Why does not `MgSO_(4)` form any precipitate when it reacts with `NH_(3)` in presence of `NH_(4)Cl`?

Explanation

`NH_(3)` is a weak base. In aqueous solution. It ionises partially to produce `NH_(4)^(+) and OH^(-)` ions.
`NH_(3)(aq)+H_(2)O(l) hArr NH_(4)^(+)(aq)+OH^(-)(aq)`
In presence of `NH_(4)Cl`, owing to common ion effect of `NH_(4)^(+)`, the degree of ionisation of `NH_(3)` is further suppressed. thus, the concentration of `OH^(-)` ions decreases to a large extent. at this low concentration of `OH^(-)` ions, the product of the concentration of `Mg^(2+)` ions and square of the concentration of `OH^(-)` ions decreases to a large extent. at this low concentration of `OH^(-)` ions, the product of the concentration of `Mg^(2+)` ions and square of the concentration of `OH^(-)` ions (as `K_(sp)[Mg(OH)_(2)]=[Mg^(2+)][OH^(-)]^(2))` cannot exceed the solublity product of `Mg(OH)_(2)`, i.e., `[Mg^(2+)]xx[OH^(-)]^(2)ltK_(sp)` (solublity product). as a result, `Mg(OH)_(2)` does not get precipitated.