`Mg(OH)_2` is soluble in `NH_4`CI, but not in NaCl. Why?

Explanation

Adding `NH_4 CI` to `Mg(OH)_2`, the interaction takes place to give magnesium chloride and ammonium hydroxide.
`Mg(OH)_2 + 2NH_4 CI to MgCl_2 + 2NH_4 OH `
`NH_4 OH ` being a weak base, reduces the `[OH^(-)]` in solution. Product of `[Mg^(2+)]` and `[OH^(-)]` remains lower than `K_(sp)` of `Mg(OH)_2` to give no precipitation of `Mg(OH)_2` .
Adding NaCl to `Mg(OH)_2`, the interaction leads to give magnesium chloride and sodium hydroxide.
`Mg(OH)_2 +2NaCl to MgCl_2 + 2NaOH`
NaOH being a strong base produces free `OH^(-)` ions in solution.
Ionic product of `[Mg^(2+)] and [OH^(-) ]` exceeds the solubility product, showing precipitation of magnesium hydroxide