November 26, 2024, 09:30:12 PM
Forum Rules: Read This Before Posting


Topic: Na2S2O3.5H2O and stoichiometry question  (Read 3132 times)

0 Members and 1 Guest are viewing this topic.

Offline The Guy

  • Regular Member
  • ***
  • Posts: 34
  • Mole Snacks: +1/-0
Na2S2O3.5H2O and stoichiometry question
« on: February 22, 2013, 07:54:53 PM »
I have to make a %15 solution of Na2S2O3, but I only have Na2S2O3.5H2O

I had hard time figure out how to make that solution from the hydrated form :(

Could you please help me on this? and if you know any good refference for such calculation? I need to review these things and build up my skills.

Thank you

Offline ramboacid

  • Full Member
  • ****
  • Posts: 129
  • Mole Snacks: +19/-3
  • USNCO High Honors 2012, 2013
Re: Na2S2O3.5H2O and stoichiometry question
« Reply #1 on: February 22, 2013, 09:15:49 PM »
The ratio of the mass of the sodium thiosulfate to the mass of the water is constant per unit of hydrate. Thus, when you add x grams of hydrate, a proportion p of the mass will be the salt and the proportion of water in the hydrate will be  (1-p). You'll have to take into account the added water from the hydrate because your solution will be based on percent mass.
"Opportunity is missed by most people because it is dressed in overalls and looks like work." - Thomas Edison

Offline Arkcon

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 7367
  • Mole Snacks: +533/-147
Re: Na2S2O3.5H2O and stoichiometry question
« Reply #2 on: February 22, 2013, 09:40:39 PM »
You can consider the hydrate's water an impurity in your solid.   What mass of solid would you use to make the solution if you had the anhydrous form?  You need more, because some of you solid's mass is water.  But you know how much more, because you have the formula, you know how much "false" mass the hydrate solid has.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline The Guy

  • Regular Member
  • ***
  • Posts: 34
  • Mole Snacks: +1/-0
Re: Na2S2O3.5H2O and stoichiometry question
« Reply #3 on: February 23, 2013, 11:07:39 AM »
 :) I really appreciate your comments guys

Smart logic  ;)

Sponsored Links