August 10, 2020, 06:32:00 AM
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Topic: Volume for nuetralization  (Read 3648 times)

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  • Guest
Volume for nuetralization
« on: March 09, 2005, 05:36:15 PM »

     What volume of 0.146 M KOH is needed for the complete neutralization of 20.00 mL of 0.0942 M H2SO4?

This is how I solved it,

20 mL x 0.0942 M H2SO4 = 1.884 mmol H2SO4

XmL x 0.146 mmol = 1.884 mmol KOH
                 1 mL

I got X = 12.9 mL. However, the book says it should be double that 25.8 mL. Why is that?

Thank You,


  • Guest
Re:Volume for nuetralization
« Reply #1 on: March 09, 2005, 06:03:23 PM »
You are forgetting that H2SO4 is a diprotic acid (ie gives up 2 moles of H+ for every mole of H2SO4)!!
Your calculations took into account only half of the protons given up by H2SO4, therefore the answer was double what you calculated.  :)


  • Guest
Re:Volume for nuetralization
« Reply #2 on: March 09, 2005, 06:19:31 PM »

        I had a feeling that it had something to do with diprotic acids but since I missed that lecture I wasn't quite certain on how to apply its properties. Thank you.


Offline Donaldson Tan

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Re:Volume for nuetralization
« Reply #3 on: March 10, 2005, 09:38:58 AM »
a monoprotic acid gives out 1 proton per molecule
a diprotic acid gives 2 protons per molecule

hence, the same concentration if a strong diprotic acid gives u twice as much H+ of a strong monoprotic acid.
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