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When 0.1 mole of NH(3) is dissolved in w...

When 0.1 mole of `NH_(3)` is dissolved in water to make 1.0 L of solution , the `[OH^(-)]` of solution is `1.34 xx 10^(-3)`M. Calculate `K_(b)` for `NH_(3)`.

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`K_b=Calpha^2=0.1 xx1.34xx10^(-2)xx1.34xx10^(-2)=0.18xx10^(-4)`
`rArr 1.8xx10^(-5)=K_b`
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