Calculate the concentrations of all species in a 0.320-M solution of H2X if K1 =

Question

Calculate the concentrations of all species in a 0.320-M solution of H2X if K1 = 7.94E-06 and K2 = 3.71E-09 [H2X] = M [H3O1+] = M [HX1-] = M [X2-] = M

Explanation / Answer

H2X ------> HX- + H+ , at equi [H2X] = 0.32-x , [HX-] = x =[H+] , K= 7.94 x10^ 6 = x^2/(0.32-x) , x= 0.00159= [HX-] , HX- ----> H+ + X 2- , at equi [HX-] = 0.00159-x , [H+]=[X2-] = x. K2 = 3.71 x10^ -9 = x^2/(0.00159-x) , x = 2.427 x10^ -6 , hence at equi [H3O+] = (0.00159+2.427 x10^-6) = 0.001592427, [H2X] = 0.32-0.00159 = 0.3184, [MX-] = 0.00159-2.427 x10^ -6 = 0.0015876, [X2-] = 2.427 x10^ -6

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