December 27, 2024, 04:38:36 AM
Forum Rules: Read This Before Posting


Topic: Volume for nuetralization  (Read 4727 times)

0 Members and 2 Guests are viewing this topic.

Edher

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

     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,
Edher

Froggirl

  • 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.  :)

Edher

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

        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.

Edher

Offline Donaldson Tan

  • Editor, New Asia Republic
  • Retired Staff
  • Sr. Member
  • *
  • Posts: 3177
  • Mole Snacks: +261/-13
  • Gender: Male
    • New Asia Republic
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.
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

Sponsored Links