Equal volume of `0.2M NH_(4)OH` (or ammonia) are `0.1MH_(2)SO_(4)` are mixed.Calculated pH of final solution.Given :`K_(b)` of `NH_(3)=1.8xx10^(-5)` a

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Equal volume of `0.2M NH_(4)OH` (or ammonia) are `0.1MH_(2)SO_(4)` are mixed.Calculated pH of final solution.Given :`K_(b)` of `NH_(3)=1.8xx10^(-5)` a

Equal volume of `0.2M NH_(4)OH` (or ammonia) are `0.1MH_(2)SO_(4)` are mixed.Calculated pH of final solution.Given :`K_(b)` of `NH_(3)=1.8xx10^(-5)` at `25^(@)C`.

Monjur Alahi

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2025-01-04 04:42

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

Salim Ahmed Kawsar

`C=[NH_(4)^(+)]=0.1M`(:.volume get doubeled so concentration must have been halved)
`h=sqrt(K_(h))/(C )=sqrt((10^(-14))/(1.8xx10^(-5)xx0.1))(lt0.1)`
`:.pH=1//2{14-4.74+1}=(10.26)/(2)=5.13`

4 views Answered 2023-02-05 15:29