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

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Our tutors have indicated that to solve this problem you will need to apply the Strong Acid Strong Base Titrations concept. You can view video lessons to learn Strong Acid Strong Base Titrations. Or if you need more Strong Acid Strong Base Titrations practice, you can also practice Strong Acid Strong Base Titrations practice problems.

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