Chemistry problem dealing with solubility

Click For Summary
SUMMARY

The discussion focuses on a chemistry problem involving the precipitation of compounds when hydrochloric acid (HCl) is added to a solution containing cations such as Ag, Al, As, Ba, Bi, Ca, Cd, Co, Cr, Cu, and Fe. Participants confirm that using solubility rules is the correct approach to identify precipitated compounds, specifically noting that AgCl is insoluble and will precipitate. For the second part of the problem, the optimal reagent is hydrogen sulfide (H2S) combined with HCl, which will precipitate the fewest remaining cations by maximizing the number of dissolved compounds.

PREREQUISITES
  • Understanding of solubility rules in chemistry
  • Familiarity with common cations and their reactions with hydrochloric acid
  • Knowledge of precipitation reactions and their significance
  • Basic concepts of reagents in chemical reactions
NEXT STEPS
  • Study the solubility rules for ionic compounds in aqueous solutions
  • Learn about the properties and reactions of hydrogen sulfide in precipitation reactions
  • Explore the concept of selective precipitation and its applications in analytical chemistry
  • Investigate the behavior of specific cations in acidic environments
USEFUL FOR

Chemistry students, particularly those enrolled in introductory chemistry courses, educators teaching chemical reactions, and anyone interested in understanding precipitation reactions and solubility principles.

ACLerok
Messages
194
Reaction score
0
I'm taking Chem II over the summer at a junior college and was given this problem on one of my homework assignments:

I am told I have a sol'n containing the cations Ag, Al, As, Ba, Bi, Ca, Cd, Co, Cr, Cu, and Fe.

My objective is to write all the compounds that would be precipitated if hydrochloric acid is added to the above sol'n.

My instincts are telling me to just use the solubility rules. For example, when Ag and hydrochloric acid combine, AgCl is formed but according to the solubility rules AgCl is insoluble so AgCl would be one of the compounds that would be precipitated. Am I going about this all wrong or what?

The 2nd part says that a reagent is now needed tha will precipitate the smalest number of remaining cations and that the best reagent for this is hydrogen sulfide plus hydrochloric acid. It then asks me to find all the compounds that would be precipitated.

Do I just go about this the same way for the first part? I don't understand when they say 'smallest number of cations'. Please can anyone help me with this?
 
Physics news on Phys.org
Yea you are doing it right. If a compound is precipitated than it did not dissolve after combining with the HCl. All you have to do for the first part is find all the compounds that won't dissolve afterwards.

Same thing goes for the second part. The smallest number of cations simply means that a reagant is necessary so that the maximum number of compounds dissolve. In other words, this reagant will make as many compounds dissolve as possible.
 



Your instincts are correct in using the solubility rules to determine which compounds would be precipitated when hydrochloric acid is added to the sol'n. However, it is important to remember that not all compounds will necessarily precipitate. Some may remain in solution, depending on their solubility in the acidic environment.

For the second part, when it says "smallest number of remaining cations," it means that you should choose a reagent that will precipitate the fewest number of cations possible. In this case, hydrogen sulfide plus hydrochloric acid is the best choice because it will precipitate the most cations at once.

To find all the compounds that would be precipitated, you can follow the same method as in the first part. Use the solubility rules to determine which compounds are insoluble in the acidic environment and therefore will be precipitated. Remember to also consider the products of the reaction between the cations and the reagent (in this case, hydrogen sulfide).

I hope this helps and good luck with your Chem II studies!
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
10K
  • · Replies 3 ·
Replies
3
Views
9K
Replies
7
Views
6K
  • · Replies 9 ·
Replies
9
Views
6K
Replies
1
Views
5K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 2 ·
Replies
2
Views
26K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
8K
  • · Replies 19 ·
Replies
19
Views
11K