# 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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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!

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

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

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.

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.

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.

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

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

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

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

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%

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

Continue with Buffer notes see previous ppt.

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