At `18^(@)C`, the conductivities at infinite dilution of `NH_(4)Cl, NaOH` and `NaCl` are 129.8, 217.4 and 108.9 mho respectively. If the equivalent co

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At `18^(@)C`, the conductivities at infinite dilution of `NH_(4)Cl, NaOH` and `NaCl` are 129.8, 217.4 and 108.9 mho respectively. If the equivalent co

At `18^(@)C`, the conductivities at infinite dilution of `NH_(4)Cl, NaOH` and `NaCl` are 129.8, 217.4 and 108.9 mho respectively. If the equivalent conductivity of N/100 solution of `NH_(4)OH` is 9.93 mho, calculate the degree of dissociation of `NH_(4)OH` at this dilution.

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

`Lambda_(oo NH_(4)Cl)=lambda_(NH_(4)^(+))+lambda_(Cl^(-))=129.8` ...(i)
`Lambda_(oo NaOH)=lambda_(Na^(+))+lambda_(OH^(-))=217.4` ...(ii)
`Lambda_(oo NaCl)=lambda_(Na^(+))+lambda_(Cl^(-))=108.9` ...(iii)
Adding eqs. (i) and (ii) and substracting eq. (iii),
`lambda_(NH_(4)^(+))+lambda_(Cl^(-))+lambda_(Na^(+))+lambda_(OH^(-))-lambda_(Na^(+))- lambda_(Cl^(-))`
`=lambda_(NH_(4)^(+))+lambda_(OH^(-))=129.8+217.4-108.9`
`Lambda_(oo NH_(4)OH)=238.3` mho
Degree of dissociation, `alpha=Lambda_(v)/Lambda_(oo)=9.93/238.3=0.04167`
or `4.17 %` dissociated.