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(a) Explain why \(\mathrm{Mg}(\mathrm{OH})_{2}\) precipitates when \(\mathrm{CO}_{3}^{2-}\) ion is added to a solution containing \(\mathrm{Mg}^{2+}\). . (b) Will \(\mathrm{Mg}(\mathrm{OH})_{2}\) precipitate when \(4.0 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is added to \(1.00 \mathrm{~L}\) of a solution containing 125 ppm of \(\mathrm{Mg}^{2+}\) ?

Short Answer

Expert verified
(a) The precipitation of Mg(OH)鈧 occurs when CO鈧兟测伝 ion is added to a solution containing Mg虏鈦 because the reaction shifts in the direction that consumes CO鈧兟测伝 and Mg虏鈦 ions and forms the solid Mg(OH)鈧. This happens as the concentration of Mg虏鈦 and CO鈧兟测伝 exceeds the equilibrium solubility product (Ksp) of Mg(OH)鈧. (b) Since the ion product Q (9.59 x 10鈦宦孤) is greater than Ksp (5.61 x 10鈦宦孤), the precipitation of Mg(OH)鈧 will occur when 4.0 g of Na鈧侰O鈧 is added to 1.00 L of a solution containing 125 ppm of Mg虏鈦.

Step by step solution

01

(a) Precipitation Explanation

The precipitation of Mg(OH)2 occurs when CO3^2- ion is added to a solution containing Mg^2+ because of the following chemical reaction: \[Mg^{2+}(aq) + CO_{3}^{2-}(aq) + 2OH^{-}(aq) \rightarrow Mg(OH)_{2}(s) + CO_{3}^{2-}(aq)\] The addition of CO3^2- ions to a solution containing Mg^2+ ions causes the reaction to shift in the direction that consumes CO3^2- and Mg^2+ ions and forms the solid Mg(OH)2. The precipitate forms because the concentration of Mg^2+ and CO3^2- exceeds the equilibrium solubility product (Ksp) of Mg(OH)2.
02

(b) Calculating the concentration of CO3^2- and Mg^2+

First, we need to calculate the concentration of CO3^2- and Mg^2+ in the solution: The given mass of Na2CO3 is 4.0 g. To find the moles of Na2CO3, we need to know its molar mass: Molar mass of Na2CO3 = (2 x 22.99) + 12.01 + (3 x 16.00) = 105.99 g/mol Moles of Na2CO3 = mass / molar mass = 4.0 g / 105.99 g/mol = 0.0377 mol The volume of the solution is 1 L, so the concentration of CO3^2-(aq) in the solution is: \[CO_{3}^{2-}=\frac{0.0377 ~mol}{1 ~L}=0.0377 ~M\] The given concentration of Mg^2+ is 125 ppm, which can be converted to molarity: \[Mg^{2+} = \frac{125 ~mg}{1 ~L} \cdot \frac{1 ~g}{1000 ~mg} \cdot \frac{1 ~mol}{24.305 ~g} = 0.00514 ~M\]
03

(b) Checking for precipitation using Ksp

The solubility product constant (Ksp) for Mg(OH)2 is 5.61 x 10^-12. The Ksp expression for Mg(OH)2 is: \[Ksp = [Mg^{2+}][OH^-]^2\] Since we have CO3^2- reacting with Mg^2+ to form the precipitate, the concentration of OH^- ions also depends on the concentration of CO3^2-. The unbalanced reaction can be expressed as: \[CO_{3}^{2-}(aq) + xMg^{2+}(aq) \rightarrow xOH^-(aq)\] Thus, the concentration of OH- (y) can be expressed as: \[ OH^- = x\cdot y = 0.0377 * 0.00514 = 1.937 x 10^{-4} M\] Now, we check whether the precipitation will occur or not by calculating the ion product (Q) and comparing it to Ksp: \[Q = [Mg^{2+}][OH^-]^2 = (0.00514 ~M)(1.937 x 10^{-4} ~M)^2 = 9.59 x 10^{-13}\] Since Q (9.59 x 10^-13) is greater than Ksp (5.61 x 10^-12), the precipitation of Mg(OH)2 will indeed occur when 4.0 g of Na2CO3 is added to 1.00 L of a solution containing 125 ppm of Mg^2+.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solubility Product Constant (Ksp)
The solubility product constant, abbreviated as Ksp, is a crucial concept in understanding chemical precipitation. It provides us with a numerical value that indicates the solubility of a compound under equilibrium conditions in a saturated solution. Specifically, the Ksp value represents the maximum amount of a substance that can dissolve in a solution before it starts precipitating.

