Question

The concentrations of various cations in seawater, in moles per liter, are $$ \begin{array}{llllll} \hline \text { Ion } & \mathrm{Na}^{+} & \mathrm{Mg}^{2+} & \mathrm{Ca}^{2+} & \mathrm{A}^{3+} & \mathrm{Fe}^{3+} \\ \text { Molarity }(M) & 0.46 & 0.056 & 0.01 & 4 \times 10^{-7} & 2 \times 10^{-7} \\ \hline \end{array} $$ (a) At what \(\left[\mathrm{OH}^{-}\right]\) does \(\mathrm{Mg}(\mathrm{OH})_{2}\) start to precipitate? (b) At this concentration, will any of the other ions precipitate? (c) If enough \(\mathrm{OH}^{-}\) is added to precipitate \(50 \%\) of the \(\mathrm{Mg}^{2+}\), what percentage of each of the other ions will precipitate? (d) Under the conditions in (c), what mass of precipitate will be ob- tained from one liter of seawater?

Short answer

The concentration of OH- at which Mg(OH)2 starts to precipitate is approximately \(3.54 \times 10^{-6}M\). b) Will any other ions precipitate at this OH- concentration? No other ions will precipitate at this OH- concentration. c) What is the percentage of each ion that will precipitate if enough OH- is added to precipitate 50% of the Mg2+ ions? Only Mg2+ ions will precipitate, and 50% of them will do so. No other ions will precipitate at this OH- concentration. d) What is the mass of precipitate obtained from one liter of seawater under the conditions in part (c)? The mass of precipitate obtained from one liter of seawater under the conditions in part (c) is approximately 1.63 grams.

Step-by-step solution

Calculate the solubility product constant (Ksp) for Mg(OH)2

The Ksp values for Mg(OH)2 is given as \(K_{sp} = [Mg^{2+}][OH^-]^2\). We are given the concentration of Mg2+ as 0.056 M. To find the concentration of OH- at which Mg(OH)2 starts to precipitate, we need to solve the Ksp expression for [OH-]: $$ [OH^-] = \sqrt{\frac{K_{sp}}{[Mg^{2+}]}}$$ Using \(K_{sp} = 5.61 \times 10^{-12}\) for Mg(OH)2 and [Mg2+] = 0.056 M: $$ [OH^-] = \sqrt{\frac{5.61 \times 10^{-12}}{0.056}} \approx 3.54 \times 10^{-6} M$$ So, the concentration of OH- at which Mg(OH)2 starts to precipitate is approximately \(3.54 \times 10^{-6}M\). (b)

Determine if other ions will precipitate

We need to compare the Ksp values of the precipitates for other ions (assuming they all form hydroxides) with their actual ion product (Q) at the OH- concentration found in part (a). If Q > Ksp, then the ion will precipitate. Using the given concentrations, we can calculate Q for each ion as follows: $$Q_{NaOH} = [Na^+][OH^-] = 0.46\times 3.54\times 10^{-6}$$ $$Q_{Ca(OH)_2} = [Ca^{2+}][OH^-]^2 = 0.01\times (3.54\times 10^{-6})^2$$ $$Q_{A(OH)_3} = [A^{3+}][OH^-]^3 = 4\times 10^{-7}\times (3.54\times 10^{-6})^3$$ $$Q_{Fe(OH)_3} = [Fe^{3+}][OH^-]^3 = 2\times 10^{-7}\times (3.54\times 10^{-6})^3$$ Now, we compare these values with their respective Ksp values, for example: $$K_{sp} ({Ca(OH)_2}) = 6.5\times 10^{-6}$$ $$K_{sp} ({Fe(OH)_3}) = 2.79\times 10^{-39}$$ $$K_{sp} ({A(OH)_3}) = 3\times 10^{-34}$$ The Ksp value for NaOH is not relevant, as it is a strong base and does not have a significant tendency to precipitate. We compare Q with Ksp and find that none of these values exceeds their respective Ksp values, which means that no other ions will precipitate at this OH- concentration. (c)

Find the percentage of other ions precipitated

When enough OH- is added to precipitate 50% of Mg2+, the concentration of Mg2+ in solution will decrease by half: $$[Mg^{2+}]_{new} = \frac{1}{2} \times 0.056 M$$ At this point, we should recalculate Q for each ion with the new concentration of Mg2+ and considering that the [OH-] required to do this transformation is the same as before (from part (a)): $$Q_{NaOH} = [Na^+][OH^-] = 0.46\times 3.54\times 10^{-6}$$ $$Q_{Ca(OH)_2} = [Ca^{2+}][OH^-]^2 = 0.01\times (3.54\times 10^{-6})^2$$ $$Q_{A(OH)_3} = [A^{3+}][OH^-]^3 = 4\times 10^{-7}\times (3.54\times 10^{-6})^3$$ $$Q_{Fe(OH)_3} = [Fe^{3+}][OH^-]^3 = 2\times 10^{-7}\times (3.54\times 10^{-6})^3$$ We find that still none of these values exceeds their respective Ksp values, which means that no other ions will precipitate at this OH- concentration. (d)

