Question

Calculate the [C3H2O42-] and the pH of a 0.10 M solution of malonic acid (H2C3H2O4) (Ka1 = 1.5 x 10-3 ; Ka2 = 2.0 x 10-6).

Is there a way to do this without the ICE tables? Please show work. Thanks!

Answer 1

YES you can avoid ICE tables; just learn this way:

H2A <-> HA- + H+

Ka1 = [H+][HA-]/[H2A]

[H+]= [HA-]= x

[H2A] = M-x = 0.1-x

1.5*10^-3 = x*x/(0.1-x)

x = 0.01152

[H+] 1st = 0.01152

BUT also

[HA-] = x = 0.01152

continue as before:

Ka2 = [H+][A-2]/[HA-]

[H+] = [A-2] = y

[HA-] = x-y =  0.01152-y

2*10^-6 = y*y/(0.01152-y)

y = 1.5*10^-4

Total H+ = H1+H2+ = x + y =0.01152+1.5*10^-4 = 0.01167

pH = -log(0.01167)= 1.93

note that

[C3H2O42-] is actually "x" or "HA-"

then x = 0.01152