Physical and analytical chemistry

Click For Summary
SUMMARY

The discussion focuses on creating a saturated solution containing 45g of solid substance X per liter by mixing solutions S1 (5.1g/L), S2 (39g/L), and S3 (57g/L) at 20 degrees Celsius, where the solubility of X is 42g/L. It is established that a mixture of S1 and S2 cannot achieve the desired concentration due to both being below 45g/L. Therefore, a combination of S1 and S3 is necessary, with the calculation involving the variable "x" representing the volume of S3 needed to reach the target concentration.

PREREQUISITES
  • Understanding of solubility concepts and saturation
  • Familiarity with the lever rule in physical chemistry
  • Basic algebra for solving equations
  • Knowledge of concentration calculations in solutions
NEXT STEPS
  • Study the lever rule in detail for solution mixing scenarios
  • Learn about solubility product constants (Ksp) and their applications
  • Explore concentration calculations using dilution equations
  • Investigate methods for increasing solubility, such as temperature adjustments
USEFUL FOR

Chemistry students, educators, and professionals involved in physical and analytical chemistry, particularly those focusing on solution chemistry and solubility dynamics.

rngz
Messages
6
Reaction score
0

Homework Statement


at 20 degree celsius, the solubility of a solid substance x in water is 42g/L. There are 3 solutions containing x kept at this temperature: S1 contains 5.1g/L,S2 contains 39g/L, and S3 contains 57g/L.so, how to make a saturated solution containing 45g x per litre overall for every 1L of S1. how much of S2 or S3 should I add??what is the critical assumption that must be taken??


Homework Equations


[n1L1=n2L2]


The Attempt at a Solution


by using the lever rules..
 
Physics news on Phys.org
Do you think that taking any mixture of S1 (5.1 g/L) and S2 (39 g/L), both with lower than 45 g/L, will ever add up to 45 g/L without doing something like boiling off or removing some of the liquid?

If not, then that means it must be some mixture of S1 and S3. How much S3? We don't know, so call it "x". The problem says you have 1 Liter of S1. What if I told you there were x = 5 Liters of S3 (by the way, that's not the correct answer), how would you go about calculating the final concentration? Try that problem first. Use dimensions to help guide you through it.

Now, do the same calculation except set the final concentration to 45 and replace 5 Liters with "x" and do the algebra to find "x".
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
4K
Replies
9
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
4
Views
4K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 5 ·
Replies
5
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K