When a salt dissolves in water, it dissociates into its constituent ions. For example, magnesium hydroxide (Mg(OH)2) dissociates into magnesium ions (Mg^2+) and hydroxide ions (OH^-). The Ksp expression for this compound is thus the product of the concentrations of these ions raised to the power of their stoichiometric coefficients: \[Ksp = [Mg^{2+}][OH^-]^2\]
The smaller the value of Ksp, the less soluble the compound is. When the product of the ion concentrations in a solution exceeds the Ksp, the excess ions will start to form a precipitate, reverting to the solid form.
Equilibrium
Equilibrium in chemistry refers to the state where the rate of the forward reaction equals the rate of the backward reaction, resulting in no net change in the concentration of reactants and products over time. It's a dynamic process, meaning that even though we don't see any macroscopic changes, molecules continuously react in both directions.

In the context of solubility, equilibrium is reached when a solute's dissolution and precipitation occur at the same rate, which means the solution becomes saturated with the solute. This equilibrium is disturbed when additional ions are introduced into the solution, as seen in the original exercise where adding carbonate ions to a solution containing magnesium ions leads to the precipitation of magnesium hydroxide. This shift in equilibrium is governed by Le Chatelier's Principle, which predicts that the equilibrium will adjust in response to a change in conditions, such as concentration, temperature, or pressure, to partially oppose the change.
Molar Mass
Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is a fundamental concept used to convert between the mass of a substance and the number of moles, which is essential for chemical calculations involving reactions and stoichiometry.

In the given problem, we used the molar mass of sodium carbonate (Na2CO3) to determine the number of moles present in a 4.0 g sample. The calculation involved summing the atomic masses of all atoms present in the formula: \[Molar \, mass \, of \, Na2CO3 = (2 \times 22.99) + 12.01 + (3 \times 16.00)\]
Understanding molar mass is crucial for calculating concentrations, which are needed to determine the solubility and potential precipitation of substances in a solution.
Molarity
Molarity, symbolized by M, is the measure of concentration of a solution, representing the number of moles of solute per liter of solution. It's a key concept when dealing with chemical reactions in solution as it allows us to express concentrations in a uniform way, providing a basis for calculating how much of a substance is present in a given volume of solution.

In the solution provided, molarity was used to calculate the concentration of the carbonate ions in the solution after dissolving 4.0 g of Na2CO3. This value was then used, along with the concentration of magnesium ions, to determine if the product of these concentrations would exceed the solubility product constant (Ksp) and result in precipitation. The concept helps students understand how different amounts of substances relate and react with one another in a solution.
Ion Product (Q)
The ion product (Q) is a parameter that indicates the current product of the molar concentrations of the ions involved in a reaction at any point in time, which may or may not be at equilibrium. It is similar in form to the Ksp but is not limited to equilibrium conditions and can be used to predict whether a precipitate will form in a solution.

By comparing Q to Ksp, we can determine if a solution is unsaturated (Q < Ksp), at equilibrium (Q = Ksp), or supersaturated (Q > Ksp). In the supersaturated case, the solution has more dissolved solute than it can theoretically hold at equilibrium, and precipitation is likely to occur. In the original exercise, we calculated Q for the magnesium and hydroxide ions. Since the value of Q was greater than the Ksp, it indicated that the solution was supersaturated with magnesium hydroxide, and precipitation would occur.

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