The titration process is
NaHCrO4 + NaOH --> Na2CrO4 + HOH
This is a hydrolysis problem of the salt, Na2CrO4.
CrO4^-2 + HOH ==> HCrO4^- + OH^-
Kb = (Kw/k2) = (HCrO4^-)(OH^-)/(CrO4^-2)
You know Kw, k2(that's k2 for H2CrO4), and you know the concn of CrO4^-2. Let x = HCrO4^- and x = OH^-.
Solve for x and convert to pH.
Note: concn of the CrO4^- is moles/L solution. Don't forget for volume to add the volume you started with to the volume added to reach the equivalence point.
Calculate the pH at the equivalence point in titrating a 0.120 M solution of NaHCrO4 (sodium hydrogen chromate) with 7.0×10−2 NaOH.
Please help!
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