Problem: 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

FREE Expert Solution

Step 1

20.0 mL ×10-3 L1 mL×0.150 mol KOH1 L = 0.003 mol KOH

23.0 mL ×10-3 L1 mL×0.125 mol HClO41 L = 0.002875 mol HClO4

Step 2

mol KOH = 0.003 mol - 0.002875 mol = 0.000125 mol

mol HClO4 = 0.002875 mol - 0.002875 mol = 0


Step 3


total volume = 20.0 mL + 23.0 mL = 43.0 mL

[OH-] = 0.000125 mol KOH ×1 mol OH-1 mol KOH43.0 mL ×103 L1 mL = 2.9070x10-3 M OH-


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Problem Details

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