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.

  • $\begingroup$ Is it magnesium(II) hydroxide when that is the only known magnesium hydroxide? $\endgroup$ – Oscar Lanzi Dec 27 '19 at 18:23
  • 1
    $\begingroup$ I don't know mag( II ) hydroxide...someone has edited my question...I only typed magnesium hydroxide. $\endgroup$ – 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:

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

| improve this answer | |

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.

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