Molar conductivities at infinite dilution (at 298 K) of `NH_(4)CI`, NaOH and NaCI are 129.8, 217.4 and 108.9 `Omega^(-1)cm^(-1)mol^(-1)` respcetively. If the molar conductivity of a centimolar solution of `NH_(4)OH` is 9.33 `Omega^(-1) cm^(-1)`, what is percentage dissociation of `NA_(4)OH` at this concentation ? Also calculte the dissociation constant for `NH_(4)OH`.

Answer

`3.90% ; 1.57xx10^(-5) mol L^(-1)`

Explanation

Step I. Calcualtion of `Lambda_(m)^(oo)` of `NH_(4)OH`
`Lambda_(m(NH_(4)OH))^(oo)=Lambda_(m(NH_(4)Cl))^(oo)+Lambda_(m(NaOH))^(oo)-Lambda_(m(NaCl))^(oo)`
`=(129.8+217.4-108.9)" ohm"^(-1)cm^(2)mol^(-1)=238.3" ohm"^(-1)cm^(2)mol^(-1)`.
Step II. Calculation of percent dissociation of `NH_(4)OH`
Degree of dissociation `(alpha)=(Lambda_(m)^(C))/(Lambda_(m)^(oo))=((9.33" ohm"^(-1) cm^(2)mol^(-1)))/((238.3" ohm"^(-1)cm^(2)mol^(-1)))=0.039 or 3.9%`
Step III. Calculation of dissociation constant
`NH_(4)OH hArrNH_(4)^(+)(aq)+OH^(-)(aq)`
`{:("Initial molar conc".:," "C,""0,""0),("Eqm.molar conc".:,C(1-alpha),Calpha,Calpha):}`
`k_(C)=([NH_(4)^(+)(aq)][OH^(-)(aq)])/([NH_(4)OH^(-)])=(CalphaxxCalpha)/(C(1-alpha))=(Calpha^(2))/(1-alpha)`
`C=M//100=0.01" mol "L^(-1), alpha=3.9xx10^(-2)`
`:. " " k_(C)=((0.01" mol L "^(-1))xx(3.9xx10^(-2))^(2))/((1-3.9xx10^(-2)))=1.57xx10^(-5)" mol L"^(-1)`.