Chemistry
posted by ASood on .
25.0 ml of a 0.100 M solution of the weak acid, CH3OOH is titrated with a 0.10 M solution of the strong base, NaOH. Calculate the pH of the equivalence point.

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...0.05x............x......x
Kb for Ac^ = (Kw/Ka for HAc) = (x)(x)/(0.05x)
Solve for x, convert to pH.