At 291 K, the molar conductivities at infinite of NH_(4)Cl, NaOH and NaCl are 128.8,217.4 and 108.9 S cm^(2) respectively. If the molar conductivity of a centrinormal solution of NH_(4)OH is 9.33 S cm^(2), what is the percentage dissociation of NH_(4)OH at this dilution? Also calculate the dissociation constant of NH_(4)OH.

At 291 K, the molar conductivities at infinite of NH_(4)Cl, NaOH and N

1.	At 291 K, the molar conductivities at infinite of NH_(4)Cl, NaOH and NaCl are 128.8,217.4 and 108.9 S cm^(2) respectively. If the molar conductivity of a centrinormal solution of NH_(4)OH is 9.33 S cm^(2), what is the percentage dissociation of NH_(4)OH at this dilution? Also calculate the dissociation constant of NH_(4)OH.
Answer» Solution :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)`.