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We’ll use equilibrium concepts to explain how a buffer solution minimizes the change in pH when a small amount of base is added. Buffer Solutions -How.

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Presentation on theme: "We’ll use equilibrium concepts to explain how a buffer solution minimizes the change in pH when a small amount of base is added. Buffer Solutions -How."— Presentation transcript:

1 We’ll use equilibrium concepts to explain how a buffer solution minimizes the change in pH when a small amount of base is added. Buffer Solutions -How They Work Part 2 – Adding a Base

2 We’ll start off by considering a buffer solution in which the concentration of acetic acid and the concentration of the acetate ion are both 1 M. Buffer Solution 1 M CH 3 COOH (aq) 1 M CH 3 COO – (aq)

3 Here is the equilibrium equation describing the buffer. 1 M low Buffer Solution 1 M CH 3 COOH (aq) 1 M CH 3 COO – (aq)

4 Because this is in an aqueous solution, any changes in the amount of water in this equilibrium are insignificant. We’ll just colour the water blue and we won’t be concerned with it. 1 M low

5 The concentration of hydronium in this equilibrium is much lower than the 1 M concentrations of acetic acid and acetate. 1 M low

6 Now, we’ll construct graphs showing how the concentrations of the three species in the equilibrium will vary when we add a base to this buffer and it re-adjusts. [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] 1 M low ~ ~ ~ ~ ~ ~ Time 

7 This symbol shows that the y axes on these graphs is actually much longer than portrayed in the diagram. [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] 1 M low ~ ~ ~ ~ ~ ~ Time 

8 The 1 molar concentrations of acetic acid and acetate ion are considerably higher than that of the hydronium ion. [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] 1 M low ~ ~ ~ ~ ~ ~ Time 

9 First, we’ll consider the concentrations of the three species before we add the base [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] 1 M low ~ ~ ~ ~ ~ ~ Time  Before base is added

10 The concentrations of all three species remain constant. [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] 1 M low ~ ~ ~ ~ ~ ~ Time  Before base is added Time 

11 At this point, we add a base, like NaOH to this buffer. [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] ~ ~ ~ ~ ~ ~ Time  A base is added Time 

12 A Base will react with H 3 O + and Neutralize it, as shown in this (click) net ionic equation. [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] ~ ~ ~ ~ ~ ~ Time  A base is added A Base will react with H 3 O + and Neutralize it. This will decrease [H 3 O + ]

13 In this neutralization reaction, hydronium is consumed so this will decrease the [H 3 O + ] [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] ~ ~ ~ ~ ~ ~ Time  A base is added A Base will react with H 3 O + and Neutralize it. This will decrease [H 3 O + ]

14 So we add the base and the concentration of H3O+ (click) quickly decreases [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] ~ ~ ~ ~ ~ ~ Time  A base is added Time 

15 Adding a base had no immediate effect on the concentration of acetic acid. [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] ~ ~ ~ ~ ~ ~ Time  A base is added

16 Or on the concentration of the acetate ion. [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] ~ ~ ~ ~ ~ ~ Time  A base is added

17 But the sudden decrease in the hydronium ion concentration that took place when the base was added, causes a stress on the equilibrium system. Because hydronium is a product whose concentration has been decreased… [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] ~ ~ ~ ~ ~ ~ Time  A base is added Sudden Decrease in [H 3 O + ] Time 

18 The equilibrium will (click) shift to the right in order to compensate. [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] ~ ~ ~ ~ ~ ~ Time  A base is added Shift to the Right Time  Sudden Decrease in [H 3 O + ]

19 As the shift to the right occurs, (click), the concentration of hydronium gradually increases as it partially compensates for its sudden drop. [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] ~ ~ ~ ~ ~ ~ Time  Shift to the Right Time  Shifts to the Right

20 The acetate ion is also a product, so as the shift to the right occurs, (click), its concentration will gradually increase like the hydronium did. [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] ~ ~ ~ ~ ~ ~ Time  Shift to the Right Time  Shifts to the Right

21 Acetic acid is a reactant, so as the shift to the right occurs (click), its concentration will gradually decrease. [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] ~ ~ ~ ~ ~ ~ Time  Shift to the Right Time  Shifts to the Right

22 When the shift to the right is complete, a new equilibrium is established, so as time continues, the concentrations of all three species [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] ~ ~ ~ ~ ~ ~ Time  New Equilibrium Time 

23 Remains constant and the lines on the graph are horizontal. [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] ~ ~ ~ ~ ~ ~ Time  New Equilibrium Time 

24 In the overall process, the concentration of acetic acid (click) showed a net decrease. [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] ~ ~ ~ ~ ~ ~ Time  Net Decrease

25 And the concentration of the acetate ion (click) showed a net increase. [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] ~ ~ ~ ~ ~ ~ Time  Net Increase

26 The concentration of hydronium (click) dropped rapidly when the base was added. [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] ~ ~ ~ ~ ~ ~ Time  Net Increase Rapid Drop

27 As the equilibrium shifted to the right and buffering took place, the hydronium concentration (click) gradually increased in order to partially compensate for the drop [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] ~ ~ ~ ~ ~ ~ Time  Net Increase Gradual Increase

28 So in the overall process, the hydronium ion concentration showed a very small net decrease. [CH 3 COOH] [CH 3 COO – ] [H 3 O + ] ~ ~ ~ ~ ~ ~ Time  Net Increase Very Small Net Decrease

29 Now, we’ll consider the changes in pH as the base is added and the system adjusts. Remember pH is the Negative log of the hydronium ion concentration. [H 3 O + ] ~ ~ Time  pH Time  pH = – log[H 3 O + ]

30 The negative sign means as the concentration of H3O+ goes down, the pH goes up. [H 3 O + ] ~ ~ Time  pH Time  pH = – log[H 3 O + ] As [H 3 O + ], pH

31 So when the base was first added, as the hydronium ion concentration goes down (click), [H 3 O + ] ~ ~ Time  pH Time  As [H 3 O + ], pH A base is added

32 the pH goes up. [H 3 O + ] ~ ~ Time  pH Time  As [H 3 O + ], pH A base is added

33 As the shift to the right occurs and the concentration of hydronium gradually increases, [H 3 O + ] ~ ~ Time  pH Time  As [H 3 O + ], pH Shift to the right Shift to the Right

34 the pH gradually decreases [H 3 O + ] ~ ~ Time  pH Time  As [H 3 O + ], pH Shift to the right Shift to the Right

35 So in the overall process, as the hydronium ion concentration (click) shows a very small net decrease. [H 3 O + ] ~ ~ Time  pH Time  Very Small Net Decrease

36 The pH shows a very small net Increase. [H 3 O + ] ~ ~ Time  pH Time  pH shows a Very Small Net Increase Very Small Net Decrease

37 If there was no buffer solution present, and we added a base to water, (click) the pH would quickly rise. pH Time  Un-buffered Base is Added

38 But there is nothing to buffer this rise, so the pH will (click) remain at this level. pH Time  Base is Added Un-buffered

39 And show a large net increase in the overall process… pH Time  pH would show a Large Net Increase Base is Added

40 Instead of the small net increase experienced by the buffer solution. The buffer has accomplished it’s purpose. It has minimized the change in pH resulting from the addition of a base. pH Time  pH shows a Very Small Net Increase


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