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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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Answer» 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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