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The limiting molar conductivities of HCl...

The limiting molar conductivities of `HCl, CH_(3) COONa and NaCl` are respectively 425, 90 and 125 mho `cm^(2) mol^(-1)` at `25^(@)C`. The molar conductivity of 0.1 M `CH_(3)COOH` solution is 7.8 mho `cm^(2) mol^(-1)` at the same temperature. The degree of dissociation of 0.1 M acetic acid solution at the same temperature is

A

0.1

B

0.02

C

0.15

D

0.03

Text Solution

Verified by Experts

The correct Answer is:
B
According to Kohlrausch.s law,
`^^.^(@)` for `CH_(3)COOH = lambda_(CH_(3)COO^(-))^(@) + lambda_(H^(+))^(@)`
`^^.^(@)` for `NaCl = lambda_(Na^(+))^(@) + lambda_(Cl^(-))^(@)`
`= "125 mho cm"^(2) mol^(-1)" "...(i)`
`^^.^(@)` for `HCl = lambda_(H^(+))^(@) + lambda_(Cl^(-))^(@)`
`= "425 mho cm"^(2) mol^(-1)" "...(ii)`
`^^.^(@)` for `CH_(3)COONa = lambda_(CH_(3)COO^(-))^(@) + lambda_(Na^(+))^(@)`
`= "90 mho cm"^(2) mol^(-1)" "...(iii)`
Adding Eqs. (ii) and (iii) and subtracting (i), we get
`lambda_(H^(+))^(@) + lambda_(Cl^(-))^(@) + lambda_(CH_(3)COO^(-))^(@) + lambda_(Na^(+))^(@) - lambda_(Na^(+))^(@) - lambda_(Cl^(-))^(@)`
= 425 + 90 - 125
= 390 mho `cm^(2) mol^(-1)`
or `overset(o)(^^) CH_(3)COOH = lambda_(CH_(3)COO^(-))^(@) + lambda_(H^(+))^(@)`
= 390 mho `cm^(2) mol^(-1)`
Thus, the molar conductivity of `CH_(3)COOH` at infinite dilution `overset(o)(^^) = "390 mho cm"^(2) mol^(-1)`
The molar conductivity of 0.1 M `CH_(3)COOH` solution `(^^_(m)^(c))`
= 7.8 mho `cm^(2) mol^(-1)`
Degree of dissociation `(alpha) = (^^_(m)^(c))/(^^_(m)^(o)) = (7.8)/(390) = 0.02`
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