Question

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?

Answer 1

Explanation:

The given data is as follows.

    Concentration = 0.1 $mol/dm^{3}$

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

                             = $6.022 \times 10^{25} ions/m^{3}$

               T = $30^{o}C$ = (30 + 273) K = 303 K

Formula for electric double layer thickness ($\lambda_{D}$) is as follows.

            $\lambda_{D}$ = $\frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

where, $n^{o}$ = concentration = $6.022 \times 10^{25} ions/m^{3}$

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

                 $\lambda_{D}$ = $\sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$                    

                          = $\sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}$  

                         = $9.669 \times 10^{-10}$ m

or,                     = $9.7 A^{o}$

                          = 1 nm (approx)

Also, it is known that $\lambda_{D}$ = $\sqrt \frac{1}{n^{o}}$

Hence, we can conclude that addition of 0.1 $mol/dm^{3}$ of KCl in 0.1 $mol/dm^{3}$ of NaBr "$\lambda_{D}$" will decrease but not significantly.