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To a solution of 0.01M Mg^(2+) and 0.8M ...

To a solution of `0.01M 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)`.

Text Solution

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Suppose `VmL` of solution contains `0.1M Mg^(2+)` and `0.8M NH_(4)CI`.
Now `V mL` of `a` molarity `NH_(3)` is added which just givens precipitate of `Mg(OH)_(2)`, then
`[Mg^(2+)] [overset(Theta)OH]^(2) = K_(sp) Mg(OH)_(2)`
`[(0.1V)/(2V)] [overset(Theta)OH]^(2) = 1.4 xx 10^(-11)`
`:. [overset(Theta)OH] = 1.67 xx 10^(-5)M ([Mg^(2+)] = ("Millimoles")/("Total volume")`
Now if the `[overset(Theta)OH] =- log K_(b) + "log" ([NH_(4)CI])/([NH_(3)])`
`- log (1.67 xx 10^(-5)) = - log (1.8 xx 10^(-5)) + "log" ((0.8 xxV)//2V)/((axxV)//2V)`
`:. a = 0.7421M`
`:. [NH_(3)]` in solution `= (0.7421 xx V)/(2V) = 0.3710M`
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