Question

Be sure to answer all parts. Calculate the concentrations of H2C204, HC204, C204-, and I' ions in a 0.171 M oxalic acid solution at 25°C. (Ka, = 6.5 x 10-2 and Ka, = 6.1 x 10-5 for oxalic acid.) [H2C2041 = [HC204)= [C20,27= [H"]= M. M * 10 M M (Enter your answer in scientific notation.)

Answer 1

H C,0,- (aq) + H2O*(aq) Concentration of H,C,04 = 0.171 M ICE table constructed is as follows; H,C,0.(aq) + H2O(l) 2 I(M): 0.171 C(M): * E(M) (0.171- x) (HC,0,H0] 1st aci dionization constant, ku = (H.C.O.] 6.5x102- (*) (*) (0.171-x) x=0.0778 M From the equilibrium table, (H20]= (x) M = 0.0778M [HC,0,-]= (x) M = 0.0778 M [H,C,0)= (0.171-x) M = (0.171-0.0778) M = 0.0932 M ICE table constructed for the 2nd ionization of H,C,, is as follows, HC, (aq) + H20(1) 2 0,0,2- (aq) + H, 0* (aq) I(M): 0.0778 C(M): -* +x +x E(M): (0.0778- x) 0.0778+x 0.0778 + 2nd aci dionization constant, ky = [H3O+][0,0,2-] (HC,0;'] 6.1x10- - (0.0778+x) () 0.0778- x=6.1x10 From the equilibrium table, [H20*]= (0.0778+x)M = 0.0778+6.1x10%) M 0.0778M K, 1.0x10-14 -= 1.28x10-13 * 0.0778 pH = -log[H,0* ] = -log(0.0778) = 1.11 [0,0]= (x) M = 6.1x10ʻM HC,07] = (0.0778- x) M = (0.0778-6.1x10-“) M = 0.0778 M