Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) csci electrolyte solution at 30 °c. how are these numbers affected by the addition of 0.1 mol/dm3 of kcl? at what distance from the particle surface (r) has the potential decayed to 1% of its initial value?

Explanation:

The given data is as follows.

    Concentration = 0.1

                             = 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

                             =

               T = = (30 + 273) K = 303 K

Formula for electric double layer thickness () is as follows.

            =

where, = concentration =

Hence, putting the given values into the above equation as follows.

                  =                    

                          =  

                         = m

or,                     =

                          = 1 nm (approx)

Also, it is known that =

Hence, we can conclude that addition of 0.1 of KCl in 0.1 of NaBr "" will decrease but not significantly.

pv=nrt

explanation: