Given that ( `ohm^(-1) cm^(2)eq^(-1)), T=298 K`
`{:(lambda_(E)^(oo) f o r Ba(OH)_(2)=228.8),(lambda_(E)^(oo) fo r BaCl_(2)=120.3),(lambda_(E)^(oo) for NH_(4)Cl=129.8):}|{:("specific conductance"),("for "0.2N NH_(4)OH" solution"),(is 4.766xx10^(-4)ohm^(-1)cm^(-1)):}`
then value of pH of the solution of `NH_(4)OH` will be nearly

Options

A. `9.2`
B. `11.3`
C. `12.1`
D. `7.9`

Answer

2

Explanation

`lambda_(E)^(m)`for `NH_(4)OH=228.8-120.3+128.8=238.3`
Now, `lambda_(E)=(1000xxk)/(N)=(1000xx4.766xx10^(-4))/(0.2)=2.383`
`alpha=(lambda_(E))/(lambda_(E)^(oo))=0.01`
`OH^(-)=alphaxxC=0.1xx0.2=2xx10^(-3)`
`p^(OH)=30log2=3-0.3=2.7`
`pH=14-2.7=11.3`