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To a solution of 0.1M Mg^(2+) and 0.8M N...

To a solution of `0.1M Mg^(2+)` and `0.8M NH_(4)CI`, and equal volume of `NH_(3)` is added which just gives precipitates. Calculate `[NH_(3)]` in solution.
`K_(sp) of Mg(OH)_(2) = 1.4 xx 10^(-11)` and `K_(b) of NH_(4)OH = 1.8 xx 10^(-5)`.

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`0.3710M;`
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