Question

In the production of magnesium, \(\mathrm{Mg}(\mathrm{OH})_{2}\) is precipitated by using \(\mathrm{Ca}(\mathrm{OH})_{2},\) which itself is "insoluble." (a) Use \(K_{\mathrm{ip}}\) values to show that \(\mathrm{Mg}(\mathrm{OH})_{2}\) can be precipitated from seawater in which \(\left[\mathrm{Mg}^{2+}\right]\) is initially \(0.052 \mathrm{M}\). (b) If the seawater is saturated with \(\mathrm{Ca}(\mathrm{OH})_{2},\) what fraction of the \(\mathrm{Mg}^{2+}\) is precipitated?

Short answer

Most of the \( \text{Mg}^{2+} \) precipitates from seawater with a remaining fraction of approximately 7.56 \times 10^{-7}.

Step-by-step solution

Write the Given Data

Given the initial concentration of \(\text{Mg}^{2+}\) in seawater is 0.052 M. The solubility product constant (\(K_{\text{sp}}\)) for \(\text{Mg(OH)}_2\) is \[K_{\text{sp}}(\text{Mg(OH)}_2) = 1.8 \times 10^{-11}\]. The solubility product constant for \(\text{Ca(OH)}_2\) is \[K_{\text{sp}}(\text{Ca(OH)}_2) = 5.5 \times 10^{-6}\].

Determine Hydroxide Ion Concentration from \( \text{Ca(OH)}_2 \)

Since \( \text{Ca(OH)}_2 \) is insoluble, it dissociates to a small extent in water. The dissociation reaction is \[ \text{Ca(OH)}_2(s) \rightarrow \text{Ca}^{2+}(aq) + 2 \text{OH}^-(aq) \]. Assume that the concentration of \( \text{Ca}^{2+} \) in seawater is very low compared to \( \text{OH}^- \) ions. Thus, the hydroxide ion concentration is given by \[ \text{[OH]^-} = \text{(solubility of } \text{Ca(OH)}_2\text{)} \times 2 \]. Solving for hydroxide ion concentration: \[ K_{\text{sp}} = [\text{Ca}^{2+}][\text{OH}^{-}]^2 = (s)(2s)^2 = 4s^3 \]. Solving for \(s\): \[ s = \frac{\text{[OH]}^-}{2} = \text{(solubility of } \text{Ca(OH)}_2\text{)} = 1.07 \times 10^{-2} \text{M} \]. Therefore, \[ [\text{OH}]^- = 2 \times 1.07 \times 10^{-2} = 2.14 \times 10^{-2} \text{M} \].

Establish the Ion Product (Q) for \( \text{Mg(OH)}_2 \)

The ion product (Q) for \( \text{Mg(OH)}_2 \) is defined by the equation: \[ Q = [\text{Mg}^{2+}][\text{OH}^-]^2 \]. Substituting the values: \[ Q = (0.052)(2.14 \times 10^{-2})^2 = (0.052)(4.58 \times 10^{-4}) = 2.38 \times 10^{-5} \].

Compare Ion Product (Q) to Solubility Product (K_sp)

Compare the calculated ion product, Q, to the solubility product constant \(K_{\text{sp}}\) of \( \text{Mg(OH)}_2 \): \[ Q = 2.38 \times 10^{-5} > K_{\text{sp}}(\text{Mg(OH)}_2) = 1.8 \times 10^{-11} \]. Since \( Q > K_{\text{sp}} \), \( \text{Mg(OH)}_2 \) precipitates from seawater.

Calculate the Fraction of \( \text{Mg}^{2+} \) Precipitated

Let \( x \) represent the moles of \( \text{Mg(OH)}_2 \) that precipitate per liter of seawater. The equilibrium concentration of \( \text{Mg}^{2+} \) ions remaining in solution is \[ (0.052 - x) \text{ M} \], and for \( \text{OH}^- \) it is \[ (2.14 \times 10^{-2} - 2x) \text{ M} \]. At equilibrium, \[ 1.8 \times 10^{-11} = (0.052 - x)(2.14 \times 10^{-2} - 2x)^2 \]. Solving this quadratic equation, \( x \) can be found. Given \[ x \text{ is very small compared to the initial concentrations} \], then considering \[ x \] negligible, \[ x \text{ approximately equates to } 1.8 \times 10^{-11}/(2.14 \times 10^{-2})^2\text{ .} = 1.8 \times 10^{-11}/(4.58 \times 10^{-4}) ≈ 3.93 \times 10^{-8} M \].

