December 22, 2024, 05:29:40 PM
Forum Rules: Read This Before Posting


Topic: Calculating concentration - common ion  (Read 2474 times)

0 Members and 1 Guest are viewing this topic.

Offline ch3mical

  • Regular Member
  • ***
  • Posts: 9
  • Mole Snacks: +0/-0
Calculating concentration - common ion
« on: August 05, 2021, 04:52:57 PM »
Hi,

I have attempted the question below.

Q) The solubility of barium sulphate after adding 1mL of 10M H2SO4 is 1.6E-8M. The volume of your solution is 1000mL. Calculate the solubility of barium sulphate in a saturated solution before the addition of H2SO4.

My attempt
BaSO4 ::equil:: Ba2+ + SO42-

We are told that [BaSO4] = 1.6E-8M. The concentration of sulphate ions from the H2SO4 is 10M×1mL/1000mL=0.01M

I know that Ksp = [Ba2+][SO42-], so I can say that
Ksp = (x)(0.01 + x)

I don't know how to proceed. Please advise.

Thanks.

Offline Orcio_87

  • Full Member
  • ****
  • Posts: 440
  • Mole Snacks: +39/-3
Re: Calculating concentration - common ion
« Reply #1 on: August 05, 2021, 05:14:44 PM »
1. Aproximate that all SO42- comes from dissociation of H2SO4, not BaSO4.
2. Calculate Ksp.
3. Calculate solubility of the BaSO4.

Offline ch3mical

  • Regular Member
  • ***
  • Posts: 9
  • Mole Snacks: +0/-0
Re: Calculating concentration - common ion
« Reply #2 on: August 05, 2021, 05:29:26 PM »
1. Aproximate that all SO42- comes from dissociation of H2SO4, not BaSO4.
2. Calculate Ksp.
3. Calculate solubility of the BaSO4.

Many thanks for the pointers! I will try to apply it as follows...

1. We have 0.01M of H2SO4, so I will assume the total [SO42-] = 0.01M

2. Ksp = [Ba2+][SO42-] = 1.6E-8 × 0.01 = 1.6E-9

3. I am not sure how to perform this step. Is it [BaSO4] = (1.6E-9)0.5 = 4E-5M ?

My approach in step 3 feels wrong as I haven't taken into account the absence of H2SO4. Could you please clarify how to do this, and why it's correct?

Offline Orcio_87

  • Full Member
  • ****
  • Posts: 440
  • Mole Snacks: +39/-3
Re: Calculating concentration - common ion
« Reply #3 on: August 06, 2021, 02:17:11 AM »
Absence of H2SO4 in don't important. For me it is the correct solution.

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27885
  • Mole Snacks: +1815/-412
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Calculating concentration - common ion
« Reply #4 on: August 06, 2021, 03:11:44 AM »
There is no such thing as

the absence of H2SO4

if the solution is said to be saturated with respect to BaSO4 - that means there is a solid BaSO4 present and the dissolution produces SO42-.

But there is another problem with this question and your solution.

Quote
1. We have 0.01M of H2SO4, so I will assume the total [SO42-] = 0.01M

While this is a common approximation, it is wrong in general. pKa2 for sulfuric acid equals 2, that means is 0.01 M solution (which has pH close to 2 - compare that with pKa2) not all HSO4- is dissociated, and the concentration of SO42- is substantially lower (actually below 0.005 M).

No idea if you are expected to take that into account, but if you ignore you are making an over 100% error in the final answer, hardly negligible.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline mjc123

  • Chemist
  • Sr. Member
  • *
  • Posts: 2074
  • Mole Snacks: +302/-12
Re: Calculating concentration - common ion
« Reply #5 on: August 06, 2021, 06:28:30 AM »
Also 1.6E-8 * 0.01 is not 1.6E-9.

Offline ch3mical

  • Regular Member
  • ***
  • Posts: 9
  • Mole Snacks: +0/-0
Re: Calculating concentration - common ion
« Reply #6 on: August 06, 2021, 12:24:21 PM »
Thank you all.

Sponsored Links