posted by curtis .
Consider the titration of 100.0 mL of 0.110 M NaOH with 1.00 M HBr. Find the pH at the following volumes of acid added and make a graph of pH versus Va:Va = 0,2,5,8,10,10.1,11,11.1, and 12mL. Compute Ve, first.
Find mols NaOH, subtract mols HBr.
If mols NaOH are greater then pOH = -log(OH^-) and convert to pH by
pH + pOH = pKw = 14.
If mols HBr are in excess, pH = -log(H^+).
Those three steps will do all of the points except for the equivalence point. That one is done by the hydrolysis of the salt, NaBr, which is 7 at the equivalence point.