Posted by ASood on Sunday, February 26, 2012 at 10:44am.
If 25.0 mL of 0.100 M soln is titrated with 0.10 M base, you know it will take 25.0 mL. So the salt at the equivalence point will be 0.05M (That's 0.025L x 0.1M to begin, you've added 0.025L so the concn is 0.025 x 0.1/0.050 = 0.05M) The pH at the equivalence point is determined by the hydrolysis of the salt. The salt is CH3COONa which I will call NaAc. It's the Ac^- that hydrolyzes.
..........Ac^- + HOH ==> HAc + OH^-
initial...0.05............0.....0
change....-x..............x......x
equil...0.05-x............x......x
Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.05-x)
Solve for x, convert to pH.
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