Cracking the SAT Chemistry Subject Test

(Marvins-Underground-K-12) #1

The half-equivalence point is an important location on the titration curve
because at this point, equal amounts of the unknown acid or base and its
conjugates exist in the solution. That means the solution is a buffer, so the
Henderson-Hasselbalch equation applies.


For titration   of  weak    acid    with    strong  base:

pH  =   pKa +   log 

For titration   of  weak    base    with    strong  acid:

pOH =   pKb +   log 

Furthermore, since [A−] = [HA] at the half-equivalence point, the Henderson-
Hasselbalch equation simplifies to the following:


For titration   of  weak    acid    with    strong  base:

pH(at   the half-eq point)  =   pKa

For titration   of  weak    base    with    strong  acid:

pOH(at  the half-eq point)  =   pKb

Therefore, the solution’s pH at the half-equivalence point is actually the pKa of


the unknown weak acid. For a weak base, the pKb can be quickly calculated as


follows:

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