Respuesta :
Explanation:
The given data is as follows.
[tex]\Lambda^{o}_{m}[/tex](NaCl) = [tex]1.264 \times 10^{-2}[/tex]
[tex]\Lambda^{o}_{m}[/tex](H-O=C-ONO) = [tex]1.046 \times 10^{-2}[/tex]
[tex]\Lambda^{o}_{m}[/tex](HCl) = [tex]4.261 \times 10^{-2}[/tex]
Conductivity of monobasic acid is [tex]5.07 \times 10^{-2} S m^{-1}[/tex]
Concentration = 0.01 [tex]mol/dm^{3}[/tex]
Therefore, molar conductivity ([tex]\Lambda_{m}[/tex]) of monobasic acid is calculated as follows.
[tex]\Lambda_{m} = \frac{conductivity}{concentration}[/tex]
= [tex]\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}[/tex]
= [tex]\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}[/tex]
= [tex]5.07 \times 10^{-3} S m^{2} mol^{-1}[/tex]
Also, [tex]\Lambda^{o}_{m}[/tex] = [tex]\Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}[/tex]
= [tex]4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}[/tex]
= [tex]4.043 \times 10^{-2} S m^{2} mol^{-1}[/tex]
Relation between degree of dissociation and molar conductivity is as follows.
[tex]\alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}[/tex]
= [tex]\frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}[/tex]
= 0.1254
Whereas relation between acid dissociation constant and degree of dissociation is as follows.
K = [tex]\frac{c \times \alpha^{2}}{1 - \alpha}[/tex]
Putting the values into the above formula we get the following.
K = [tex]\frac{c \times \alpha^{2}}{1 - \alpha}[/tex]
= [tex]\frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}[/tex]
= [tex]0.017973 \times 10^{-2}[/tex]
= [tex]1.7973 \times 10^{-4}[/tex]
Hence, the acid dissociation constant is [tex]1.7973 \times 10^{-4}[/tex].
Also, relation between [tex]pK_{a}[/tex] and [tex]K_{a}[/tex] is as follows.
[tex]pK_{a} = -log K_{a}[/tex]
= [tex]-log (1.7973 \times 10^{-4})[/tex]
= 3.7454
Therefore, value of [tex]pK_{a}[/tex] is 3.7454.
