What is the pH at the equivalence point in
the titration of 20.0 mL of 0.200 M HA with
0.200 MNaOH? The Ka for HA is 5.04×10−6.
I know that at equilibrium there should be equal moles of acid and base. I also know that you have to use a "Kb" equation at equilibrium because the hasselhoff equation doesnt work, which is what i wrote down in my notes. Unfortunately, I can't figure out exactly what to do...
At equivalence, all of the acid is converted into base. Get the Kb (Kw/Ka), and the new concentration (moles of base/total volume) and then to the Kb = x2/[A-]. Remember, x in this problem is [OH-] and you'll need to subtract from 14.
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