Magnesium hydroxide, \(\operatorname{Mg}(\mathrm{OH})_{2},\) is the active ingredient in the antacid TUMS and has a \(K_{\mathrm{sp}}\) value of $8.9 \times 10^{-12}\( . If a 10.0 -g sample of \)\mathrm{Mg}(\mathrm{OH})_{2}$ is placed in \(500.0 \mathrm{mL}\) of solution, calculate the moles of OH -ions present. Because the \(K_{\mathrm{sp}}\) value for \(\mathrm{Mg}(\mathrm{OH})_{2}\) is much less than \(1,\) not a lot solid dissolves in solution. Explain how \(\mathrm{Mg}(\mathrm{OH})_{2}\) works to neutralize large amounts of stomach acid.

First, we need to calculate the molarity of \(\mathrm{Mg(OH)_{2}}\). The molar mass of \(\mathrm{Mg(OH)_{2}}\) is: $$ \mathrm{Mg(OH)_{2}} = 1 \times \mathrm{(24.31\, g/mol)}\,+(2\,x\,{\mathrm{O\,(16.00\,g/mol)} + \mathrm{H\,(1.01\,g/mol)}) = 58.33\,g/mol $$ Now, we can find the moles of \(\mathrm{Mg(OH)_{2}}\) in the solution: $$ \mathrm{moles\, of\,Mg(OH)_{2}} = \frac{\mathrm{10.0\,g}}{\mathrm{58.33\,g/mol}} = 0.1716\,\mathrm{mol} $$ Finally, we can calculate the molarity of \(\mathrm{Mg(OH)_{2}}\) as: $$ \mathrm{Molarity\, of\, Mg(OH)_{2}} = \frac{\mathrm{0.1716\,mol}}{\mathrm{0.500\,L}} = 0.343\, \mathrm{mol/L} $$

The solubility product constant, \(K_\textrm{sp}\), is given as: $$ K_\textrm{sp} = [\textrm{Mg}^{2+}] [\mathrm{OH}^-]^2 $$ Since one mole of magnesium ion is released for each mole of magnesium hydroxide, and two moles of hydroxide ions are released for each mole of magnesium hydroxide, we have: $$ [\textrm{Mg}^{2+}] = [\textrm{Mg(OH)}_2] $$ and $$ [\textrm{OH}^-] = 2[\textrm{Mg(OH)}_2] $$ Substituting these values in the \(K_\textrm{sp}\) equation: $$ 8.9 \times 10^{-12} = [\textrm{Mg(OH)}_2] \times (2[\textrm{Mg(OH)}_2])^2 $$ Solve for \([\textrm{Mg(OH)}_2]\): $$ [\textrm{Mg(OH)}_2] = 1.35 \times 10^{-4}\, \mathrm{mol/L} $$ Now, find the hydroxide ion concentration: $$ [\textrm{OH}^-] = 2[\textrm{Mg(OH)}_2] = 2 \times 1.35 \times 10^{-4}\, \mathrm{mol/L} = 2.7 \times 10^{-4}\, \mathrm{mol/L} $$

Now, we can use the concentration of \(\mathrm{OH}^-\) in the solution to find the moles of \(\mathrm{OH}^-\) ions present: $$ \mathrm{moles\, of\,OH^-} = \mathrm{concentration\, of\,OH^-} \times \mathrm{volume\,of\,solution} = 2.7 \times 10^{-4}\, \mathrm{mol/L} \times 0.500\, \mathrm{L} = 1.35 \times 10^{-4}\, \mathrm{mol} $$

Magnesium hydroxide, \(\mathrm{Mg(OH)_{2}}\), works to neutralize stomach acid, primarily composed of hydrochloric acid (\(\mathrm{HCl}\)), by the following reaction: $$ \mathrm{Mg(OH)_2(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + 2H_2O(l)} $$ The magnesium hydroxide dissociates into magnesium ions and hydroxide ions in the solution. The \(\mathrm{OH}^-\) ions combine with the \(\mathrm{H^+}\) ions from the stomach acid to form water, thus reducing the acidity of the stomach and providing relief from heartburn or indigestion. Although the solubility of \(\mathrm{Mg(OH)_{2}}\) is low, the relatively high concentration of \(\mathrm{HCl}\) in the stomach leads to a continuous reaction between the solid \(\mathrm{Mg(OH)_{2}}\) and the gastric acid, ensuring a sustained neutralizing effect.