Posted by Kaleen on Wednesday, November 23, 2011 at 7:11pm.
........H2CO3 --> H^+ + HCO3^-
initial.0.019....0........0
change...-x.......x.......x
equil.0.019-x.....x........x
k1 = (H^+)(HXO3^-)/(H2CO3).
Substitute into k1 expression and solve for H^+. I obtained approximately 1E-4 but you should do it more accurately.
Then HCO3^- ==> H^+ + CO32^2-
init.1E-4.......0.......0
change.-x.......x.......x
equil...1E04-x...x.......x
I have show the two above as if the H^+ from equation 1 is not the same as in equation 2. Actually, they are the same for both equation; however, I have done it this way to show the different.
k2 = (H^+)(CO3^2-)/(HCO3^2-)
But from equation 1 we see that (H^+) = (HCO3^-), which makes from equation 2 that (C03^2-) = k2 = about 5E-11
So what percentage of the H^+ comes from k2? Very small.
[(H^+)from k2/(H^+)from k1]*100 =% from the second ionization.
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