Question

The solubility product of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is \(1.2 \times 10^{-11} .\) What minimum \(\mathrm{OH}^{-}\) concentration must be attained (e.g., by adding \(\mathrm{NaOH}\) ) to decrease the Mg concentration in a solution of \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)\), to less than \(1.0 \times 10^{-10} \mathrm{M} ?\)

Short answer

The minimum \([\text{OH}^-]\) concentration required is approximately 0.346 M.

Step-by-step solution

Understand the Ksp Equation

The solubility product constant (Ksp) expression for \(\text{Mg(OH)}_2\) dissolution is given by \[\text{Mg(OH)}_2 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + 2\text{OH}^- (aq)\] and the corresponding Ksp expression is \[K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2\]. Here, \(K_{sp} = 1.2 \times 10^{-11}\).

Substitute the Required Mg Concentration

We need the concentration of \(\text{Mg}^{2+}\) to be less than \(1.0 \times 10^{-10} \text{ M}\). Substitute \([\text{Mg}^{2+}]=1.0 \times 10^{-10}\) into the Ksp expression: \[1.2 \times 10^{-11} = (1.0 \times 10^{-10}) [\text{OH}^-]^2\].

Solve for \([\text{OH}^-]\)

Rearrange the equation to find \([\text{OH}^-]\): \[[\text{OH}^-]^2 = \frac{1.2 \times 10^{-11}}{1.0 \times 10^{-10}}\]. Simplify to \[[\text{OH}^-]^2 = 1.2 \times 10^{-1}\]. Take the square root of both sides to get \([\text{OH}^-] = \sqrt{1.2 \times 10^{-1}}\).

Compute the \([\text{OH}^-]\) Concentration

Calculate \(\sqrt{1.2 \times 10^{-1}}\): \[[\text{OH}^-] = \sqrt{0.12} \approx 0.346\]. Therefore, the minimum concentration required is \([\text{OH}^-] = 0.346 \text{ M}\).

Key concepts

Solubility Product Constant

The solubility product constant, often denoted as **Reactant Balance**: Often in equilibrium reactions, adding more one side can affect the overall concentration and shift the equilibrium position. When working with Ksp, it represents the point at which a solution reaches its maximum concentration of dissolved ions without precipitation. If the ion products exceed this constant, the excess tends to precipitate out. This is the principle behind manipulating concentrations to form precipitates or dissolve substances.

Mg(OH)2

**Magnesium Hydroxide** Magnesium hydroxide, **Application** Insolubility plays a significant role when it comes to treating wastewater or managing metal concentrations in solutions. By knowing how Additionally, Mg(OH) Typically, when dealing with substances like Mg(OH)

OH- Concentration

**Understanding OH- Role** Increased **Calculation Process** The calculation begins with the Ksp expression of a given compound. For example, in Mg(OH) There is a direct relationship in the formula between the hydroxide ions and the metallic ion concentrations: If one aims to decrease the concentration of the metallic ion, they can do so by increasing the hydroxide ions, effectively pushing the equilibrium to the left, forming more solid precipitate and reducing the soluble metal ion.

Equilibrium Expressions

**Basics of Equilibrium Expressions** Equilibrium expressions provide a formulaic representation of a reaction at equilibrium and factor in the concentrations of reactants and products. For instance, in dissolving Mg(OH) **Solving Techniques** When faced with a problem, set up your expression carefully. Plug in known values and solve for the unknown, always paying attention to the stoichiometry. By mastering equilibrium expressions, you can predict how changes in conditions will affect your reaction, including shifts caused by changes in concentration, pressure, or temperature.