Using the method of successive approximations, what is [H3O+] in 0.500 M HClO2 (

Question

Using the method of successive approximations, what is [H3O+] in 0.500 M HClO2 (Ka=1.1*10^-2) and how many iterations (n) are needed to validate that a constant value had been obtained? Express [H3O+] using two significant figures, followed by a comma and the number of iterations as an integer.

Explanation / Answer

HClO2 + H2O -----> H3O+ + ClO2 - 0.5 ---- intial conc of HClO2 0.5-x-----final conc of HClO2 so x ---- conc of H3O+ = conc. of ClO2- Ka = [H3O+][ ClO2 -]/[HClO2] water is in excess :) so we get Ka = x^2/(0.5-x) Ka is given as 1.1 x 10^-2 solving this equation we get 1.1x10^-2 (0.5 - x) - x^2 = 0 x = 0.06886 M = 0.07 M

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