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lecture notes2 - Chem 102 D Miller Solutions to Review...

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Chem 102 D. Miller Solutions to Review Problems for Acid/Base Chemistry 1. Glacial acetic acid, pure HC 2 H 3 O 2 (FW = 60.0), has a concentration of 17.54 M. If 85.5 mL of glacial acetic acid are diluted to 250 mL, what is the acetic acid concentration? This is a dilution problem; use M 1 V 1 = M 2 V 2 . (17.54 M)(85.5 mL) = M 2 (250 mL) M 2 = 6.0 0 M 2. If 26 mL of this diluted acetic acid (see Prob. 1) are further diluted to exactly 800 mL, the solution pH is 2.74. Calculate K a for acetic acid. First, there is a dilution, followed by an equilibrium calculation involving a solution of a weak acid. For the dilution, (6.0 0 M)(26 mL) = M 2 (800. mL) and M 2 = 0.19 5 M For the weak acid solution, HC 2 H 3 O 2 + H 2 O X H 3 O + + C 2 H 3 O 2 - init 0.19 5 M ~10 -7 M 0 equil 0.19 5 -x ~x x where x = increase in [C 2 H 3 O 2 - ]. Since the pH is known, the [H 3 O + ], and hence x, is known. pH = 2.74 Y [H 3 O + ] = 1.8 2 x 10 -3 M = x
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Chem 102 D. Miller Solutions to Review Problems for Acid/Base Chemistry 3. If 13.2 g NaC 2 H 3 O 2 (FW = 82.0) are added to the 800 mL of solution in Problem 2, what is the resulting pH? The addition of C 2 H 3 O 2 - to a solution of HC 2 H 3 O 2 creates a HC 2 H 3 O 2 / C 2 H 3 O 2 - buffer.
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