph of a solution which is M in HCN. The K a of HCN is HCN + H O º H O + + CN [HCN] [H O + ] [CN ] equilibrium substituting: and assume << so: Ans: ) Calculate the ph of a solution which is 0.10 M in HCN and 0.10 M in NaCN. The K a of HCN is (Notice that the important chemical species here is the CN - ion. Na + is a spectator ion) HCN + H O º H O + + CN [HCN] [H O + ] [CN ] equilibrium ( ) substituting: and assume << Ans: 9.8 7) Calculate the ph of a solution which is 0.10 M in HCN and 0.0 M in NaCN. The K a of HCN is (What is Na +? Ans: spectator ion ) HCN + H O º H O + + CN [HCN] [H O + ] [CN ] equilibrium ( ) substituting: and assume << 0.10 (and << 0.0 ) Ans: 9.85

3 8) Calculate the ph of a solution which is M in NH. The for NH is NH + H O º OH + + NH [NH ] [OH ] [NH + ] equilibrium substituting: and assume << so:. 10 ([OH ] ) poh.7 Ans: ) Calculate the ph of a solution which is M in NH and M in NH +. The for NH is NH + H O º OH + + NH [NH ] [OH ] [NH + ] equilibrium substituting: and assume << ([OH ] ) poh.7 Ans: ) Calculate the ph of a solution which is 0.10 M in NH and M in NH +. The for NH is NH + H O º OH + + NH [NH ] [OH ] [NH + ] equilibrium substituting: and assume << ( and << 0.10 ) ([OH ] ) poh.7 Ans: 10.6 Problems 11, 1 and 1 are dilution problems. If you do not know how to do dilution problems, do some review first. These are in preparation for subsequent questions.

4 11) If ml of a solution is originally 0.10 M CH COOH and one adds 50.0 ml of water to this solution, what is the final concentration of the CH COOH? For dilutions: C 1 V 1 C V C M V ml C? V 150 ml (volume of dilute solutions add with little error.) substitute: (0.10 M)(100.0 ml) C (150 ml) Ans: M 1) If ml of a solution is originally 0.10 M CH COOH and one adds 50.0 ml of 0.10 M NaCl to this solution, what is the final concentration of the CH COOH? C 1 V 1 C V For CH COOH: (0.10 M)(100.0 ml) C (150.0 ml) (NaCl is irrelevant) Ans: M 1) If ml of a solution is originally 0.10 M CH COOH and one adds 50.0 ml of 0.10 M NaCH COO to this solution, what is the final concentration of the CH COOH? What is the final concentration of the NaCH COO? (At this point, ignore the establishment of the Brønsted-Lowry equilibrium.) C 1 V 1 C V For CH COOH: NaCH COO: (0.10 M)(100.0 ml) C (150.0 ml) (0.10 M)(50.0 ml) C (150.0 ml) For M of CH COOH: Ans: M CH COOH For M of NaCH COO Ans: 0.0 M NaCH COO (What is Na + in NaCH COO?) What is the important ion? Ans: the CH COO (acetate) ion

5 1) What are the phs of the solutions described in questions 11, 1 and 1? The K a for CH COOH For 11: CH COOH + H O º H O + + CH COO [CH COOH] [H O + ] [CH COO ] initially equilibrium substituting: and assume << 0.10 (and << 0.0 ) Ans #11: ph.96 For 1: Same as for 11 since Na + and Cl are a spectator ions. Ans #1: ph.96 For 1: CH COOH + H O º H O + + CH COO [CH COOH] [H O + ] [CH COO ] initially equilibrium ( ) substituting: and assume << 0.0. (and << ) Ans #1: ph. 15) A solution is created by miing ml of a solution which is 0.10 M CH COOH with 50.0 ml of a solution which is 0.10 M NaOH. (Confused? - Look up Arrhenius Acid/Base and neutralization.) a) Write the Arrhenius acid-base reaction for this: CH COOH + NaOH! NaCH COO + H O b) What is the concentration of the CH COOH if one does not take into account the Brønsted-Lowry reaction with water? By definition C n/v where V is in liters. and so n CV

