posted by George .
A 50.0 mL sample of 0.23 M propanoic acid, CH3CH2COOH, a weak monoprotic acid, is titrated with 0.14 M KOH. Ka of CH3CH2COOH = 1.4 multiplied by 10-5.
What is the pH at equivalence point?
The pH will be determined by the hydrolysis of the salt, sodium propionate.
If we abbreviate propanoic acid as HPr, then the titration is
HPr + NaOH ==> NaPr + H2O
The salt is NaPr and it's the Pr^- that is hydrolyzed.
Pr^- + HOH ==> HPr + OH^-
Set up an ICE chart and substitute into the below.
Kb = (Kw/Ka) = (HPr)(OH^-)/(Pr^-)
Kw = 1E-14
Ka = the value for HPr
(HPr) = (OH^-) = x
(Pr^-) = (moles/L pr millimoles/mL) which you will need to calculate.
You had 50 mL x 0.23 M HPr = ?millimoles and that will require mmoles/0.14 = about 80 mL NaOH to reach the equivalence point (you need to do the 80 more accurately). Then total volume is 80 or so + 50 = ??.