Ka=[H3O+][C3H5O3-]/[acid]
= .067^2*conc^2/conc= .067^2*.0284
check my thinking.
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.
3 answers
Just wondering, why do you square the numerators?
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.
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.
5 answers