What concentration of `OH^(-)` ions will reduce `NH_(4)^(+)` ion to `2xx10^(-5)M` in `0.4M` solution of `NH_(4)OH ? K(b)(NH_(4)OH)=1.8xx10^(-5)`

Question

What concentration of `OH^(-)` ions will reduce `NH_(4)^(+)` ion to `2xx10^(-5)M` in `0.4M` solution of `NH_(4)OH ? K(b)(NH_(4)OH)=1.8xx10^(-5)`

What concentration of `OH^(-)` ions will reduce `NH_(4)^(+)` ion to `2xx10^(-5)M` in `0.4M` solution of `NH_(4)OH ? K(b)(NH_(4)OH)=1.8xx10^(-5)`
A. `0.36M`
B. `0.036M`
C. `2xx10^(-5)M`
D. None of these

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

Salim Ahmed Kawsar

`{:(,NH_(4)OHhArr,NH_(4)^(+),+,OH^(+)),(t=eq,0.4(1-alpha),2xx10^(-5),,),(,,(=0.4alpha),,):}`
So,`alpha=5xx10^(-5)(ltlt1):.0.4(1-alpha)~~0.4`
`K_(b)=(2xx10^(-5)xx[OH^(-)])/(0.4)=1.8xx10^(-5)rArr [OH^(-)]=0.36M`

4 views Answered 2023-02-05 15:29