Problem 72 - General Chemistry: Principles and Modern Applications

Question

Calculate the molar solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}\) in \(1.00 \mathrm{M}\) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq})\).

Step-by-Step Solution

Answer

The molar solubility solution involves setting up the equilibrium based on the problem, using the given \(K_{sp}\), and solving for \(s\). The exact numeric value will depend on the approximation or solution method used for the quadratic equation, but should be in the molar range.

Step 1: Write the Chemical Equation and Expressions for Molar Solubility and Solubility Product

The balance chemical equation for the dissolution of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is: \(\mathrm{Mg(OH)_{2}(s)} \rightleftharpoons \mathrm{Mg^{2+}(aq)} + 2\mathrm{OH^{-}(aq)}\). Let \(s\) represent the molar solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}\) in \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq})\). The concentration of \(\mathrm{Mg}^{+2}(aq)\) will be \(s\) and \(\mathrm{OH}^{-}\) will be \(2s\). The solubility product expression for \(\mathrm{Mg(OH)_{2}\) is: \(K_{sp} = [\mathrm{Mg^{2+}}][\mathrm{OH}^-]^{2}\)

Step 2: Use the \(K_{sp}\) value of \(\mathrm{Mg(OH)_{2}\)

The \(K_{sp}\) value of \(\mathrm{Mg(OH)_{2}\) at room temperature is \(5.61 \times 10^{-12}\). Substitute these values into the \(K_{sp}\) equation: \(5.61 \times 10^{-12} = (s)(2s)^2\)

Step 3: Solve for Molar Solubility

Now simplify the equation, reduce it to a quadratic form, and solve for \(s\) to find the molar solubility of \(\mathrm{Mg(OH)_{2}\) in \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq})\). Remember to check if the quadratic approximation is valid; if not, use the quadratic formula to solve the equation.

Key Concepts

Chemical Equilibrium

In chemistry, equilibrium refers to the state where the forward and reverse reactions occur at the same rate. Hence, the concentrations of reactants and products remain constant over time. This concept is fundamental when dealing with reactions that do not go to completion, such as the dissolution of sparingly soluble salts.

For example, when \(\mathrm{Mg(OH)_2(s)}\) dissolves in water, an equilibrium is established between solid \(\mathrm{Mg(OH)_2}\) and its ions in solution: \(\mathrm{Mg^{2+}}\) and \(\mathrm{OH}^-\). The balanced equation for this process is:

• \(\mathrm{Mg(OH)_{2}(s)} \rightleftharpoons \mathrm{Mg^{2+}(aq)} + 2\mathrm{OH^{-}(aq)}\)

This equilibrium can be shifted by changing conditions, like the concentration of ions in the solution or temperature. In the given exercise, \(\mathrm{NH}_4\mathrm{Cl}\) can affect this equilibrium by influencing the dissolved ion concentration. Therefore, understanding chemical equilibrium helps predict and calculate the impact on molar solubility.

Solubility Product Constant (Ksp)

The solubility product constant, \(K_{sp}\), is a vital concept in understanding the solubility of ionic compounds. It is a special type of equilibrium constant that applies to the dissolution of sparingly soluble salts. \(K_{sp}\) represents the level at which a compound will dissolve in water to form a saturated solution.

For \(\mathrm{Mg(OH)_2}\), the \(K_{sp}\) expression is:

• \(K_{sp} = [\mathrm{Mg^{2+}}][\mathrm{OH}^-]^2\)

This expression shows that the solubility product is the product of the molar concentrations of its ions, each raised to the power of their respective coefficients in the dissolution equation. The known \(K_{sp}\) for \(\mathrm{Mg(OH)_{2}}\) at room temperature, \(5.61 \times 10^{-12}\), helps calculate its molar solubility in different conditions, such as in the presence of other aqueous ions.

Quadratic Equations in Chemistry

Solving quadratic equations is a common mathematical necessity in chemistry, especially when calculating the concentrations of ions involved in equilibrium processes. In this context, the molar solubility equation involving \(K_{sp}\) often reduces to a quadratic form.

In the exercise provided, the dissolution of \(\mathrm{Mg(OH)_2}\) results in the expression:

• \(5.61 \times 10^{-12} = (s)(2s)^2\)

Simplifying, we get a quadratic equation in terms of \(s\):

• \(5.61 \times 10^{-12} = 4s^3\)

To solve for \(s\), the molar solubility, one might first check if a simplification can avoid using the quadratic formula. If not, apply the quadratic formula:

• \(s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where a, b, and c are coefficients from the equation in standard form \(ax^2 + bx + c = 0\). Mastery of quadratic equations thus enables chemists to solve diverse solubility and equilibrium problems efficiently.