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Lecture16_S08 - Feb 25 2008 Lecture 16 Chapter 17 Further...

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Feb. 25, 2008 Lecture 16, Chapter 17: Further Equilibria Typical Exam Question, 3 parts: (a) pH of 0.30 M H(OAc) soln if K a = 1.8x10 -5 (b) pH of 0.30 M Na(OAc) soln (c) pH of 0.30 M H(OAc) and 0.30 M Na(OAc) in same container H(OAc) (aq) H + (aq) + (OAc)- (aq) K a = [H + ][OAc - ] [H(OAc)] (a) I 0.30 M 0 M 0 M C - x + x + x E (0.30 –x) (x) (x) (x) 2 (0.30-x) = (x) 2 (0.30) = (0.30)(1.8x10 -5 ) (x) 2 K a = [H + ][OAc - ] [H(OAc)] If x small = 1.8x10 -5 pH = -log (2.3x10 -3 ) = 2.3x10 -3 = [H+] x pH = 2.64 (acidic) (OAc) - (aq) OH - (aq) + H(OAc) (aq) K b = [OH - ][H(OAc)] [(OAc)-] (b) I 0.30 M 0 M 0 M C - x + x + x E (0.30 –x) (x) (x) (x) 2 (0.30-x) = (x) 2 (0.30) = (0.30)(5.6x10 -10 ) (x) 2 If x small = 5.6x10 -10 = 1.3x10 -5 = [OH - ] x pOH = -log (1.3x10 -5 ) pOH = 4.90 K b = K w K a 1.0x10 -14 1.8x10 -5 K b = 5.6x10 -10 K b = K b = [OH - ][H(OAc)] [(OAc)-] pH = 9.10 (basic) pH + pOH = 14 pH = 14- pOH = 14 - 4.90
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By mixing the two solutions it seems like we should probably end up with an acidic solution Acid is farther from neutral pH, 7.00 – 2.64 = 4.36 pH units Than the base is from neutral, 9.10 – 7.00 = 2.10 pH units Difference = 2.26 pH units So maybe the pH will be acidic by 2.26 units, or pH = 7.00-2.26 = 4.74?? Calculator Party! Lets do this more rigorously….
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