Question

(a) Describe a preparation for magnesium hydroxide \(\left[\mathrm{Mg}(\mathrm{OH})_{2}\right]\) and predict its solubility. (b) Milk of magnesia contains mostly \(\mathrm{Mg}(\mathrm{OH})_{2}\) and is effective in treating acid (mostly hydrochloric acid) indigestion. Calculate the volume of a \(0.035 \mathrm{M} \mathrm{HCl}\) solution (a typical acid concentration in an upset stomach) needed to react with two spoonfuls (approximately \(10 \mathrm{~mL}\) ) of milk of magnesia [at \(\left.0.080 \mathrm{~g} \mathrm{Mg}(\mathrm{OH})_{2} / \mathrm{mL}\right]\).

Short answer

Prepare Mg(OH)2 by precipitation; it's sparingly soluble. 783 mL of 0.035 M HCl is needed.

Step-by-step solution

Describe Preparation of Mg(OH)2

Magnesium hydroxide, \(\mathrm{Mg(OH)}_2\), can be prepared through a precipitation reaction between a soluble magnesium salt such as magnesium chloride \(\mathrm{MgCl}_2\) and a sodium hydroxide \(\mathrm{NaOH}\) solution. The reaction is as follows:\[\mathrm{MgCl_2(aq)} + 2\mathrm{NaOH(aq)} \rightarrow \mathrm{Mg(OH)_2(s)} + 2\mathrm{NaCl(aq)}\]Upon mixing, magnesium hydroxide precipitates out of the solution as a white solid.

Predict Solubility of Mg(OH)2

Magnesium hydroxide is considered to be sparingly soluble in water. It has a low solubility product constant \(K_{sp}\) value, which indicates that only a small amount dissolves to form \(\mathrm{Mg^{2+}}\) and \(\mathrm{OH^-}\) ions. Therefore, \(\mathrm{Mg(OH)_2}\) will not dissolve significantly, existing mainly as a solid.

Calculate moles of Mg(OH)2

First, calculate the mass of \(\mathrm{Mg(OH)_2}\) in 10 mL of milk of magnesia:\[\text{Mass of } \mathrm{Mg(OH)_2} = 10\, \mathrm{mL} \times 0.080\, \mathrm{g/mL} = 0.800\, \mathrm{g}\]Next, convert this mass to moles using the molar mass \( (24.31 + 2(16.00 + 1.01) = 58.32) \) g/mol:\[\text{Moles of } \mathrm{Mg(OH)_2} = \frac{0.800\, \mathrm{g}}{58.32\, \mathrm{g/mol}} \approx 0.0137\, \mathrm{mol}\]

Calculate moles of HCl needed

From the reaction:\[\mathrm{Mg(OH)_2(s)} + 2\mathrm{HCl(aq)} \rightarrow \mathrm{MgCl_2(aq)} + 2\mathrm{H_2O(l)}\]It is seen that 1 mole of \(\mathrm{Mg(OH)_2}\) reacts with 2 moles of \(\mathrm{HCl}\). Therefore, 0.0137 moles of \(\mathrm{Mg(OH)_2}\) will react with:\[2 \times 0.0137 = 0.0274\, \mathrm{mol}\, \mathrm{HCl}\]

Calculate volume of HCl solution needed

Using the molarity equation \(M = \frac{n}{V}\), rearrange to find \(V\):\[V = \frac{n}{M} = \frac{0.0274\, \mathrm{mol}}{0.035\, \mathrm{mol/L}} \approx 0.783\, \mathrm{L}\]Converting liters to milliliters, we get:\[0.783\, \mathrm{L} \times 1000\, \mathrm{mL/L} = 783\, \mathrm{mL}\]

Key concepts

Solubility

The concept of solubility is fundamental when discussing substances like magnesium hydroxide. Solubility describes how well a solute can dissolve in a solvent to form a homogeneous solution. In the case of magnesium hydroxide, it is only sparingly soluble in water. This means that only a small amount can dissolve before the solution becomes saturated.

The solubility product constant, often denoted as \(K_{sp}\), quantifies this solubility. Magnesium hydroxide has a low \(K_{sp}\) value, which explains why it does not dissolve significantly in water. As a result, magnesium hydroxide tends to exist mainly as a solid precipitate instead of a dissolved solute. Understanding solubility helps predict how substances behave in solutions and is crucial for applications in chemistry and medicine.

Precipitation Reaction

Precipitation reactions are a type of chemical reaction that results in the formation of an insoluble solid, known as a precipitate, from the solution. This occurs when the products of the reaction have a low solubility in water. When preparing magnesium hydroxide, a precipitation reaction is used.

In the specific reaction between magnesium chloride \(\text{MgCl}_2\) and sodium hydroxide \(\text{NaOH}\), magnesium hydroxide \(\text{Mg(OH)}_2\) is the precipitate formed. The chemical equation for this reaction is:

• \(\text{MgCl}_2(\text{aq}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{Mg(OH)}_2(\text{s}) + 2\text{NaCl}(\text{aq})\)

The mixing of the two aqueous solutions results in magnesium hydroxide appearing as a white solid, which distinctly separates from the solution. This type of reaction is commonly used in purification processes and in preparations of materials in a laboratory setting.

Molar Calculations

Molar calculations are essential for determining the quantities of substances involved in chemical reactions. They relate the mass of a substance to its number of moles, using its molar mass as a conversion factor.

Let's explore how to find the number of moles of magnesium hydroxide in milk of magnesia. Consider that milk of magnesia contains 0.080 g of \(\text{Mg(OH)}_2\) per milliliter. Given 10 mL, the total mass can be calculated as follows:

• Mass of \(\text{Mg(OH)}_2 = 10\, \text{mL} \times 0.080\, \text{g/mL} = 0.800\, \text{g}\)

Using the molar mass of magnesium hydroxide (58.32 g/mol), we convert the mass to moles:

• Moles of \(\text{Mg(OH)}_2 = \frac{0.800\, \text{g}}{58.32\, \text{g/mol}} \approx 0.0137\, \text{mol}\)

Understanding molar calculations enables us to predict how much of a given product can be obtained from certain reactants, or how much reactant is required to produce a certain amount of product.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It ensures that reactions adhere to the law of conservation of mass, meaning matter is neither created nor destroyed.

In the reaction involving magnesium hydroxide and hydrochloric acid \(\text{HCl}\), stoichiometry tells us how much HCl is needed to completely react with \(\text{Mg(OH)}_2\). The balanced equation is:

• \(\text{Mg(OH)}_2(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{MgCl}_2(\text{aq}) + 2\text{H}_2\text{O}(\text{l})\)

This indicates that 1 mole of \(\text{Mg(OH)}_2\) reacts with 2 moles of \(\text{HCl}\). If we have 0.0137 moles of \(\text{Mg(OH)}_2\), we need \(2 \times 0.0137 = 0.0274\) moles of HCl. Using the molarity equation \(M = \frac{n}{V}\), where \(n\) is moles and \(V\) is volume in liters, we can find the volume required:

• \(V = \frac{0.0274\, \text{mol}}{0.035\, \text{mol/L}} \approx 0.783\, \text{L}\)
• Convert liters to milliliters: \(0.783\, \text{L} \times 1000\, \text{mL/L} = 783\, \text{mL}\)

Thus, stoichiometry allows for precise measurement and resource management in practical applications, such as pharmaceuticals and industrial processes.