Calculate the concentrations of all species in a 0.270-M solution of H2X if K1 = 6.31E-06 and K2 = 1.74E-08
[H2X] =
[H3O1+] =
[HX1-] =
[X2-] =
H2O + H2X <====> H3O+ + HX- Ka1 = 6.31 x 10-06
Intial Con 0.270 10-7 0
Change Con -x +x +x
Eq Con 0.27-x 10-7+x +x
Ka1 = (10-7+x)(x) / ( 0.27-x ) => 6.31 x 10-06 ( x >> 10-7 or simply x = [ Ka1*C]1/2 )
x = 1.31 x 10-3 M
x = [H3O+] = [ HX-] = 1.31 x 10-3 M
[H2X ] = 0.270 - 0.00131 => 0.2687 M
H2O + HX- <====> H3O+ + X2- Ka2 = 1.74 x 10-08
Intial Con 1.31 x 10-3 1.31 x 10-3 0
Change Con -x +x +x
Eq Con 1.31 x 10-3 -x 1.31 x 10-3 +x +x
Ka2 = (1.31 x 10-3 +x)(x) / ( 1.31 x 10-3 -x ) => 1.74 x 10-08
x = 1.74 x 10-08 M
x = [ X2-] = 1.74 x 10-08 M
( Note in diprotic acid [ X2-] =Ka2 )
[H2X] = 0.2687 M
[H3O1+] = 1.31 x 10-3 M
[HX1-] = 1.31 x 10-3 M
[X2-] = 1.74 x 10-08 M
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