First you must calculate the M of the salt, NaAc (sodium acetatate) at the equivalence point. That will be millimols/mL. mmols NaAc = 25 x 0.1056 = approx 2.6 but you should do it more accurately. Since the M NaOH is the same you know it will take 25 mL of the NaOH; therefore, M NaAc = about 2.6/50 = approx 0.05.
The pH at the equivalence point is determined by the hydrolysis of the NaAc.
......Ac^- + HOH --> HAc + OH^-
I....0.05............0......0
C.....-x.............x......x
E....0.05-x..........x......x
Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.05-x).
Solve for x = (OH^-) and convert t pH.
Determine the pH at the equivalence point when 25.00 mL of 0.1056 M CH3COOH is titrated with 0.1056 M NaOH.
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