May 24, 2024, 02:53:58 PM
Forum Rules: Read This Before Posting

Topic: Solubility equilibrium  (Read 9978 times)

0 Members and 1 Guest are viewing this topic.


  • Guest
Solubility equilibrium
« on: January 06, 2005, 08:49:26 PM »
Hi, there. Just a little problem with some senior chemistry in highschool. Only 2 questions.

1.) A solution contains 0.0100 M TlNO3 and 0.0100 M AgNO3.

a) Which compound precipitates first when solid NaI is slowly added to 100. mL of this solution? (Ksp TlI = 8.90 x 10^-8 ; Ksp AgI = 1.50 x 10^-16)

b) What mass of this ion remains unprecipitated when the second compound begins to precipitate (Answer: 1.81 x 10^-1 mg...we have to show our work, don't know what to do)

2.) A 1.00 L solution contains 100. mg of Ba+2 and 10.0 g of Sr+2. Within what range must the concentration of CrO4-2 be in order to precipitate Ba+2 without precipitating any Sr+2?
(Ksp BaCrO4 = 1.20 x 10^-10 ; Ksp SrCrO4 = 3.60 x 10^-5)
(Answer: 1.65 x 10^-7 < [CrO4-2] < 3.16 x 10^-4)

So, our teacher does give us the answers, however, we need to show the work and how we obtain the answer. For some, I'm sure those problems were very easy. The thing is, she hasn't gone into depth like this, and our textbooks do not cover solubility problems like these ones, just the basic ones.

For the first one, I assumed that the nitrate ions just act as spectators and do not get involved in the reaction when the Tl and Ag react with the iodide ion. I then did something like this:
ksp = [Ag+][I-]
1.50 x 10^-16 = (1.00 x 10^-2)[I-]
[I-] = 1.50 x 10^-14

ksp = [Tl+][I-]
8.90 x 10^-8 = (1.00 x 10^-2)[I-]
[I-] = 8.90 x 10^-6
I figured that either of those I values would tell me which precipitates first..but then I kind of lost myself in where I was going. For the second question, well I am kind of lost. I fooled around with a bunch of calculations for a good hour and ended up with a headache...if you can help, great. Thanks


  • Guest
Re:Solubility equilibrium
« Reply #1 on: January 06, 2005, 09:41:17 PM »
1a) The solubility product equation tells you how much soluble ion can exitst for a system. Obviously, the compound that can tolerate the least [I-] in solution will precipitate first.

1b) Knowing the [I-] for the second compound to start precip., that [I-] also exists in common for the first compound. Therefore plug it in for the first compound's equation to solve for it's [ion] and convert to mass.

You may be able to solve Q2, if you understand Q1.

Offline AWK

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 7979
  • Mole Snacks: +555/-93
  • Gender: Male
Re:Solubility equilibrium
« Reply #2 on: January 07, 2005, 01:57:21 AM »
1a. We have the same concentration of both compounds and the same formula of solubility product since both iodides are binary compounds. Hence that compound precipitates first which show lowest solubility product.
1b. Calculate molar solubility of AgI which is equal to molar concentration of Ag(+)  at the moment when TlI start to precipitate ([I-]=8.9x10^-8 / 0.01 = 8.9x10^-6
Then molar solubility of AgI = 1.5x10^-16 / 8.9x10^-6 =1.685x10^-11
Solubility of AgI and  concentration of Ag(+) are equal since AgI is a binary compound hence
mass of Ag(+)
m=1000 x 0.1 x 108 x 1.685x10^-11=1.82x10^-7 mg

Your answer seems to be wrong

1. Calculate molar concentration of Sr(2+) and Ba(2+)
2. From solubility products calculate the concentrations od CrO4(2-) at which corresponding chromates start to precipitate. You can selectively precipitate the first one between both concentrations. At the highest value the second chromate start to precipitate.
« Last Edit: January 07, 2005, 07:30:26 AM by AWK »

Sponsored Links