# chemistry

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Calculate the pH at the equivalence point for the titration of 0.25 M CH3COOH with 0.25
-5
M NaOH. (For CH3COOH, Ka= 1.8 â‹… 10 )

• chemistry -

If we let HAc stand for acetic acid (CH3COOH), then the titration of acetic acid and NaOH is
HAc + NaOH ==>NaAc + H2O
At the equivalence point we have exactly neturalized HAc and NaOH and neither is in excess. All we have is the salt, NaAc, in water; therefore the pH is determined by the hydrolysis of the salt.
The (Ac^-) at the equivalence point is 0.125M (that's half the starting M of acetic acid and NaOH).
.......Ac^- + HOH ==> HAc + OH^-
I.....0.125............0.....0
C.......-x.............x.....x
E....0.125-x...........x.....x

Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.125-x)
Solve for x = (OH^-) and convert to pH.

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