Question

\(\mathrm{Mg}(\mathrm{OH})_{2}\) is the main ingredient in the antacid TUMS and has a \(K_{\text {sp }}\) value of \(8.9 \times 10^{-12}\). If a \(10.0-\mathrm{g}\) sample of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is placed in \(500.0 \mathrm{~mL}\) of solution, calculate the moles of \(\mathrm{OH}^{-}\) ions present. Because the \(K_{\mathrm{sp}}\) value for \(\mathrm{Mg}(\mathrm{OH})_{2}\) is small, not a lot of solid dissolves in solution. Explain how \(\mathrm{Mg}(\mathrm{OH})_{2}\) works to neutralize large amounts of stomach acids.

Short answer

The moles of OH⁻ ions present in the solution are 2.64 × 10⁻⁴ moles. Mg(OH)₂ works to neutralize stomach acids by reacting with hydrochloric acid (HCl) to form new compounds and release water. This removes H⁺ ions and reduces acidity, thus neutralizing large amounts of stomach acids.

Step-by-step solution

Write the balanced dissociation equation for Mg(OH)₂

The balanced dissociation equation for Mg(OH)₂ in water is: \( Mg(OH)_{2}(s) \rightleftarrows Mg^{2+}(aq) + 2OH^-(aq) \)

Calculate the moles of Mg(OH)₂

Given a 10.0 g sample of Mg(OH)₂, first we must find the number of moles of the substance. Molar mass of Mg(OH)₂ = 24.31 (Mg) + 2 (16.00 + 1.008) = 58.32 g/mol Moles of Mg(OH)₂ = mass of Mg(OH)₂ / molar mass = 10.0 g / 58.32 g/mol = 0.1715 moles

Write the dissociation reaction in terms of moles and concentration

Suppose x moles of Mg(OH)₂ dissociate upon dissolving in 500.0 mL of water. Hence, Moles of Mg²⁺ = x moles Moles of OH⁻ = 2x moles Since the total volume of solution is 500.0 mL, the concentration of the ions can be calculated: Concentration of Mg²⁺: x/0.5000 L Concentration of OH⁻: 2x/0.5000 L

Write the expression for Ksp and solve for x

The expression for Ksp is given by: \(K_{sp} = [Mg^{2+}][OH^-]^2 \) Using the concentrations derived in step 3: \(K_{sp} = (\frac{x}{0.5000})(\frac{2x}{0.5000})^2 \) Given Ksp = 8.9 × 10⁻¹², we have: \(8.9 \times 10^{-12} = x(2x)^2\)

Calculate the moles of OH⁻ ions

Solve the cubic equation obtained in step 4 for x: \(x = 1.32 \times 10^{-4} \) Now, we can find the moles of OH⁻ ions, using the relationship from step 3: Moles of OH⁻ = 2x = 2(1.32 × 10⁻⁴) = 2.64 × 10⁻⁴ moles

Explain how Mg(OH)₂ works to neutralize stomach acids

Despite having a low Ksp value, Mg(OH)₂ can effectively neutralize stomach acids. This is because when Mg(OH)₂ reacts with stomach acids, such as hydrochloric acid (HCl), it forms new compounds and releases water. The removal of H⁺ ions from the stomach acid helps to neutralize it, while the formation of water and Mg²⁺ ions contributes to the overall reduction in acidity. The reaction between Mg(OH)₂ and HCl is given by: \(Mg(OH)_{2}(s) + 2HCl(aq) \rightarrow MgCl_{2}(aq) + 2H_{2}O(l)\) This process of removing H⁺ ions and releasing water continues, allowing Mg(OH)₂ to neutralize large amounts of stomach acids effectively.

Key concepts

Chemical Equilibrium

When a substance like magnesium hydroxide \(\mathrm{Mg}(\textrm{OH})_{2}\) is dissolved in water, it dissociates to form magnesium \(\mathrm{Mg}^{2+}\) and hydroxide ions \(\mathrm{OH}^{-}\). This process reaches a point where the rate of dissociation equals the rate of recombination, leading to a state of chemical equilibrium. The equilibrium can be expressed with the equation \(\mathrm{Mg}(\textrm{OH})_{2}(\textrm{s}) \leftrightarrows \mathrm{Mg}^{2+}(\textrm{aq}) + 2\mathrm{OH}^{-}(\textrm{aq})\).

At this stage, the concentrations of the ions in the solution remain constant. The equilibrium constant for the dissolution is called the solubility product constant, abbreviated as \(K_{sp}\), which indicates the product of the ion concentrations raised to the power of their stoichiometric coefficients. The \(K_{sp}\) value helps us understand how much of a compound will dissolve in a solution. A low \(K_{sp}\) implies that very little of the compound dissolves, resulting in a saturated solution with excess undissolved solid.

Ion Concentration

Ion concentration is a measure of the amount of an ion in a solution and is usually expressed in moles per liter (Molarity). The problem requires calculating the concentration of hydroxide ions \(\mathrm{OH}^{-}\) given that magnesium hydroxide \(\mathrm{Mg}(\textrm{OH})_{2}\) has a low solubility in water.

To figure out the concentration of hydroxide ions, the molar solubility of the compound, which indicates the moles of solute that dissolve to form a liter of saturated solution, is calculated using the known \(K_{sp}\). As seen in the solution steps, dissociation leads to the formation of twice as many \(\mathrm{OH}^{-}\) ions as \(\mathrm{Mg}^{2+}\) ions, because each unit of \(\mathrm{Mg}(\textrm{OH})_{2}\) releases two hydroxide ions. The concentration of ions dictates the solution’s properties and can be used to predict the outcome of reactions, like the neutralization of acids.

Acid-Base Neutralization

Acid-base neutralization is a chemical process in which an acid and a base react to form water and a salt. This reaction is essential for substances like magnesium hydroxide \(\mathrm{Mg}(\textrm{OH})_{2}\) used in antacids such as TUMS. Even with low solubility, the base \(\mathrm{Mg}(\textrm{OH})_{2}\) effectively neutralizes stomach acid by reacting with hydrochloric acid \(\mathrm{HCl}\) to produce magnesium chloride \(\mathrm{MgCl}_{2}\) and water.

The equation for the neutralization reaction is \(\mathrm{Mg}(\textrm{OH})_{2}(\textrm{s}) + 2\mathrm{HCl}(\textrm{aq}) \rightarrow \mathrm{MgCl}_{2}(\textrm{aq}) + 2\mathrm{H}_{2}O(\textrm{l})\). The \(\mathrm{OH}^{-}\) ions from magnesium hydroxide combine with the \(\mathrm{H}^{+}\) ions of hydrochloric acid, thus neutralizing the acid while decreasing its concentration in the stomach, leading to relief from acidity symptoms. The efficiency of this neutralization process allows magnesium hydroxide to be an effective active ingredient in antacids, despite its low solubility in water.