For `NH_(3)`, `K_(b)=1.8xx10^(-5)`. `K_(a)` for `NH_(4)^(+)` would beA. `1.8xx10^(5)`B. `5.56xx10^(5)`C. `1.8xx10^(10)`D. `5.56xx10^(-10)`

For `NH_(3)`, `K_(b)=1.8xx10^(-5)`. `K_(a)` for `NH_(4)^(+)` would beA

1.	For `NH_(3)`, `K_(b)=1.8xx10^(-5)`. `K_(a)` for `NH_(4)^(+)` would beA. `1.8xx10^(5)`B. `5.56xx10^(5)`C. `1.8xx10^(10)`D. `5.56xx10^(-10)`
Answer» Correct Answer - D `K_(a)xxK_(b)=K_(w)` `K_(a)=(K_(w))/(K_(b))= (10^(-14))/(1.8xx10^(-5))=5.56xx10^(-10)`

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For `NH_(3)`, `K_(b)=1.8xx10^(-5)`. `K_(a)` for `NH_(4)^(+)` would beA. `1.8xx10^(5)`B. `5.56xx10^(5)`C. `1.8xx10^(10)`D. `5.56xx10^(-10)`

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