दिया गया है कि (`S cm^(2)eq^(-1)` में ) T = 298 K पर:
`Ba(OH_(2)), BaCl_(2)` और `NH_(4)Cl` के लिए `wedge_(eq)^(@)` क्रमश: 228.8, 120.3 और 129.8 हैं। 0.2 N `NH_(4)` OH विलयन के लिए विशिष्ट चालकता `4.766 xx 10^(-4)S cm^(-1)` है, तो `NH_(4)` OH के दिए गए विलयन का pH मान लगभग होगा। (log 2 = 0.3 लीजिए)

Answer

`11.3`

Explanation

`wedge_(eq)^(@) Ba(OH)_(2)=lambda^(@)Ba^(2+)+lambda_(eq)^(@)OH^(-) " " ......(I)`
`lambda_(eq)^(@) BaCl_(2)=lambda_(eq)^(@)Ba^(2) +lambda_(eq)^(@)Cl^(-) " ".....(II)`
`lambda_(eq)^(@)NH_(4)Cl=lambda_(eq)^(@) NH_(4)^(+)+lambda_(eq)^(@)Cl^(-1) " " .....(III)`
`lambda_(eq)^(@) NH_(4)OH=lambda_(eq)^(@) NH_(4)^(+) +lambda_(eq)^(@) OH^(-)`
`I+III-II`
`lambda_(eq)^(@) NH_(4)OH =(228.8+129.8)-120.3`
`=238.33 cm^(2)eq^(-1)`
`lambda_(eq)NH_(4)OH =(4.766xx10^(-4)xx1000)/(0.2) =2.383`
`alpha=(lambda_(eq)NH_(4)OH)/(lambda_(eq_(1))^(@)NH_(4)OH) =10^(-2)`
`underset(c(1-alpha))(NH_(4)OH)hArr underset(calpha)(NH_(4)^(+))+underset(c alpha)(OH^(-))`
`[OH^(-)]=0.2 xx 10^(-2) =2xx10^(-3)`
`pOH =3-log 2 rArr pH =14 -(3-log2)=11.3`