exam3reviewAs - Solutions to exam review questions 1 a no...

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Solutions to exam review questions: 1) a. no base has been added HF initial = 0.025 L (0.4 M) = 0.01 mol HF ** K a ice table** HF + H 2 O H 3 O + + F - K a = 3.5 x 10 -4 = (x 2 )/ (0.4 - x) I 0.4 -- 0 0 **can use cheat b/c (100)3.5 x 10 -4 < 0.4 C -x -- +x +x K a = 3.5 x 10 -4 = (x 2 )/ (0.4) E 0.4 - x -- x x x = 0.0118 M = [H 3 O + ] pH = -log [H 3 O + ] = -log (0.0118) = 1.9 b. 20 mL base added HF initial = 0.01 mol Base added = 0.02 L (0.25 M) = 0.005 mol < HF initial (buffer formed) **moles ice table** OH - + HF H 2 O + F - [HF] ±nal = 0.005 moles/ 0.045 L = 0.111 M I 0.005 0.01 -- 0 [F - ] ±nal = 0.005 moles/ 0.045 L = 0.111 M C -0.005 -0.005 -- +0.005 pH = -log K a + log ([F - ]/[HF]) E 0 0.005 -- 0.005 pH = -log (3.5 x 10 -4 ) + log (1) = 3.46 c. 40 mL base added HF initial = 0.01 mol [F - ] = 0.01 mol/0.065 L = 0.154 M Base added = 0.04 L (0.25 M) = 0.01 mol = HF initial (eq. point K b table) ** K b ice table** F - + H 2 O OH - + HF K b = K w /(3.5 x 10 -4) = (x 2 )/ (0.154 - x) = 2.86 x 10
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exam3reviewAs - Solutions to exam review questions 1 a no...

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