Chem: Ksp & deltaG for Ag2CO3 at 20C and 100C

  • Thread starter Thread starter Doobwa
  • Start date Start date
  • Tags Tags
    Chem Ksp
Click For Summary
SUMMARY

The discussion focuses on calculating the solubility product constant (Ksp) and Gibbs free energy change (ΔG) for silver carbonate (Ag2CO3) at 20°C and 100°C. The participant initially calculated Ksp at 20°C as 6.25e-12, while the correct value is 2.77e-11. The solubility of Ag2CO3 was determined using the formula Solubility = (mass)/(volume)(molar mass), with the molar mass calculated as 275.75 g/mol. The participant's error lies in the calculation of the concentration of ions, which directly affects the Ksp value.

PREREQUISITES
  • Understanding of solubility product constant (Ksp) calculations
  • Knowledge of Gibbs free energy and its relation to Ksp
  • Familiarity with molar mass calculations
  • Basic principles of chemical equilibrium
NEXT STEPS
  • Learn how to calculate Ksp for different salts at varying temperatures
  • Study the relationship between Ksp and ΔG using the equation ΔG = -RTlnKsp
  • Explore the impact of temperature on solubility and Ksp values
  • Review examples of solubility product calculations from reliable chemistry resources
USEFUL FOR

Chemistry students, educators, and anyone involved in chemical equilibrium studies will benefit from this discussion, particularly those focusing on solubility and thermodynamic properties of compounds.

Doobwa
Messages
3
Reaction score
0
Hello! I've just completed this homework problem but got ALL answers incorrect. I really thought I knew what I was doing so I'm kind of surprised. Could someone help point out what I had done wrong?

Homework Statement


Ag2CO3, is considerably more soluble in hot than in cold water. At 20 °C, 0.0032 g of this compound dissolves in 0.100 L of water, but 0.050 g dissolves at 100.0 °C. Calculate Ksp and G° in kJ/mol at each temperature.

Homework Equations


Ksp=[Ag^{+}]^{2}[CO3^{-}]
deltaG=-RTlnKsp.
Solubility=(mass)/(volume)(molar mass)

The Attempt at a Solution


I think I went wrong with figuring out Ksp. I believe if I knew what I had done wrong here, I could easily solve the rest, so I'll just show my work for this portion.

At 20C (293.15K) 0.0032g of Ag2CO3 dissolves in 0.100L of water. Molar mass of Ag2CO3 found to be 275.75g/mol.

Solubility of Ag2CO3= (0.0032g)/(0.100L)(275.75g/mol)=1.160471442e-4M
Ag2CO3\searrow2Ag^{+}+CO3^{-}
[Ag^{+}]=2.320942884e-4M
[CO3^{-}]=1.160471442e-4M

Ksp=[Ag^{+}]^{2}[CO3^{-}]
=(2.320942884e-4)^{2}(1.160471442e-4)
=6.25e-12

Actual answer: 2.77e-11

I really appreciate any guidance you can give me!
 
Physics news on Phys.org
I don't see anything wrong with your working. For comparison I referred to a very similar exercise here: http://www.chem.purdue.edu/gchelp/howtosolveit/equilibrium/solubility_products.htm"
 
Last edited by a moderator:

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 37 ·
2
Replies
37
Views
7K
Replies
7
Views
3K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
3K
Replies
4
Views
12K
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
6K
Replies
6
Views
4K