Calculate the solubility product of magnesium fluoride MgF2, if the specific electrical conductivity of its saturated solution is 0.02538 S/m, with the specific conductivity of the solvent (water) 5.0*10-5 S/m. Assume that the molar conductivity of magnesium fluoride in saturated solution does not differ from the limiting molar conductivity and that the activity coefficients are equal to 1 (limiting conductivities of Mg2 and F- ions are 0.010610 and 0.00554 S*m2*mol-1, respectively).

If you do not open the file - install the RAR archiver. Inside the archive you will find a solution in Word format.