Problem 48 - Chemistry An Atoms First Approach

Question

The concentration of \(\mathrm{Mg}^{2+}\) in seawater is \(0.052 \mathrm{M}\). At what \(\mathrm{pH}\) will \(99 \%\) of the \(\mathrm{Mg}^{2+}\) be precipitated as the hydroxide salt? \(\left[K_{\mathrm{sp}} \text { for } \mathrm{Mg}(\mathrm{OH})_{2}=8.9 \times 10^{-12} .\right]\)

Step-by-Step Solution

Answer

When \(99\%\) of the \(\mathrm{Mg}^{2+}\) is precipitated as the hydroxide salt in seawater, the remaining concentration of \(\mathrm{Mg}^{2+}\) is \(5.2 \times 10^{-4} \mathrm{M}\). At the point of precipitation, the ion product (Q) must be equal to the solubility product (\(K_{\mathrm{sp}}\)). Solving for the concentration of \(\mathrm{OH}^{-}\) and using the relationship between pH and \([\mathrm{H}^{+}]\), we find that the pH at which \(99 \%\) of the magnesium ions will be precipitated is approximately \(10.62\).

Step 1: Determine the concentration of \(\mathrm{Mg}^{2+}\) after precipitation

We are given that \(99\%\) of the magnesium ions will be precipitated. We need to find the remaining concentration of \(\mathrm{Mg}^{2+}\) in the solution after the precipitation. Let's calculate it as follows: After precipitation, Remaining concentration of \(\mathrm{Mg}^{2+} = 0.052 \mathrm{M} \times (1 - 0.99) = 0.052 \mathrm{M} \times 0.01 = 5.2 \times 10^{-4} \mathrm{M}\).

Step 2: Write the solubility product expression (\(K_{\mathrm{sp}}\)) and ion product (Q) for Mg(OH)2

The balanced equation for the dissolution of \(\mathrm{Mg(OH)_2}\) is: \(\mathrm{Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2 OH^{-}}\) Then, the solubility product expression (\(K_{\mathrm{sp}}\)) for \(\mathrm{Mg(OH)_2}\) is: \(K_{\mathrm{sp}} = [\mathrm{Mg}^{2+}][\mathrm{OH}^{-}]^2\) We are given that \(K_{\mathrm{sp}} = 8.9 \times 10^{-12}\). Now, let's find the ion product (Q), calculated as: \(Q = [\mathrm{Mg}^{2+}][\mathrm{OH}^{-}]^2\) We don't know the concentration of \(\mathrm{OH^-}\) ions directly, but we can use pH to represent it indirectly. Since we know \([\mathrm{Mg}^{2+}] = 5.2 \times 10^{-4} \mathrm{M}\), \(Q = (5.2 \times 10^{-4})([\mathrm{OH}^{-}])^2\) Note that \(Q\) must be equal to \(K_{\mathrm{sp}}\) when the precipitation occurs.

Step 3: Relate the concentration of OH- to pH and solve for pH

Since we know the relationship between \([\mathrm{OH}^{-}]\) and \([\mathrm{H}^{+}]\): \(K_{\mathrm{w}} = [\mathrm{H}^{+}][\mathrm{OH}^{-}]\), where \(K_{\mathrm{w}} = 1 \times 10^{-14}\) at \(25 ^{\circ}\mathrm{C}\), and \(pH = -\log_{10}{[\mathrm{H}^{+}]}\) We can solve for \([\mathrm{OH}^{-}]\) and substitute it into the Q equation. First, express \([\mathrm{H}^{+}]\) in terms of \([\mathrm{OH}^{-}]\): \([\mathrm{H}^{+}] = \dfrac{K_w}{[\mathrm{OH}^{-}]}\) Substitute this into the pH equation: \(pH = -\log_{10}{\left(\dfrac{K_w}{[\mathrm{OH}^{-}]}\right)}\) Now, let's plug the value of Q in the equation \(Q = K_{\mathrm{sp}}\): \((5.2 \times 10^{-4})([\mathrm{OH}^{-}])^2 = 8.9 \times 10^{-12}\) Solve for \([\mathrm{OH}^{-}]\): \([\mathrm{OH}^{-}] = \sqrt{\dfrac{8.9 \times 10^{-12}}{5.2 \times 10^{-4}}} = 1.3 \times 10^{-4} \mathrm{M}\) Now plug this into the pH equation: \(pH = -\log_{10}{\left(\dfrac{1 \times 10^{-14}}{1.3 \times 10^{-4}}\right)} \approx 10.62\) Therefore, the pH when \(99\%\) of the \(\mathrm{Mg}^{2+}\) will be precipitated as hydroxide salt is approximately \(10.62\).

