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- To get the pH: Do a RICE problem with the initial [ ] of acid
point 1
- Before the titration begins
point 1
- Before the equivalence point
Region A
- The middle part of this region doesn’t change much because it is a buffer
region A
- To get the pH: determine the moles of acid remaining and the moles of conjugate base produced, divide both by the new total volume, plug into the Henderson-Hasselbach equation to get the pH.
region A
- [HA] > [A-] in this region before the midpoint
region A
[A-] > [HA] after midpoint
region A
- Moles of base added equals half the moles of acid you started with
point 2
- [HA] = [A-]
point 2
- Volume of titrant added equals half the volume of titrant added at equivalence point
point 2
- Half-equivalence point
point 2
pH = pKa
point 2
midpoint
point 2
can be used to determine identity of unknown acid based on Ka
point 2
indicator should have pKa equal to pH at this point
point 3
- To get the pH: determine the moles of conjugate base produced, divide by the new total volume, use a RICE problem of the conjugate base with water to determine the pH
point 3
- Moles of acid you started with equals the moles of base added
point 3
- The indicator should (if chosen correctly) change colour here
point 3
- Equivalence point
point 3
- Can be seen on a pH curve by picking a point halfway up the rapid-rise region
point 3
- Can be used to calculate the molarity or weight % of the analyte
point 3
- To get the pH: determine the moles of base/titrant remaining, divide by the new total volume, and take the –log[base] and subtract from 14
region B
[A-] can be ignored, but is present
region B
- Excess base has been added beyond the point at which all the acid has been neutralized
region B
beyond the equivalence point
region B
[HA] = 0 (pretty much)
region B