= 0.003 mol KOH
= 0.002875 mol HClO4
mol KOH = 0.003 mol - 0.002875 mol = 0.000125 mol
mol HClO4 = 0.002875 mol - 0.002875 mol = 0
total volume = 20.0 mL + 23.0 mL = 43.0 mL
= 2.9070x10-3 M OH-
A 20.0-mL sample of 0.150 M KOH is titrated with 0.125 M HClO4 solution. Calculate the pH after the following volumes of acid have been added.
23.0 mL of the acid
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