Calculate the Fraction Precipitated

The fraction of the initial \( \text{Mg}^{2+} \) concentration that is precipitated is given by: \[ \text{Fraction} = \frac{x}{\text{Initial} \text{[Mg}^{2+}\text{]}} ≈ \frac{3.93 \times 10^{-8} \text{ M}}{0.052 \text{ M}} ≈ 7.56 \times 10^{-7} \]. This implies that almost all of the \( \text{Mg}^{2+} \) has precipitated.

Key concepts

Solubility Product Constant (Ksp)

The solubility product constant, denoted as \(K_{\text{sp}}\), is a measure of the solubility of a compound. Specifically, it is an equilibrium constant for the dissociation of a solid substance into its aqueous ions. For a general salt \( \text{AB}_2 \), which dissociates as \( \text{AB}_2(s) \rightarrow \text{A}^{2+}(aq) + 2\text{B}^-(aq) \), the \(K_{\text{sp}}\) expression would be:
\[ K_{\text{sp}} = [\text{A}^{2+}][\text{B}^-]^2 \]
The value of \( K_{\text{sp}} \) indicates how much of the solid will dissolve in water. A higher \( K_{\text{sp}} \) means the compound is more soluble. In the given exercise, the \( K_{\text{sp}} \) values of \[ K_{\text{sp}}(\text{Mg(OH)}_2) = 1.8 \times 10^{-11} \text{ and } K_{\text{sp}}(\text{Ca(OH)}_2) = 5.5 \times 10^{-6} \] are used to determine the solubility and potential for precipitation.

Ion Product (Q)

The ion product, \( Q \), is similar to the solubility product constant but applies to any given set of initial concentrations, not just at equilibrium.
For a compound like \( \text{Mg(OH)}_2 \), which dissociates as \( \text{Mg(OH)}_2(s) \rightarrow \text{Mg}^{2+}(aq) + 2\text{OH}^-(aq) \), the ion product is calculated as:
\[ Q = [\text{Mg}^{2+}][\text{OH}^{-}]^2 \]
By comparing \( Q \) to \( K_{\text{sp}} \), one can predict whether precipitation will occur:

• If \( Q < K_{\text{sp}} \): the solution is unsaturated, and no precipitation occurs.
• If \( Q = K_{\text{sp}} \): the solution is saturated, and at equilibrium.
• If \( Q > K_{\text{sp}} \): the solution is supersaturated, leading to precipitation.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the backward reaction, and the concentrations of the reactants and products remain constant over time. This concept is crucial in understanding both the solubility product constant (\( K_{\text{sp}} \)) and the ion product (\( Q \)).
In a saturated solution of \( \text{Mg(OH)}_2 \), the following equilibrium exists:
\[ \text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\text{OH}^-(aq) \]
At equilibrium, \[ K_{\text{sp}} = [\text{Mg}^{2+}][\text{OH}^{-}]^2 \]
This concept helps explain the balance between dissolved ions and undissolved solid, which is effectively used to solve problems of precipitation.

Molar Concentration

Molar concentration, also known as molarity (M), is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution.
In the context of precipitation reactions, molar concentration helps in calculating the ion product \( Q \) and comparing it to the solubility product constant \( K_{\text{sp}} \). For instance, the initial molar concentration of \( \text{Mg}^{2+} \) in seawater given in the problem is 0.052 M.
The concentration of hydroxide ions, \( \text{OH}^- \), from saturated \( \text{Ca(OH)}_2 \) can be found using its solubility:
\[ [\text{OH}^-] = 2 \times (\text{solubility of } \text{Ca(OH)}_2) \ \]
Thus, the molarity of various ions is crucial in calculating the likelihood of precipitation.

Precipitation Reactions

Precipitation reactions occur when ions in a solution combine to form an insoluble compound or precipitate. This happens when the ion product, \( Q \), of the insoluble compound exceeds its solubility product constant, \( K_{\text{sp}} \).
For example, in the given exercise, \( \text{Mg(OH)}_2 \) precipitates from seawater when hydroxide ions are introduced from \( \text{Ca(OH)}_2 \). The steps are:

• Calculate the hydroxide ion concentration from \( \text{Ca(OH)}_2 \)
• Calculate the ion product \( Q \) for \( \text{Mg(OH)}_2 \)
• Compare \( Q \) to \( K_{\text{sp}} \) of \( \text{Mg(OH)}_2 \)
• If \( Q > K_{\text{sp}} \), \( \text{Mg(OH)}_2 \) will precipitate.

Understanding precipitation reactions is vital in predicting the formation of precipitates in chemical solutions.