Calculate the mass of precipitate obtained from one liter of seawater

At this point, we know that only Mg(OH)2 is precipitating and that 50% of the Mg2+ ions have precipitated. To find the mass of precipitate, we will use the stoichiometry of the reaction. $$Mg^{2+} (aq) + 2OH^{-} (aq) \rightarrow Mg(OH)_2 (s)$$ Moles of Mg2+ ions precipitated = 0.5 × moles of Mg2+ ions in solution $$= 0.5 \times 0.056 \, mol$$ Since the molar mass of Mg(OH)2 is 58.32 g/mol, the mass of precipitate obtained is: $$Mass \, of \, Mg(OH)_2 = 0.5 \times 0.056 \, mol \times 58.32 \, g/mol = 1.63g$$ Therefore, the mass of precipitate obtained from one liter of seawater under the conditions in part (c) is approximately 1.63 grams.

Key concepts

Ksp calculation

Understanding the solubility product constant, or Ksp, is essential in determining when a compound will begin to precipitate from a solution. This is especially important in the case of sparingly soluble salts like magnesium hydroxide, Mg(OH)\(_2\). The Ksp is an equilibrium constant specific to each compound that describes the level at which the compound will remain dissolved in a solution.
When calculating Ksp for Mg(OH)\(_2\), the equation is given as follows:
\[ K_{sp} = [Mg^{2+}][OH^-]^2 \]
Given the molarity of \([Mg^{2+}] = 0.056 \ M\) and the known Ksp value \(5.61 \times 10^{-12}\), we can determine the concentration of \([OH^-]\) necessary for precipitation. By solving
\[ [OH^-] = \sqrt{\frac{K_{sp}}{[Mg^{2+}]}} \approx 3.54 \times 10^{-6} \ M \]
we discover the threshold concentration of hydroxide ions where Mg(OH)\(_2\) begins to form a solid precipitate.

Ion precipitation

Ion precipitation occurs when ions in a solution exceed their solubility, causing a solid to form. This process is intricately tied to the solubility product constant (Ksp) of a substance. In exercises involving seawater, determining which ions precipitate is crucial.
To identify potential precipitation of other ions in a seawater mixture when magnesium hydroxide starts to precipitate, we calculate the ion product, Q, for each relevant ion. If a calculated Q exceeds the known Ksp value, it suggests precipitation.

• For NaOH: The ion product is typically ignored due to the strong base characteristic of NaOH.
• For Ca(OH)\(_2\): Using coefficients of \([Ca^{2+}]\) and \([OH^-]^2\), we find Q to be less than its Ksp, indicating no precipitation.
• For species like Al\(^{3+}\) and Fe\(^{3+}\), the hydroxide concentrations are small enough that no precipitation occurs unless under special conditions.

Thus, before achieving a significant shift in conditions, these ions will not precipitate at the initial hydroxide ion concentration level.

Molarity in seawater

Molarity, a key measure in chemistry, represents the concentration of a solute dissolved in a solvent, in this case, seawater. Different ions in seawater have different molarities, affecting solubility and precipitation behaviors.
For seawater, the molarity of magnesium ions is set at \(0.056 \ M\), a primary concern here for Mg(OH)\(_2\) precipitation. The molarity directly impacts the solubility product calculations and the extent of ion interactions in the solution.
When discussing precipitation, changes in molarity due to removal of certain ions (such as when half of the magnesium ions precipitate) directly influence the solution chemistry. This understanding helps in predicting the mass of precipitates and the solution's subsequent concentration profile as reactions progress.

Chemical reactions in solution

Chemical reactions in solution are driven by the interactions of dissolved ions. In the real-world setting of seawater, these interactions can initiate reactions like precipitation.
Consider the reaction:
\[ Mg^{2+} (aq) + 2OH^{-} (aq) \rightarrow Mg(OH)_2 (s) \]
This equation summarizes the precipitation of magnesium hydroxide. As hydroxide ions are introduced, reaching a critical concentration, the dissolved magnesium starts to convert into a solid, causing the solution to shift towards equilibrium.
As OH\(^-\) ions increase in seawater, they interact with magnesium cations to form a solid precipitate - modeling real-world occurrences such as scale formation or water treatment processes. Understanding the stoichiometry of this reaction clarifies how much solid forms from a known quantity of initial ions, thus connecting mineral deposition with concentration measurements directly.