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Entry Task: Feb 11 th Monday Question: 0.080 M in sodium formate, NaCHO 2, and 0.200 M formic acid, HCHO 2 You have 5 minutes!

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Presentation on theme: "Entry Task: Feb 11 th Monday Question: 0.080 M in sodium formate, NaCHO 2, and 0.200 M formic acid, HCHO 2 You have 5 minutes!"— Presentation transcript:

1 Entry Task: Feb 11 th Monday Question: 0.080 M in sodium formate, NaCHO 2, and 0.200 M formic acid, HCHO 2 You have 5 minutes!

2 Agenda Discuss Common ion-Effect ws Finish Buffer notes and in-class practice HW: Buffer ws #1

3

4 1. a) Consider the equilibrium B(aq) + H 2 O (l)  HB + (aq) + OH - (aq). In terms of LeChatelier’s principle, explain the effect of the presence of a salt of HB+ on the ionization of B. b) Give an example of a salt that can decrease the ionization of NH 3 in solution. Refresh yourself on LeChatelier’s Principle B(aq) + H 2 O (l)  HB + (aq) + OH - (aq) a. By adding more salt of HB+, like Hb salt, it would decrease the OH- on the product side- shifting reaction left. This would also increase substance B

5 2. Does the pH increase, decrease, or remain the same on addition of each of the following? EXPLAIN!!!!! a)NaNO 2 to a solution of HNO 2 NaNO 2 came from a strong base/weak acid so the solution would become more basic- increasing pH. b) (CH 3 NH 3 )Cl to a solution of CH 3 NH 2 (CH 3 NH 3 )Cl came from a weak base and strong acid so the solution would become more acidic – decrease in pH.

6 2. Does the pH increase, decrease, or remain the same on addition of each of the following? EXPLAIN!!!!! c) sodium formate to a solution of formic acid sodium formate came from a strong base/weak acid so the solution would become more basic- increasing pH. d) potassium bromide to a solution of hydrobromic acid potassium bromide came from a strong base and strong acid so the solution would remain the same.

7 2. Does the pH increase, decrease, or remain the same on addition of each of the following? EXPLAIN!!!!! e) HCl to a solution of NaC 2 H 3 O 2 HCl is a strong acid and NaC 2 H 3 O 2 came from a strong base and weak acid but this is hydolyzed and HCl would move the equilibrium toward acidic- decreasing pH.

8 Calculate the pH of the following solutions: a) 0.060M in potassium propionate, KC 3 H 5 O 2, and 0.085 M in propionic acid, HC 3 H 5 O 2 Ka= 1.3x10 -5 [x] [0.060 + x] [0.085 - x] 1.3x10 -5 = HC 3 H 5 O 2 (aq) H + (aq) + C 3 H 5 O 2 − (aq) [H + ] [C 3 H 5 O 2 − ] [HC 3 H 5 O 2 ] K a = = 1.3 x 10 -5 (1.3 x 10 -5 )(0.085) 0.060 1.11 x 10 -6 0.060 X = 1.84 x10 -5 x = [H + ] pH=–log( 1.84 x10 -5 ) = 4.73

9 Calculate the pH of the following solutions: b) 0.090 M in sodium formate, NaCHO 2, and 0.100 M formic acid, HCHO 2 Ka= 1.8x10 -4 [x] [0.090 + x] [0.100 - x] 1.8x10 -4 = HCHO 2 (aq) H + (aq) + CHO 2 − (aq) [H + ] [CHO 2 − ] [HCHO 2 ] K a = = 1.8 x 10 -4 (1.8 x 10 -4 )(0.100) 0.090 1.8 x 10 -5 0.090 X = 2.0 x10 -4 x = [H + ] pH=–log( 2.0 x10 -4 ) = 3.70

10 Calculate the pH of the following solutions: c) 0.075M in trimethylamine, (CH 3 ) 3 N, and 0.10M trimethylammonium chloride, (CH 3 ) 3 NHCl Kb= 6.4 x10 -5 [x] [0.10 + x] [0.075 - x] 6.5x10 -5 = (CH 3 ) 3 N (aq) OH - (aq) + (CH 3 ) 3 NH + (aq) [OH _ ] [(CH 3 ) 3 NH + ] [(CH 3 ) 3 N] K b = = 6.5 x 10 -5 (6.5 x 10 -5 )(0.075) 0.10 4.875 x 10 -6 0.10 X = 4.875 x10 -5 x = [OH - ] pOH=–log( 4.875 x10 -5 ) = 4.31 then subtract from 14 = 9.69

11 Calculate the pH of the following solutions: d) 0.0750M pyridine, C 5 H 5 N, and 0.0850 M in pyridinium chloride, C 5 H 5 NHCl Kb= 1.7 x10 -9 [x] [0.0850 + x] [0.0750 - x] 1.7x10 -9 = C 5 H 5 N (aq) OH - (aq) + C 5 H 5 NH + (aq) [OH - ] [C 5 H 5 NH + ] [C 5 H 5 NH] K b = = 1.7 x 10 -9 (1.7 x 10 -9 )(0.0750) 0.0850 1.275 x 10 -10 0.0850 X = 1.5 x10 -9 x = [OH - ] pOH=–log( 1.5 x10 -9 ) = 8.8 then subtract from 14 = 5.18

12 4. a) Calculate the percent ionization of 0.050M butanoic acid (Ka= 1.5x10 -5 ) b) Calculate the percent ionization of 0.050M butanoic acid in a solution containing 0.070M of sodium butanoate. x 2 0.050 1.5x10 -5 = (1.5 x 10 -5 )(0.050) = x 2 7.5 x 10 -7 = x 2 8.66 x 10 -4 0.050 X 100 = 1.7%

13 4. a) Calculate the percent ionization of 0.050M butanoic acid (Ka= 1.5x10 -5 ) b) Calculate the percent ionization of 0.050M butanoic acid in a solution containing 0.070M of sodium butanoate. 0.070 0.050 1.5x10 -5 = 7.5x 10 -7 0.070 X 100 = 0.021% (1.5 x 10 -5 )(0.050) 0.070 1.07 x 10 -5 0.050

14 Continue with Buffer notes see previous ppt.


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