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

`$\lambda_D =964 \ nm$`

Step-by-step explanation:

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

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

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

Since, pH = -log
`$H^+$`

`$[H^+]=10^(-7)$`

`$\lambda_D = \left(93.05 * 10^_(-14)\right)^(1/2)$`

`$\lambda_D = √(93.05) * 10^(-7)$`

`$\lambda_D =9.64 * 10^(-7)$`

`$\lambda_D =964 * 10^(-9) \ m$`

`$\lambda_D =964 \ nm$`