Posted by alyssa on Saturday, November 19, 2011 at 2:23am.
This is not nearly as easy as it looks. The simple way to do it is to add H^+ from 0.016M HCl, 0.02M H2SO4(first H^+ ONLY) and ignore the H^+ from k2 for H2SO4 and H^+ from HCN (weak acid).
You can go through the math but the second H from H2SO4 contributes only 0.0046 but compared to 0.036 from above probably should be added in. If you want to do that use
k2 = (H^+)(SO4^-)/(HSO4^-). For (H^+) substitute x+0.036 and for (SO4^2-) substitute x and for (HSO4^-) substitute 0.02-x. I get 0.00456M for k2 contribution from H2SO4. For the HCN; that is
HCN ==> H^+ + CN^-
Ka = (H^+)(CN^-)/(HCN) = 2.1E-9
(x+0.04056)/(0.018) = about 3E-6 which can be ignored.
Then pH = -log(H^+).
Post your work if you get stuck.
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