Question

Lactic acid, which builds up in muscle tissue upon strenuous exercise, is partially dissociated in aqueous solution: The following equilibrium concentrations were measured at 25 ∘C: Lactic acid [C3H6O3]=9.64×10−2M;[H+]=3.65×10−3M; Lactate ion [C3H5O3−]=3.65×10−3M. Calculate the value of the equilibrium constant.

Answer 1

lactic acid +H2O $\rightleftharpoons$lactate ion+H3O+

given : Lactic acid [C3H6O3]=9.64*10−2M;[H+]=3.65*10−3M; Lactate ion [C3H5O3−]=3.65*10−3M

equilibrium constant ,K=[lactate][H3O+]/[lactic acid]=(3.65*10−3M)(3.65*10−3M)/(9.64*10−2M)=1.382*10^-4 M