Sunday
May 19, 2013

# Homework Help: Chemistry

Posted by Alice on Sunday, January 4, 2009 at 2:02pm.

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.

• Chemistry - DrBob222, Sunday, January 4, 2009 at 5:47pm

Check me out on this.
kappa*R = K(cell contant).

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.

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