Chp6_C0vsXaprox

# Chp6_C0vsXaprox - Approximations and iterative solutions to...

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Approximations and iterative solutions to non-linear (i.e. quadratic) equations We often encounter the following sort of expression in problems dealing with aqueous solutions of weak acids or bases: x c x K b a - = 0 2 / where x is the degree of dissociation and c 0 is the “make-up” concentration. Formally, this is a quadratic expression if the unknown is “x” which can represent either the [H 3 O 1+ ] or the [OH 1– ] depending on whether we have a solution of a weak acid or weak base. Consider an arbitrary specific case: K = 8.80 × 10 –4 and c 0 = 0.0300. We are going to solve this iteratively, namely, in cycles; so I shall define the value of x for each cycle with a subscript viz: x 1 is the value of x for cycle or iteration one. Cycle 1 : 1 2 1 4 03 . 0 10 8 . 8 x x - = × - If x 1 << 0.03, we approximate 0.03 – x 1 ≈ 0.03 and obtain x 1 = 5.14 × 10 –3 . This is about a 17% relative error in terms of x 1 /c 0 . Cycle 2 : To use iteration, we now substitute the value of the approximated quantity into

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Unformatted text preview: the place were the approximation was made, viz: 3 2 2 4 10 14 . 5 03 . 10 8 . 8--×-= × x where, note, we have now written x 2 as the second iteration and substituted the value of x 1 only in the place where we approximated it to be negligible – in the denominator. Solving yields x 2 = 4.68 × 10 –3 . Cycle 3 : We now repeat the process to obtain: 3 2 3 4 10 68 . 4 03 . 10 8 . 8--×-= × x yielding x 3 = 4.72 × 10 –3 . Note that the difference between x3 and x2 is smaller than that between x2 and x1. Cycle 4 : Repeat and get x 4 = 4.72 × 10 –3 . To the number of significant digits, the value of x is now constant and said to be “self-consistent” or to have “converged.” The exact value obtained from solving the quadratic formula is x = 4.717 × 10 –3 ....
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