Key Concepts

Precipitation

In seawater chemistry, precipitation occurs when a solid substance forms and separates from a liquid solution. In this context, it refers to the process of magnesium ions (\(\mathrm{Mg^{2+}}\)) combining with hydroxide ions (\(\mathrm{OH^{-}}\)) to form a solid compound called magnesium hydroxide (\(\mathrm{Mg(OH)_2}\)). This phenomenon happens when the concentration of ions exceeds what the solution can hold in a dissolved state.

• When \(99\%\) of the \(\mathrm{Mg^{2+}}\) ions are precipitated, it signifies a near-total removal from the solution.
• The remaining \(\mathrm{Mg^{2+}}\) concentration helps us determine the conditions needed for this process.
• Understanding precipitation is crucial as it impacts the availability of these ions for biological and chemical processes in ocean water.

The remaining concentration is calculated as \(5.2 \times 10^{-4}\, \mathrm{M}\), which tells us how much \(\mathrm{Mg^{2+}}\) is left in the solution after \(99\%\) is removed as a solid.

Solubility Product

The solubility product, denoted as \(K_{\mathrm{sp}}\), is an equilibrium constant that provides insight into the solubility of a sparingly soluble compound in a solution. It reflects the product of the concentrations of the ions each raised to the power of their stoichiometric coefficients in the balanced dissolution equation.
For \(\mathrm{Mg(OH)_2}\):

• The dissolution can be expressed as: \(\mathrm{Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2 OH^{-}}\)
• The solubility product expression is: \(K_{\mathrm{sp}} = [\mathrm{Mg^{2+}}][\mathrm{OH^{-}}]^2\)
• The given \(K_{\mathrm{sp}}\) is \(8.9 \times 10^{-12}\)

Using the \(K_{\mathrm{sp}}\) equation helps determine the necessary concentration of \(\mathrm{OH^{-}}\) for precipitation to occur by comparing it to the ion product (\(Q\)). When \(Q = K_{\mathrm{sp}}\), the solution is perfectly saturated, and precipitation begins. This balance is essential for predicting solubility under varying conditions such as pH changes.

pH Calculation

The pH value indicates the acidity or basicity of an aqueous solution, a crucial factor when observing precipitation in seawater chemistry. Calculating pH involves determining the concentration of hydrogen ions (\(\mathrm{H^+}\)) in the solution. In the case of magnesium precipitation, pH influences the concentration of hydroxide ions (\(\mathrm{OH^{-}}\)):

• pH is calculated using the formula: \(pH = -\log_{10}[\mathrm{H^+}]\)
• The relationship between \(\mathrm{H^+}\) and \(\mathrm{OH^{-}}\) is given by \(K_{\mathrm{w}} = [\mathrm{H^+}][\mathrm{OH^{-}}]\)
• At \(25^\circ\mathrm{C}\), \(K_{\mathrm{w}} = 1 \times 10^{-14}\)

In this scenario, the calculation shows that the pH must reach approximately \(10.62\) for \(99\%\) of \(\mathrm{Mg^{2+}}\) ions to precipitate as \(\mathrm{Mg(OH)_2}\). This involves solving for \(\mathrm{OH^{-}}\) concentration using \(Q = K_{\mathrm{sp}}\) and then finding \(\mathrm{H^+}\) concentration to determine the pH. It demonstrates how pH manipulation can control precipitation processes in seawater.