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