# How does ammonium chloride increase the solubility of magnesium hydroxide in water?

In my textbook it is written that $$\ce{Mg(OH)2}$$ is slightly soluble in water, and it forms a milky solution which is called "milk of magnesia".

Its solubility can be increased by the addition of $$\ce{NH4Cl}.$$ But how? There seems to be no reason.

• Is it magnesium(II) hydroxide when that is the only known magnesium hydroxide? – Oscar Lanzi Dec 27 '19 at 18:23
• I don't know mag( II ) hydroxide...someone has edited my question...I only typed magnesium hydroxide. – Yaseen Khan Dec 28 '19 at 12:59

There is a strong reason. But it is the chemical reaction, not just a better dissolution.

$$\ce{Mg(OH)2}$$ is a base with the limited solubility, defined by $$K_\mathrm{sp}=[\ce{Mg^2+}][\ce{OH-}]^2$$

$$\ce{NH4+}$$ ion, created by $$\ce{NH4Cl}$$ dissolution, acts as a weak acid:

$$\ce{NH4+ + H2O <<=> NH3 + H3O+}$$

with $$\mathrm{p}K_\mathrm{a}=9.25$$

$$\ce{OH-}$$ ions formed by dissolution of $$\ce{Mg(OH)2}$$ are eliminated by recombination

$$\ce{OH- + H3O+ <=>> 2 H2O}$$

what supports the dissolution by keeping the product of ion concentrations below the $$K_\mathrm{sp}$$.

Effectively, there is ongoing equilibrium:

$$\ce{Mg(OH)2 v + 2 NH4+ <=> Mg^2+ + 2 NH3 + 2 H2O}$$

This is a well known problem in qualitative analysis. When you add $$\ce{NH4Cl}$$ to a solution containing $$\ce{OH-}$$ ions, you produce the reaction: $$\ce{NH4+ + OH- -> NH3 + H2O}$$ The result is that the concentration of $$\ce{OH-}$$ decreases. If this operation was done in a saturated solution of $$\ce{Mg(OH)2}$$, the solubility product is no more obtained. A greater amount of $$\ce{Mg(OH)2}$$ can pass into solution. As a consequence, the solubility of $$\ce{Mg(OH)2}$$ is increased.