A conductivity cell was calibrated using KCl solution (0.01 mol dm^-3) with a conductivity of 0.141 S m^-1 and the resistance was found to be 520 Ohms. The resistance of a saturated solution of silver chloride was found to be 3.8 x 10^5 Ohms. Calculate the solubility product of silver chloride given that the molar ionic conductivities of the silver and chloride ions are 6.2 and 7.6 mS m^2 mol^-1 respectively.
I have worked out the cell constant to be 73.3 m^-1, but I am not sure where to go from there or how to use the ionic conductivities.
Check me out on this.
kappa*R = K(cell contant).
0.141*520 = 73.3 (your value).
Use kappa*R = K to determine kappa for AgCl where R = 3.8 x 10^5.
Then use lambda = molar conductivity = kappa/C.
You have molar conductivity of each of the ions, add them together to obtain the molar conductivity of AgCl, and that allows you to calculate C for for (Ag^+) and of course (Cl^-) which is the same number. Then
(Ag^+)(Cl^-) = Ksp.
I don't know if these data are experimental data or not but the value is close to what I remember Ksp being. I hope this helps.
To calculate the solubility product of silver chloride (AgCl), we need to use the ionic conductivities and resistance measurements of the KCl solution and the saturated AgCl solution. Here's how we can proceed:
1. Calculate the conductance of the KCl solution:
- Conductivity (σ) = 0.141 S m^-1 (given)
- Resistance (R) = 520 Ω (given)
- Cell constant (C) = 73.3 m^-1 (calculated)
The conductance (G) of the KCl solution can be calculated using the formula:
G = σ × C
Substituting the given values, we get:
G(KCl) = 0.141 S m^-1 × 73.3 m^-1 = 10.34 S
2. Calculate the molar conductivities (λ) of Ag+ and Cl- ions:
- Molar ionic conductivity of Ag+ ion (λAg+) = 6.2 mS m^2 mol^-1 (given)
- Molar ionic conductivity of Cl- ion (λCl-) = 7.6 mS m^2 mol^-1 (given)
3. Calculate the resistance (R) of the saturated AgCl solution:
- Resistance of AgCl solution (R(AgCl)) = 3.8 × 10^5 Ω (given)
4. Calculate the conductance (G) of the AgCl solution:
The conductance (G) of the AgCl solution can be calculated using the formula:
G(AgCl) = 1 / R(AgCl)
Substituting the given value, we get:
G(AgCl) = 1 / (3.8 × 10^5 Ω)
5. Calculate the conductance ratio (K) between AgCl and KCl solutions:
The conductance ratio (K) can be calculated using the formula:
K = G(AgCl) / G(KCl)
Substituting the previously calculated values, we get:
K = G(AgCl) / G(KCl)
6. Calculate the molar conductance of AgCl (λAgCl):
The molar conductance (λAgCl) can be calculated using the formula:
λAgCl = K × λCl- / λAg+
Substituting the previously calculated values, we get:
λAgCl = K × λCl- / λAg+
7. Calculate the solubility product of AgCl (Ksp):
The solubility product (Ksp) of AgCl can be calculated using the formula:
Ksp = (λAgCl)^2
Substituting the previously calculated value, we get:
Ksp = (λAgCl)^2
By following these steps, you should be able to calculate the solubility product of silver chloride (AgCl).