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An 0.0284 aqueous solution of lactic acid is found to be 6.7% ionized. Determine Ka for lactic acid.

HC3H5O3 + H2O <-- --> H3O^+ + C3H5O3^- Ka =?

Thanks.

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3 answers
  1. Ka=[H3O+][C3H5O3-]/[acid]

    = .067^2*conc^2/conc= .067^2*.0284

    check my thinking.

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    👤
    bobpursley
  2. Just wondering, why do you square the numerators?

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  3. Ka = [H3O^+][C3H5O3^- ] / [HC3H5O3]
    Let [H3O^+] = [C3H5O3^- ] = x
    Ka = [x][x ] / [conc. - x] (exact)
    Ka = [x][x ] / [conc.] (approximate if x is much smaller than cocnetration)
    I assume conc. = 0.0284 is moles/liter
    [H3O^+] = [C3H5O3^- ] = (0.067)(0.0284)
    Ka = {[(0.067)(0.0284)][(0.067)(0.0284)]} / [0.0284] (approx)
    Pay close attention how the values were substituted in the last expression.

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