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A solution has 0.05M Mg^(2+) and 0.05M N...

A solution has `0.05M Mg^(2+)` and `0.05M NH_(3)`. Calculate the concentration of `NH_(4)CI` required to prevent the formation of `Mg(Oh)_(2)` in solution. `K_(SP)` for `Mg(OH)_(2)=9.0xx10^(-12)` and ionisation constant of `NH_(3)` is `1.8xx10^(-5)`.

Text Solution

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The minimum [ OH] at which there will be no precipitation of `Mg(OH)_(2)` in `0.05` M `Mg^(2+)` . These hydroxyl ions are to be derived by basic buffer of `NH_(4)Cl` and `NH_(4)OH`
`pH = pK_(b) + log. (["Salt"])/(["Base"])`
`pH = pK_(b) + log. ([NH_(4)])/([NH_(4)OH])`
`NH_(4)OH hArr NH_(4)^(+) + OH^(-)`
In presence of `[NH_(4)Cl]` all the `NH_(4)^(+)` ions provided by `NH_(4)Cl` as due to common ion effect , dissociation of `NH_(4)OH` will be supressed
` -log [OH^(-)] = - log 1.8 xx 10^(-5) + log . ([NH_(4)^(+)])/([0.05])`
` :. NH_(4)^(+) = 0.067 M `
or `[NH_(4)CL] = 0.067 ` M
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