L13-correction-explanation-cobalt-demo

L13-correction-explanation-cobalt-demo - leaving the red 4...

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Initial 0.05 M 6.0 M ~0 Change x 4x + x Eqm 0.05 – x 6.0 – 4x x Test tube #1 = 2.5 x 10 4 = x (0.05 – x) (6.0 – 4 x) 4 The magnitude of K is small enough that reactants are favored over products. Thus, we estimate that x (and 4x) is small compared to 6.0 but that x may not be small compared to 0.05, and so we simplify solving for x using this approximation [in attempting such an approximation, it is always a good idea to check, after you solve for a value of x, that your approximation is a good one!]. [Note also that the 9 am class discovered error in
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Unformatted text preview: leaving the red 4 multiplier out, while the 11 am class may not have yelled loud enough about the omission to catch my attention!]. x (0.05 x) (6.0) 4 So: x/(64.8 1296x) = 2.5x10 4 x = 0.0162 0.324x 1.324x = 0.0162, so x = 0.0122 Now use x to calculate the equilibrium concentrations in the ICE table above: [Co(H 2 O)6 2+ ] = 0.038 M, [Cl ] = 5.95 M, and [CoCl 4 2 ] = 0.012 M ( l )...
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This note was uploaded on 04/03/2008 for the course CHEM 1A taught by Professor Nitsche during the Fall '08 term at University of California, Berkeley.

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