Acids and Bases 4

Acids and Bases 4 - Some worked examples 02/29/08 Acids...

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02/29/08 1 Some worked examples
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02/29/08 2 Old Exam Question You are given that the value of K a for acetic acid is 1.77 x 10 -5 . (a) Calculate the pH of 0.3 M acetic acid (b) Calculate the resulting pH after adding 5 mL of a solution containing 1 M sodium hydroxide to 50 mL of the solution in (a). (c) Repeat the calculation of (b) so that a total of 10 mL of the NaOH solution has been added. (d) Compare the change in pH for the two stages and comment on your answer.
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3 Answer (a) CH 3 CO 2 H + H 2 O CH 3 CO 2 - + H 3 O + Assume we can neglect the protons from water. This solves to give x = 2.3 x 10 -3 M and pH = 2.64 . (b) Originally 50 mL @ 0.3 M = 15 mmoles of acetic acid. First addition is 5 mmoles of NaOH. This leaves 10 mmoles of acid and makes 5 mmoles of conjugate base. Apply HH equation: pH = pK a + log [Base]/[Acid] = pK a + log 0.5 = 4.75 - 0.3 = 4.45 (c) After second addition, we have 5 mmoles of acid and 10 mmoles of conjugate base. pH = pK
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Acids and Bases 4 - Some worked examples 02/29/08 Acids...

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