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^(2)mol^(-1)` respcetively. If the molar conductivity of a centimolar solution of `NH_(4)OH` is 9.33 `Omega^(-1) cm^(2)mol^(-1)`, what is percentage dissociation of `NH_(4)OH` at this concentation ? Also calculte the dissociation constant for `NH_(4)OH`.

Explanation

Here, we are given: `wedge^(@)` for `NA_(4)Cl=129.8" S "cm^(2),wedge^(@)` for `NaOH=217.4" S "cm^(2)`,
`wedge^(@)` for `NaCl=108.9" S "cm^(2)`,
Kohlrausch's law, `wedge^(@)` for `NH_(4)OH=lamda_(NH_(4)^(+))^(@)+lamda_(OH^(-))^(@)=wedge^(@)(NH_(4)Cl)+wedge^(@)(NaOH)-wedge^(@)(NaCl)`
`=129.8+217.4-108.9=238.3" S "cm^(2)`
`wedge_(c)=9.33" S "cm^(2)` (given)
`therefore`Degree of dissociation `(alpha)=(wedge_(c))/(wedge^(@))=(9.33)/(238.3)=0.0392` or % age dissociation `=0.0392xx100=3.92%`
Calculation of dissociation constant
`{:(,NH_(3)OH,hArr,NH_(4)^(+),+,OH^(-)),("Initial conc.",c,,,,),("Equilibrium conc.",c-calpha=c(1-alpha),,calpha,,calpha" "K=(calphaxxcalpha)/(c(1-alpha))=(calpha^(2))/(1-alpha)):}`
Putting c=0.01N=0.01M and `alpha=0.0392, ` we get
`K=((0.01)(0.0392)^(2))/(1-0.0392)=(10^(-2)xx(3.92xx10^(-2))^(2))/(0.9608)=1.559xx10^(-5)`.