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For NH(3), K(b)=1.8xx10^(-5). K(a) for N...

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

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To find the \( K_a \) for \( NH_4^+ \) given that \( K_b \) for \( NH_3 \) is \( 1.8 \times 10^{-5} \), we can use the relationship between the \( K_b \) of a base and the \( K_a \) of its conjugate acid: ### Step-by-Step Solution: 1. **Understand the relationship between \( K_a \), \( K_b \), and \( K_w \)**: The relationship is given by: \[ K_a \times K_b = K_w ...
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