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Calculate the ph of a 0.10m solution of hypochlorous acid, hocl. ka of hocl is .3.5x10^-8
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_____Acidity _____HOCl -------------------------> H+___+___ OCl- _init__0.1M________________0M_________… _eq___0.1-xM______________xM_________x… _but____________________+0.1M _so___0.1-x+yM_________0.1+x-yM_____x-… Ka = 3.5E-8M = [H+][OCl-] / [HOCl] without HCl, [H+] = x where 3.5E-8 = x^2/(0.1-x) SOLVE for x (around 6E-5M) after HCl added, 3.5E-8 = (0.1+x-y)(x-y) / (0.1-x+y) SOLVE for y [H+] = 0.1+x-y = ??M NOT much different than just 0.1M HCL !!! (2 parts in 10000) Should have expected as much as HOCl is NOT an acid! This is just how you "work" the problem for things that are acids. SOLVE
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