The pH of .10M solution of formic acid(HCOOH) is 2.39. What s the Ka of the acid

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The reaction of dissociation will be: HCOOH + H2O ---> H3O+ +HCOO- Ka is defined as: Ka = [H3O+][HCOO-]/[HCOOH] pH = -log[H3O+] The pH equals 2.39; so the concentration ofH3O+ is 10-2.39 = 0.004074 M. You also know that the following conditions will occur:    HCOOH + H2O --->H3O+ + HCOO- I: 0.10M                     0 M         0 M C: 0.10-x                     +x          +x E: 0.10-xM                   x M        x M You know that x equals 0.004074 M. Plug all this in and solvefor Ka: Ka = [H3O+][HCOO-]/[HCOOH] Ka = [0.004074][0.004074]/[0.10 - 0.004074] Ka = 0.000173 = 1.73 * 10-4

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