At 291 K, the molar conductivities at infinite dilution of `NH_(4)Cl, NH_(4)OH" and "NaCl " are " 129.8 , 217" and "108.9 Scm^(2)` respectively. If the molar conductivity of a centinormal 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`.

Explanation

Given : `^^ ^(0) (NH_(4)Cl) = 129.8 S cm^(2)`,
`^^ ^(0)(NaOH) = 217.4 Scm^(2)`
By Kohlrausch .s law ,
`^^_(NH_(4)OH)^(0) = lambda_(NH_(4)^(+))^(0) + lambda_(OH^(-))^(0)`
` = ^^_(NH_(4)Cl)^(0) + ^^_(NaOH)^(0) - ^^_(NaCl)^(0)`
` = 129.8 + 217.4 - 108.9`
` = 238.3 Scm^(2)`
`^^_(C) = 9.33 S cm^(2) ` (given)
`:. ` degree of dissociation `(alpha)`
`^^_(C)/^^_(m) = (9.33)/(283.3) = 0.0392`
`%` dissociation ` = 0.0392 xx 100`
` = 39.2 %`
Calculation of dissociation constant
`{:(,NH_(4)OH,hArr,NH_(4)^(+),+,OH^(-)),("Initial concentration"," "C,,O,,O),("Equilibrium concentration",C(1-alpha),,Calpha,,Calpha):}`
Where `.alpha.` is the degree of dissociation
`K_(C) = ([NH_(4)][OH^(-)])/([NH_(4)OH]) `
` = (Calpha xx Calpha)/(C(1- alpha))`
` = (Calpha^(2))/(1 - alpha) `
Putting ` C = 0.01 N = 0.01 M`,
` alpha = 0.0392`
`K = ((0.01)(0.0392)^(2))/(1-0.0392) `
` = (10^(2) xx (3.92 xx 10^(-2))^(2))/(0.0608) `
` = 1.599 xx 10^(-5)`