At 291 K, the molar conductivities at infinite dilution of `NH_(4)Cl`, NaOH and NaCl are 129.8, 217.4 and `108.9" S " cm^(2) mol^(-1)` respectively. T

Question

At 291 K, the molar conductivities at infinite dilution of `NH_(4)Cl`, NaOH and NaCl are 129.8, 217.4 and `108.9" S " cm^(2) mol^(-1)` respectively. T

At 291 K, the molar conductivities at infinite dilution of `NH_(4)Cl`, NaOH and NaCl are 129.8, 217.4 and `108.9" S " cm^(2) mol^(-1)` respectively. The molar conductivity of a centinormal solution of `NH_(4)OH` is `9.33" S "cm^(2)mol^(-1)`. The percentage dissociation of `NH_(4)OH` at this dilution and the dissociation constant of `NH_(4)OH` are :
A. `3.92%,1.599xx10^(-5)`
B. `6.92%,3.599xx10^(-5)`
C. `3.92%,4.599xx10^(-2)`
D. `9.92%,1.599xx10^(-5)`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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

Correct Answer - A
(a) `Lambda_((NH_(4)OH))^(@)=Lambda_((NH_(4)Cl))^(@)+Lambda_((NaOH))-Lambda_((HCl))^(@)`
`=129.8+217.4-108.9=238.3" s "cm^(2)mol^(-1)`
Degree of dissociation `(alpha)=(Lambda_(C ))/(Lambda^(@))=((9.33" S "cm^(2)mol^(-1)))/((238.3" S "cm^(2)mol^(-1)))`
`=0.0392=0.0392xx100=3.92%`
Dissociation constant
`NH_(4)OH hArr NH_(4)^(+)+OH^(-)`
`{:("Initial conc".,C,0,0),("Equilibrium conc".,C-Calpha,Calpha,Calpha),(,C(1-alpha),Calpha,Calpha):}`
`K_(a)=(CalphaxxCalpha)/(C(1-alpha))=(Calpha^(2))/(1-alpha)`
`=((0.01)xx(0.0392))/((1.0.0392))=1.599xx10^(-5)`