Question

When excess solid \(\mathrm{Mg}(\mathrm{OH})_{2}\) is shaken with \(1.00 \mathrm{~L}\) of \(1.0 \mathrm{M}\) \(\mathrm{NH}_{4} \mathrm{Cl}\) solution, the resulting saturated solution has \(\mathrm{pH}=9.00\). Calculate the \(K_{\mathrm{sp}}\) of \(\mathrm{Mg}(\mathrm{OH})_{2}\).

Short answer

The solubility product constant \(K_{\mathrm{sp}}\) of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is \(5 \times 10^{-16}\).

Step-by-step solution

Write the Equilibrium Expression

Start by writing the dissociation reaction of \(\mathrm{Mg}(\mathrm{OH})_{2}\) in water: \[\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Mg}^{2+}(aq) + 2\mathrm{OH}^{-}(aq)\] Next, write the expression for the solubility product constant, \(K_{\mathrm{sp}}\): \[K_{\mathrm{sp}} = [\mathrm{Mg}^{2+}][\mathrm{OH}^{-}]^{2}\]

Calculate \(\mathrm{OH}^{-}\) Concentration

Since \(\mathrm{pH} = 9.00\), we can calculate the \(\mathrm{pOH}\) using the relationship \(\mathrm{pH} + \mathrm{pOH} = 14\). This gives us: \[\mathrm{pOH} = 14 - 9.00 = 5.00\] The concentration of hydroxide ions, \(\mathrm{OH}^{-}\), is then: \[ [\mathrm{OH}^{-}] = 10^{-\mathrm{pOH}} = 10^{-5.00} \]

Determine Solubility from Hydroxide Ion Concentration

The stoichiometry of the reaction shows that one mole of \(\mathrm{Mg}(\mathrm{OH})_{2}\) produces 2 moles of \(\mathrm{OH}^{-}\). Hence, the solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}\), which is the concentration of \(\mathrm{Mg}^{2+}\) in the solution, will be half of the concentration of \(\mathrm{OH}^{-}\): \[ [\mathrm{Mg}^{2+}] = \frac{1}{2} [\mathrm{OH}^{-}] = \frac{1}{2} \times 10^{-5.00} \]

Calculate the Solubility Product Constant \(K_{\mathrm{sp}}\)

Substitute the values of \(\mathrm{Mg}^{2+}\) and \(\mathrm{OH}^{-}\) into the \(K_{\mathrm{sp}}\) expression: \[K_{\mathrm{sp}} = [\mathrm{Mg}^{2+}][\mathrm{OH}^{-}]^{2} = (\frac{1}{2} \times 10^{-5.00})(10^{-5.00})^2\] After calculating the values, the \(K_{\mathrm{sp}}\) of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is: \[K_{\mathrm{sp}} = (0.5 \times 10^{-5})(10^{-10}) = 5 \times 10^{-16}\]

Key concepts

Equilibrium Expressions

Understanding equilibrium expressions is fundamental when dealing with the solubility of substances in chemistry. At the heart of any equilibrium situation, like a saturated solution, is the idea that the rates of the forward and reverse reactions are equal, creating a state of balance. The equilibrium expression, represented as the equilibrium constant \( K \), quantifies this state.

For a general reaction where \(aA + bB \rightleftharpoons cC + dD\), the equilibrium expression is \(K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\), with the concentration terms raised to the power of their stoichiometric coefficients. In the context of solubility, we use the solubility product constant \(K_{\text{sp}}\), which is specific to the dissolution of solids in liquids, forming a saturated solution. In our exercise, \(K_{\text{sp}} = [\text{Mg}^{2+}][\text{OH}^-]^2\) captures the relationship between the concentrations of the dissolved ions of magnesium hydroxide in water.

pH and pOH Calculations

The pH and pOH are logarithmic measures of the concentrations of hydronium \(\text{H}_3\text{O}^+\) and hydroxide \(\text{OH}^-\) ions in a solution, respectively. They are indicators of how acidic or basic a solution is. The pH scale ranges from 0 to 14, with 7 being neutral, values less than 7 indicating acidity, and values greater than 7 indicating basicity.

Mathematically, \(pH = -\log[\text{H}_3\text{O}^+]\) and \(pOH = -\log[\text{OH}^-]\). The two are related by the equation \(pH + pOH = 14\), given the self-ionization of water \(K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1 \times 10^{-14} \). In the provided exercise, we leveraged the pH value to find the pOH, which then allowed us to calculate the hydroxide ion concentration, crucial for finding the solubility product constant.

Stoichiometry in Chemical Reactions

Stoichiometry refers to the quantitative relationship between the amounts of reactants and products in a chemical reaction. It is based on the law of conservation of mass and the concept of the mole. When we write a balanced chemical equation, we are establishing a ratio of moles of reactants to moles of products.

Stoichiometry allows us to predict the amounts of substances consumed and produced in a reaction. This has direct applications in determining the solubility of substances. In our problem, knowing that one formula unit of \(\text{Mg}(\text{OH})_2\) dissociates to produce one \(\text{Mg}^{2+}\) and two \(\text{OH}^-\) ions, we could establish a link between the concentration of hydroxide ions and the solubility of magnesium hydroxide, which was fundamental to calculating \(K_{\text{sp}}\).

Dissociation Reactions in Chemistry

Dissociation reactions are processes where compounds break apart into smaller components, typically ions, when dissolved in a solution. For ionic solids like salts and some bases, this dissociation is a physical process that can be reversible, leading to the state of dynamic equilibrium we describe with solubility product constants.

In the case of \(\text{Mg}(\text{OH})_2\), we looked at the dissociation into \(\text{Mg}^{2+}\) and \(\text{OH}^-\) ions. Understanding these reactions is crucial because they help predict the behavior of substances in various chemical environments. Solubility equilibria, such as the one described by \(K_{\text{sp}}\), allow us to understand at what point a solution becomes saturated and what concentration of ions we can expect in that saturated solution.