Question

Calculate the electrical double layer length for pure water at pH 7. Assume temperature is 300K, and give your answer in units of nm.

Answer 1

Answer:

$\lambda_D =964 \ nm$

Explanation:

We know, the double layer length of pure water is given by :

$\lambda _{D}= \left(\frac{\epsilon k_B T}{2e^2z^2N_AC_i}\right)^{1/2}$

$\lambda _{D}= \left(\frac{(78.3)\times(8.85 \times 10^{-21})\times (1.38 \times 10^{-23})\times 300}{2 \times(1.6 \times 10^{19})^2 \times 1^2 \times (6.023 \times 10^{23})\times (10^{-7}) \times 1000 }\right)^{1/2}$

Since, pH = -log $H^+$

          $[H^+]=10^{-7}$

$\lambda_D = \left(93.05 \times 10^_{-14}\right)^{1/2}$

$\lambda_D = \sqrt{93.05} \times 10^{-7}$

$\lambda_D =9.64 \times 10^{-7}$

$\lambda_D =964 \times 10^{-9} \ m$

$\lambda_D =964 \ nm$