To calculate the limiting molar conductivity (Λo) of AgCl solution, we can use the concept of conductivity at infinite dilution and the principle of additivity of ionic conductivities. This principle states that the conductivity of a solution can be expressed as the sum of the conductivities of its individual ions. In this case, we need to consider the ions produced when AgCl dissociates in solution.
Understanding Ionic Dissociation
When silver chloride (AgCl) dissolves in water, it dissociates into silver ions (Ag+) and chloride ions (Cl-). The dissociation can be represented as follows:
- AgCl (s) → Ag+ (aq) + Cl- (aq)
Using Known Conductivities
To find the limiting molar conductivity of AgCl, we need the limiting molar conductivities of the ions involved. We already have the following values:
- Λo(Ag+) = 133.4 S cm2 mol-1
- Λo(Cl-) = ? (We need to calculate this using KCl and KNO3)
Calculating the Conductivity of Cl-
To find the limiting molar conductivity of the chloride ion (Cl-), we can use the known conductivity of KCl, which fully dissociates into K+ and Cl- ions:
- Λo(KCl) = Λo(K+) + Λo(Cl-)
We know that:
- Λo(KCl) = 149.9 S cm2 mol-1
- Λo(K+) = 75.0 S cm2 mol-1 (this is a known value)
Now we can rearrange the equation to solve for Λo(Cl-):
- Λo(Cl-) = Λo(KCl) - Λo(K+)
- Λo(Cl-) = 149.9 - 75.0 = 74.9 S cm2 mol-1
Calculating Limiting Molar Conductivity of AgCl
Now that we have the limiting molar conductivities of both ions, we can calculate the limiting molar conductivity of AgCl:
- Λo(AgCl) = Λo(Ag+) + Λo(Cl-)
- Λo(AgCl) = 133.4 + 74.9 = 208.3 S cm2 mol-1
Final Result
Thus, the limiting molar conductivity of AgCl solution is 208.3 S cm2 mol-1. This value reflects the combined conductivity contributions of the silver and chloride ions when they are at infinite dilution, providing insight into the behavior of AgCl in solution.