6 moles of: when: How calculated: ANS: CH COOH initially (0.10 M)( L) mol NaOH initially (0.10 M)( L) mol NaCH COOH after Arrhenius reaction n NaCHCOO /1 n NaOH /1* mol CH COOH after Arrhenius reaction n starting n used mol *limiting reactant. C CH COOH thus: Ans: 0.0 M CH COOH mol L 15 continued: c) What is the concentration of the NaCH COO if one does not take into account the Brønsted-Lowry reaction with water? C NaCH COO thus: Ans: 0.0 M NaCH COO mol L d) What is the ph of the resultant solution described above? The K a for CH COOH ph pk a log( ).7 (Why?) Ans: ph.7 16) What is the ph of a solution formed by miing 00 ml of 0.0 M NH with 50 ml of water? for NH C 1 V 1 C V C M V 1 00 ml C? V 50 ml (0.0 M)(00 ml) C (50 ml) C 0.16 M + [NH ][OH ] ( [OH ]) Ans: ph 11.

7 17) What is the ph of a solution formed by miing 00 ml of 0.0 M NH with 50 ml of 0.50 M NH Cl? The for NH C 1 V 1 C V for both [NH ] and NH [NH ]: [NH + ]: C M C M V 1 00 ml V 1 50 ml C? C? V 50 ml V 50 ml C 0.16 M C 0.10 M + [NH ][OH ] ( [OH ]) poh.5 Ans: ph ) What is the ph of a solution formed by miing 00 ml of 0.0 M NH with 50 ml of 0.50 M HCl? (Note: you must first answer questions similar to question? above. What is the concentration of NH and NH Cl?) The for NH Final volume 50 ml n NH originally (0.0 M)(00 ml) 0 mmol n HCl originally (0.50 M)(50 ml) 5 mmol (limiting reactant) n NHCl produced n HCl originally 5 mmol n NH left 0 mmol 5 mmol 15 mmol resultant [NH ] 15 mmol/50 ml resultant [NH + ] 5 mmol/50 ml [NH ][OH ] ([OH ]) poh.97 Ans: ph ) What is the ph of a solution formed by miing 00 ml of 0.0 M NH with 50 ml of 0.50 M NH Cl and 50 ml of 0.50 M HCl? (Hint: first calculate the dilutions for each of these species, then do the Arrhenius acid-base reaction.) The for NH Final Volume 00 ml n NH originally (0.0 M)(00 ml) 0 mmol n HCl originally (0.50 M)(50 ml) 5 mmol (limiting reactant) n NHCl originally (0.50 M)(50 ml) 5 mmol n NHCl produced n HCl originally 5 mmol n NHCl total 5 mmol + 5 mmol 50 mmol n NH left 0 mmol 5 mmol 15 mmol resultant [NH ] 15 mmol/00 ml resultant [NH + ] 50 mmol/00 ml [NH ][OH ] ([OH ]) poh 5.7 Ans: ph 8.7

8 Use the following K a s to answer questions 0 through : For the reaction: The equilibrium constant is: - H PO + H O º H PO + H O + K a1 (H PO ) H PO + H O º HPO + H O + K a (H PO - ) HPO + H O º PO + H O + - K a (HPO ) ) What is the ph of a solution which is 0.10 M in H PO? K a etc. + [HPO ][HO ] [HPO ] 0.10 Ans: ph 1.56 (1.6 if << 0.10 not assumed) 1) What is the ph of a solution which is 0.10 M in H PO and 0.10 M in NaH PO? (What is Na +? It is spectator ion ) ph pk a1 Ans: ph.1 ) What is the ph of a solution which is 0.10 M in NaH PO and 0.10 M in Na HPO? ph pk a Ans: ph 7.1 ) What is the ph of a solution which is 0.10 M in Na HPO and 0.10 M in Na PO? ph pk a Ans: ph 1.