If 16.0 mL of 8.5 ✕ 10-2 M HClO4 is added to 22.0 mL of 7.6 ✕ 10-4 M KOH, what is the pH of the solution?
mols HClO4 = M x L = ?
mols KOH = M x L = ?
Subtract. (concn excess acid) or (concn excss base) divided by volume in L = new M. Then pH from that. Post your work if you get stuck.