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Assignment 1

# Assignment 1 - Solutions to Assignment 1 Problem 17(Note...

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Solutions to Assignment 1 Problem 17 Part a ) [HA] [H+] + [A - ] The dissociation of the acid can be shown as: K a = [H + ] [A - ] [HA] At 2.4% dissociation there would be .024 x .22M = .0053 M H + This would also be the amount of A produced. Therefore K a = .0053 M x .0053 M The amount of HA would .22 - .0054 = .2147 M .2147 M = 1.3 x 10 -4 M Part b) Part c) 550 ml of .22M weak acid = x mls of .1 M base (Note: You may have noticed that this problem is similar to Problem 1-18 on Page 25 of the Segel book.) Just as in the case of strong acids, the number of moles of OH - required to neutralize a weak acid is equal to the number of moles of total H + available. x = 1210 mls 550 x .22 = .1x pH = - log [H + ] = - log [.0053] pH = - [- 2.28] = 2.28

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Problem 18. Part a) As discussed in class: pH = - log [H + ] pH = 4.3 Antilog of -4.3 = 5 x 10 -5 M Part b) Since you have already determined the H + concentration in part a (above), you now determine what proportion it is of the total concentration of acid. This value is the H + concentration. 5 x 10 -5 M .27 M = .0185 = 1.85 x 10 -2 Part c) K a = [H + ] [A - ] HA K a = (5 x 10 -5 M) 2 You already determined the hydrogen ion concentration in part a above. It will equal the anion concentration
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Assignment 1 - Solutions to Assignment 1 Problem 17(Note...

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