Question

The Debye-Hückel screening length in a 0.00100 molal solution of NaCl in water at 298 K is:

Answer 1

Step-by-step explanation:

Formula to calculate the Debye-Hückel screening length is as follows.

`\lambda = ((\varepsilon  _(r).\varepsilon _(o).K_(b)T)/(2(Z_(i)e)^(2)N_(A)I))^{(1)/(2)}` ......... (1)

where,
`\varepsilon  _(r)` = dielectric constt. of water = 78.5

`\varepsilon  _(o)` = permitivity of space =
`8.85 * 10^(-12) C^(2)/Jm`

T = temperature = 298 K

e = charge on electron =
`9.1 * 10^(-19) C`

`N_(A)` = Avogadro's number =
`6.02 * 10^(23)` ions/mol

I = ionic activity = 0.001 molal

`Z_(i)` = charge on the species = 1 (in case of both
`Na^(+)` and
`Cl^(-)`)

Also, I =
`(1)/(2) \sumC_(i)Z^(2)_(i)`

=
`(1)/(2)[(1)^(2) * 0.001 + (0.001) * (1)^(2)`

= 0.001 molal

Now, putting all the given values into equation (1) as follows.

`\lambda = ((\varepsilon  _(r).\varepsilon _(o).K_(b)T)/(2(Z_(i)e)^(2)N_(A)I))^{(1)/(2)}`

=
`(((78.5) * 8.85 * 10^(-12) * 1.38 * 10^(-23) * 298)/(2(1)^(2)(1.6 * 10^(-19))^(2) * 6.02 * 10^(23) * 0.01))^{(1)/(2)}`

=
`3044.5 * 10^(-10)` m

=
`3044.5 A^(o)`

Thus, we can conclude that Debye-Hückel screening length in given solution is
`3044.5 A^(o)`.