What is `[NH_(4)^(+)]` in a solution that contain `0.02` M `NH_(3)(K_(b)=1.8xx10^(-5))` and `0.01` M KOH?
A. `3.6xx10^(-5) M`
B. `31.8xx10^(-5) M`
C. `0.9xx10^(-5) M`
D. `7.2xx10^(-5) M`

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1 Answers

Correct Answer - A
`{:(NH_(4),OHhArr,NH_(4)^(+),+OH^(-),),(KOH,to,K^(+),+OH^(-),):}`
`K_(b)=([NH_(4)^(+)][OH^(-2)])/([NH_(4)OH])`
`1.8xx10^(-5)=([X][10^(-2)])/(2xx10^(-2))rArrx=3.6xx10^(-5)`

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