chemistry
posted by alyssa on .
Calculate the pH of an aqueous solution containing 1.6 x 102 M HCl, 2.0 x 102 M H2SO4, and 1.8 x 102 M HCN.

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 0.02x. I get 0.00456M for k2 contribution from H2SO4. For the HCN; that is
HCN ==> H^+ + CN^
Ka = (H^+)(CN^)/(HCN) = 2.1E9
(x+0.04056)/(0.018) = about 3E6 which can be ignored.
Then pH = log(H^+).
Post your work if you get stuck.