# Chemistry

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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?

• Chemistry - ,

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 = ??.

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