Asked by Rosemary
when a 25.0 mL sample of an unknown acid was titrated with a 0.100 M NaOH solution. Determine Ka for the unknown acid. Volume NaOH = 25 mL; pH=8.25
Answers
Answered by
DrBob222
If you have a weak acid, HA, and titrate with NaOH, the equation is
HA + NaOH ==> NaA + H2O.
There is only NaA and H2O at the equivalence point. The A^- of NaA is hydrolyzed (it is a base) as follows:
A^- + HOH ==> HA + OH^-
Kb = (Kw/Ka) = (HA)(OH^-)/(A^-).
You know Kw, solve for Ka, convert pH 8.25 to pOH then to OH^-, (HA) = (OH^-) so there is only one unknown. Substitute and solve for Ka. Post your work if you get stuck.
HA + NaOH ==> NaA + H2O.
There is only NaA and H2O at the equivalence point. The A^- of NaA is hydrolyzed (it is a base) as follows:
A^- + HOH ==> HA + OH^-
Kb = (Kw/Ka) = (HA)(OH^-)/(A^-).
You know Kw, solve for Ka, convert pH 8.25 to pOH then to OH^-, (HA) = (OH^-) so there is only one unknown. Substitute and solve for Ka. Post your work if you get stuck.
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