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Find the percent ionization of a 0.250 m solution of hc2h3o2. express your answer numerically to two significant figures.

Mindee Nelson

in Chemistry

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Megan Page on January 2, 2018

From this reaction:and by the use of ICE table: CH3COOH ↔ H + + CH3COO-initial 0.25 0 0change - X +X +XEqu (0.25 - X) X Xwhen we have Ka for acetic acid = 1.8x10^-5 and Ka = [H+][CH3COO-] / [CH3COOH]1.8x10^-5 = X*X / (0.25-X)X = 0.00211 m∴[H+] = 0.00211 m∴percentage ionization = [H+] / [initial acid] * 100 = 0.00211 / 0.25 *100 = 0.844 %


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