Question

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

Answer 1

Explanation:

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

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

where,     $\varepsilon  _{r}$ = dielectric constt. of water = 78.5

               $\varepsilon  _{o}$ = permitivity of space = $8.85 \times 10^{-12} C^{2}/Jm$

                T = temperature = 298 K

  e = charge on electron = $9.1 \times 10^{-19} C$

 $N_{A}$ = Avogadro's number = $6.02 \times 10^{23}$ ions/mol

       I = ionic activity = 0.001 molal

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

Also,     I = $\frac{1}{2} \sumC_{i}Z^{2}_{i}$

              = $\frac{1}{2}[(1)^{2} \times 0.001 + (0.001) \times (1)^{2}$

              = 0.001 molal

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

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

                 = $(\frac{(78.5) \times 8.85 \times 10^{-12} \times 1.38 \times 10^{-23} \times 298}{2(1)^{2}(1.6 \times 10^{-19})^{2} \times 6.02 \times 10^{23} \times 0.01})^{\frac{1}{2}}$    

                 = $3044.5 \times 10^{-10}$ m

                 = $3044.5 A^{o}$

Thus, we can conclude that Debye-Hückel screening length in given solution is $3044.5 A^{o}$.