Question

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

Short answer

The pH at which 99% of the Mg2+ will be precipitated as the hydroxide salt is approximately 10.12.

Step-by-step solution

Calculate the Concentration of Mg2+ Ions After 99% Precipitation

To find the concentration of Mg2+ ions after 99% precipitation, we use the original concentration and the percentage given: \(0.99 \times 0.052 M = 0.05148 M\) So, the concentration of Mg2+ ions that would remain in the solution is: \(0.052 M - 0.05148 M = 0.00052 M\)

Set Up the Ion Product Expression for Mg(OH)2

The ion product expression (Q) for Mg(OH)2 is given by \(Q = [Mg^{2+}][OH^-]^2\) Now, we know that at the point of precipitation, Q = Ksp. Therefore \(K_{sp} = [Mg^{2+}][OH^-]^2\) We know the Ksp value for Mg(OH)2 is 8.9 x 10^(-12) and the concentration of Mg2+ left in the solution is 0.00052 M. So, \(8.9 \times 10^{-12} = [0.00052 M][OH^-]^2\)

Solve for the Concentration of OH- Ions

Now, we can solve for the concentration of OH- ions. \([OH^-]^2 = \frac{8.9 \times 10^{-12}}{0.00052 M}\) \([OH^-]^2 = 1.71 \times 10^{-8}\) Taking the square root of both sides, we get: \([OH^-] \approx 1.31 \times 10^{-4}\)

Calculate the pH Using the Concentration of OH- Ions

We can now use the concentration of OH- ions to calculate the pH of the solution. First, we need to find the pOH. \(pOH = -\log_{10}(OH^-)\) \(pOH = -\log_{10}(1.31 \times 10^{-4})\) \(pOH \approx 3.88\) Now, we can find the pH using the relation, \(pH + pOH = 14\) Thus, \(pH = 14 - pOH\) \(pH = 14 - 3.88\) \(pH \approx 10.12\) #Conclusion#: The pH at which 99% of the Mg2+ will be precipitated as the hydroxide salt is approximately 10.12.

Key concepts

Ksp (Solubility Product Constant)

The solubility product constant (Ksp) helps us understand the extent to which a compound will dissolve in water. It is a specific type of equilibrium constant that applies to the dissolution of sparingly soluble compounds. This constant is crucial to predict whether a salt will dissolve or precipitate in a solution under given conditions. It is important to remember that a lower Ksp value indicates a less soluble compound.

In our exercise, we are concerned with the solubility of the salt Mg(OH)₂. Given the Ksp for Mg(OH)₂ is 8.9 x 10⁻¹², this indicates the product of the molar concentrations of the ions, \[ Mg^{2+} \] and \[ (OH^-)^2 \], in the saturated solution should equal the Ksp when equilibrium is reached. The task here is to determine how this equilibrium is achieved when 99% of the original \[ Mg^{2+} \] ions precipitate.

Precipitation Reactions

Precipitation reactions involve the formation of an insoluble solid from the reaction of two aqueous solutions. This type of reaction occurs when the ion product of the reacting ions exceeds the solubility product constant (Ksp) of the resulting compound. The precipitate forms because the ions can no longer remain dissolved.Given our example, once 99% of \[ Mg^{2+} \] ions form Mg(OH)₂, the concentration that remains dissolved is crucial for calculating precipitation conditions. When these conditions are met, the solution can no longer support the solubility of additional ions, and a solid \[ Mg(OH)₂ \] forms as a precipitate.

In this specific situation, when the concentration of \[ [Mg^{2+}] = 0.00052 \] M is remaining after 99% precipitation, the value for \[ [OH^-]^2 \] can be calculated to ensure the product of \[ [Mg^{2+}][OH^-]^2 \] does not exceed the Ksp value, indicating the precise conditions where precipitation fully occurs.

pH Calculations

The pH of a solution is a measure of its acidity or basicity. Calculating pH is critical when dealing with precipitation reactions in alkaline conditions, as seen in the example problem with Mg(OH)₂.Here, the focus shifts to the concentration of hydroxide ions \[ [OH^-] \] in the solution. First, we find \[ [OH^-] \] using the rearranged form of the Ksp expression \[ [OH^-]^2 = \frac{8.9 \times 10^{-12}}{0.00052 M} \]. Upon calculating, we find \[ [OH^-] \approx 1.31 \times 10^{-4} \], which is then used to find the pOH by using the formula \[ pOH = -\log_{10}(OH^-) \]. Meanwhile, the pH can be determined from the pOH using the relation: \[ pH + pOH = 14 \]. Thus, in this exercise, finding that a pH of approximately 10.12 will ensure 99% precipitation of \[ Mg^{2+} \] reflects basic conditions necessary for effective precipitation of Mg(OH)₂.