# Chemistry

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A 50.0 mL sample of 0.42 M benzoic acid, C6H5COOH, a weak monoprotic acid, is titrated with 0.40 M NaOH. Calculate the pH at the equivalence point. Ka of C6H5COOH = 6.5 10-5.

I know I have to set up an ICE chart, but I keep getting the wrong answer

• Chemistry -

At the equivalence point you have sodium benzoate, the salt. Hydrolyze the salt and do the ICE.
C6H5COONa + HOH ==> C6H5COOH + OH^-

Kb for the benzoate = (Kw/Ka)
(Kw/Ka) = (C6H5COOH)(OH^-)/(C6H5COONa)
(OH^-)= x = (C655COOH)
Plug in (C6H5COONa) from the titration data and solve for x which = (OH^-), then convert to pH.

• Chemistry -

Do you mind substiuting in numbers? I am still kinda confused...sorry.

• Chemistry -

nvm got it. thanks!

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