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Limiting molar conductivity of NH(4)OH [...

Limiting molar conductivity of `NH_(4)OH` [i.e., `Lambda_(m)^(@)(NH_(4)OH)`] is equal to:

A

`Lambda_(m (NH_(4) Cl))^(@) + Lambda_(m (NaCl))^(@) - Lambda_(m (NaOH))^(@)`

B

`Lambda_(m(NaOH))^(@) + Lambda_(m (NaCl))^(@) - Lambda_(m (NH_(4)Cl))^(@)`

C

`Lambda_(m (NH_(4)OH))^(@) + Lambda_(m (NH_(4) Cl))^(@) - Lambda_(m (HCl))^(@)`

D

`Lambda_(m (NH_(4)Cl))^(@) + Lambda_(m (NaOH))^(@) - Lambda_(m(NaCl))^(@)`

Text Solution

Verified by Experts

The correct Answer is:
D
`Lambda_(m)^(@) (NH_(4) OH) = Lambda_(m)^(@) (NH_(4) Cl) + Lambda_(m)^(@) (NaOH) - Lambda_(m)^(@) (NaCl)`
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Calculating The Degree of ionization: The molar conductance of an infinitely dilute solution NH_(4)Cl is 150 S cm^(2) mol^(-1) and the limiting ionic conductacnes of Cl^(-) and OH^(-) ions are 76 and 198 S cm^(2) mol^(-1) respectively. If the molar conductivity of a 0.01 M solution of NH_(4)OH is 9.6 S cm^(2) mol^(-1) , what will be its degree of ionization Strategy: At any concentration C , if alpha is the degree of ionization then it can be approximented to the ratio of molar conductivity Lambda_(m)^(c) at the concentration C to limiting molar conducting, Lambda_(m)^(0) . We are given Lambda_(m)^(c) but we need to find Lambda_(m)^(0) .

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