Respuesta :
Explanation:
The given data is as follows.
Concentration = 0.1 [tex]mol/dm^{3}[/tex]
= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions
= [tex]6.022 \times 10^{25} ions/m^{3}[/tex]
T = [tex]30^{o}C[/tex] = (30 + 273) K = 303 K
Formula for electric double layer thickness ([tex]\lambda_{D}[/tex]) is as follows.
[tex]\lambda_{D}[/tex] = [tex]\frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}[/tex]
where, [tex]n^{o}[/tex] = concentration = [tex]6.022 \times 10^{25} ions/m^{3}[/tex]
Hence, putting the given values into the above equation as follows.
[tex]\lambda_{D}[/tex] = [tex]\sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}[/tex]
= [tex]\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}}[/tex]
= [tex]9.669 \times 10^{-10}[/tex] m
or, = [tex]9.7 A^{o}[/tex]
= 1 nm (approx)
Also, it is known that [tex]\lambda_{D}[/tex] = [tex]\sqrt \frac{1}{n^{o}}[/tex]
Hence, we can conclude that addition of 0.1 [tex]mol/dm^{3}[/tex] of KCl in 0.1 [tex]mol/dm^{3}[/tex] of NaBr "[tex]\lambda_{D}[/tex]" will decrease but not significantly.
