Solutions to problems on ions

# Solutions to problems on ions - CHM 331 Solutions to...

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CHM 331 Solutions to Problems on Ions 1. K a = [(H + ) (Ac - ) / (HAC)] (γ H + γ AC -) = K a γ 2 log γ ± = -0.509 (1 x 1) ( I ) 1/2 I = ½ [(0.05) (4) + (0.10) (1)] = 0.15 From these, γ ± = 0.6351 and γ ± 2 = 0.4034 K a = 1.75 x 10 -5 /0.4034 = 4.338 x 10 -5 We let x = (H + ) = (Ac - ) so that K a = x 2 /0.001- x ). The quadratic equation in x can be solved in the usual way. However, a useful approximation method is the following. We neglect x in the denominator, so that x 2 1 = (4.338 x 10 -5 ) (0.001) and x 1 = 2.08 x 10 -4 . In the second approximation, we use x 1 in the denominator: x 2 2 = (4.338 x 10 -5 ) (0.001 – 2.08 x 10 -4 ), whence x 2 = 1.85 x 10 -4 . This is sufficient. (A third round of approximation, using x 2 in the denominator of the expression for K a gives x 3 = 1.88 x 10 -4 .) 2. The apparent K value is: K’ = = From the given and C values, K ’= (0.1) 2 0.001/0.9 = 1.11 x 10 -5 . The more correct equation is K = = ))= was essentially unity at the low ionic strength of the first calculation, so the above

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Solutions to problems on ions - CHM 331 Solutions to...

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