Posted by Ashley on Friday, March 2, 2012 at 8:41pm.
a) Aniline is a base.
pH = 9.05; pOH = 4.95; OH^- = 1.12E-5M
......C6H5NH2 + HOH ==> C6H5NH3^+ + OH^-
initial..x................0..........0
change..-1.12E-5......1.12E-5..1.12E-5
equil..x-1.12E-5....1.12E-5...1.12E-5
Kb = (C6H5NH3^+)(OH^-)/(C6H5NH2)
Substitute into the Kb expression, look up Kb for aniline, solve for x = M aniline for pH of 9.05
b) The pH is determined by the hydrolysis of the salt. pH = 5.22; (H^+) = 6.03E-6
........NH4^+ + H2O ==> H3O^+ + NH3
initial...x...............0.......0
change..-6.03E-6.....6.03E-6..6.03E-6
equil..x-6.03E-6......6.03E-6..6.03E-6
Ka for NH4^+ = (Kw/Kb for NH3) = (NH3)(H3O^+)/(NH4^+)
Substitute into the Ka expression from above and solve for x = (NH4+) = (NH4Cl)
initial.
When I do a) I get an answer of 0.298M, however, it says I'm wrong. b) worked perfectly
I obtained 0.299 but you have used 4.2E-10 for Kb which is just two significant figures. I would round the 0.298 you obtained to 0.30 and submit that. I would appreciate letting me know if this takes care of the problem. This post is getting buried so just make a new question to DrBob222 and let me know. Thanks.
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