December 26, 2024, 07:19:19 PM
Forum Rules: Read This Before Posting


Topic: Question on solubility equilibrium  (Read 2539 times)

0 Members and 1 Guest are viewing this topic.

Offline daijia

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
Question on solubility equilibrium
« on: December 15, 2019, 01:49:03 AM »
Hi everyone,

I have encountered such question on my exercice,

How many moles of Na2SO4 we have to add into a saturated 0.5 L solution of Ag2SO4 so that the concentration of Ag is 4.0 * 10^-3 mol/L?

So what I did is following:
  • According to the source I am usingz the Ksp for Ag2SO4 is 1.4×10^-5. So Ksp = (Ag)^2*(SO4), 4s^3 = 1.4×10^-5, s= 0.015.
    As a result, we have [SO4]=0.015 mol/L in a normal saturated solution of Ag2SO4
  • Then, I found what the concentration of SO4 suppose to be if [Ag] is 4.0 * 10 ^-3 mol/L in a saturated Ag2SO4 solution.
    1.4*10^-5 = (4.0*10^-3)^2 × (SO4), SO4 = 0.875.  So we need 0.875 mol/L of SO4 in order to have a saturated solution with 4.0*10^-3 of Ag
  • SO4 needed: 0.875 - 0.015 = 0.86 mol/L. So we need to add 0.86 mol/L of SO4
  • nSO4 = 0.86 mol/L × 0.5 L = 0.43 mol
    0.43 mol of SO4 is the same as the mole of Na2SO4.
    So we need to add 0.43 mol of Na2SO4
So the full answer is: We need to add 0.43 mol of Na2SO4 into a saturated solution of Ag2SO4 in order to bring down the concentration do Ag to 4.0*10^-3 mol/L.

So I would like to ask if my step is wrong? If yes, what are the right steps!

Thanks,
Jack
« Last Edit: December 15, 2019, 02:43:16 AM by daijia »

Offline AWK

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 7976
  • Mole Snacks: +555/-93
  • Gender: Male
Re: Question on solubility equilibrium
« Reply #1 on: December 15, 2019, 02:12:27 AM »
Students make a lot of calculation errors when calculating chemical equilibria - it's hard to guess from the description that you are doing everything right.
AWK

Offline daijia

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
Re: Question on solubility equilibrium
« Reply #2 on: December 15, 2019, 02:15:17 AM »
Students make a lot of calculation errors when calculating chemical equilibria - it's hard to guess from the description that you are doing everything right.

Sure yes I understand. Suppose I don't make calculation error, are my steps logic (is my method appropriate to solve this question)? Or I need to take a different approach to solve this question?

At mean time, I gonna modify my post to include my calculations, thanks for reminding me that!

Thanks!

Offline AWK

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 7976
  • Mole Snacks: +555/-93
  • Gender: Male
Re: Question on solubility equilibrium
« Reply #3 on: December 15, 2019, 02:20:46 AM »
This is a typical manual description, but I don't know if you understand it properly.
AWK

Offline daijia

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
Re: Question on solubility equilibrium
« Reply #4 on: December 15, 2019, 02:30:39 AM »
This is a typical manual description, but I don't know if you understand it properly.
Hi,

I have just updated my post with specific calculations. Would you mind taking a look?

Thanks!

Offline AWK

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 7976
  • Mole Snacks: +555/-93
  • Gender: Male
Re: Question on solubility equilibrium
« Reply #5 on: December 15, 2019, 02:35:08 AM »
OK (In the meantime you corrected a calculation error)
AWK

Offline daijia

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
Re: Question on solubility equilibrium
« Reply #6 on: December 15, 2019, 02:39:15 AM »
OK (In the meantime you corrected a calculation error)

Yes that was more of a typo😅Because I was typing on the phone, and I feel the answer doesn't seem the same as the one I got yesterday. That is when I realized I typed an extra zero

But thanks for helping me!


Offline AWK

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 7976
  • Mole Snacks: +555/-93
  • Gender: Male
Re: Question on solubility equilibrium
« Reply #7 on: December 15, 2019, 03:45:13 AM »
These calculations are based solely on the common ion effect. In fact, the effect of ionic strength is opposite and much greater than the effect of a common ion and the addition of sodium sulfate will increase the solubility of silver sulfate. Therefore the assumed reduced concentration of silver cation will probably never be reached.
AWK

Offline daijia

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
Re: Question on solubility equilibrium
« Reply #8 on: December 15, 2019, 01:19:06 PM »
Oh, okay. Thank you for your explanation!

But considering the fact that this is a high school question and the purpose is merely to practice students' knowledge on solubility, are my calculations justified in this case?

Thanks a lot!

Offline AWK

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 7976
  • Mole Snacks: +555/-93
  • Gender: Male
Re: Question on solubility equilibrium
« Reply #9 on: December 15, 2019, 03:06:57 PM »
In the last post, I noticed that the example is terribly chosen. You can find many chemical compounds for which this type of calculation can be made with an error of 10-30%. In this case, the error is several hundred %.

I wrote earlier: OK
AWK

Sponsored Links