| man on fire - remember, the original solution has [x] concentration of the C3H5O2- ion in it already. Remember - it's [x]^2 that equals 3.66*10^-6; the [H+] times [C3H5O2] = 3.55*10^-3, so [H+] = [C3H5O2-] = 1.88*10^-3M
Therefore, you need to consider the already existing C3H5O2- ion concentration already in solution; you can't just disregard it from part a. That's why I added 1.88*10^-3 (meant for C3H5O2- concentration, not H+) to the ion added from the NaC3H5O2, because some ion is already in solution before you add the solid. I'm pretty sure I'm right on that, but I'll double check again.
Also, student, for the solid Al(OH)3 and KOH solution reaction, I had:
Al(OH)3 + OH- --> [Al(OH)4]-
KOH is a very strong base (Kb > 1) therefore it is always ionized. So the first ionic equation would be:
Al(OH)3 + K+ + OH- --> [Al(OH)4]- + K+; and the K+ would cancel out on each side, leaving: Al(OH)3 + OH- --> [Al(OH)4]-
Last edited by crypto86 : 05-14-2005 at 10:50 PM.
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