To determine the volume of water required to dissolve 0.01 moles of silver chloride (AgCl) at 25 degrees Celsius, we can use the solubility product constant (Ksp) and the relationship between solubility and Ksp.
Understanding Ksp
- The Ksp of AgCl is given as 2.56 × 10^-10.
- For the dissolution of AgCl, the equilibrium can be represented as:
AgCl (s) ⇌ Ag+ (aq) + Cl− (aq)
- The expression for Ksp is:
Ksp = [Ag+][Cl−]
Calculating Solubility
- Since AgCl dissociates into one Ag+ and one Cl− ion, if 's' is the solubility of AgCl in moles per liter (mol/L), then:
- [Ag+] = s
- [Cl−] = s
- Therefore, Ksp = s², leading to:
s² = 2.56 × 10^-10
- Solving for 's', we find:
s = √(2.56 × 10^-10) = 1.6 × 10^-5 mol/L
Volume of Water Required
- To dissolve 0.01 moles of AgCl, we need to find the volume of water (V) required using the solubility (s):
- Since s = 1.6 × 10^-5 mol/L, the volume can be calculated as:
V = moles of solute / solubility = 0.01 moles / 1.6 × 10^-5 mol/L
- This results in:
V ≈ 625 L
Conclusion
- It would require approximately 625 liters of water to dissolve 0.01 moles of AgCl at 25 degrees Celsius. This highlights the low solubility of AgCl in water, necessitating a large volume for dissolution.