Question

Will a precipitate form when \(100.0 \mathrm{~mL}\) of \(4.0 \times 10^{-4} M\) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(100.0 \mathrm{~mL}\) of \(2.0 \times 10^{-4} \mathrm{M} \mathrm{NaOH} ?\)

Short answer

A precipitate will not form when \(100.0 \mathrm{~mL}\) of \(4.0 \times 10^{-4} M \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(100.0 \mathrm{~mL}\) of \(2.0 \times 10^{-4} M \mathrm{NaOH}\) as the calculated reaction quotient (Q) of \(2.0 \times 10^{-12}\) is less than the Ksp (\(1.5 \times 10^{-11}\)) for \(Mg(OH)_2\).

Step-by-step solution

Write the balanced equation

Write the balanced equation for the potential precipitation reaction: \[Mg^{2+} + 2OH^- \rightarrow Mg(OH)_2\]

Determine the initial concentrations of ions

Determine the initial concentrations of the ions before the solutions are mixed: \[Mg^{2+} = 4.0 \times 10^{-4} M\] \[OH^- = 2.0 \times 10^{-4} M\]

Calculate the final concentrations of ions

Calculate the final concentrations of the ions after mixing the two solutions. The volume has doubled, so the concentrations are halved: \[Mg^{2+} = \frac{4.0 \times 10^{-4} M}{2} = 2.0 \times 10^{-4} M\] \[OH^- = \frac{2.0 \times 10^{-4} M}{2} = 1.0 \times 10^{-4} M\]

Calculate the reaction quotient (Q)

Calculate the reaction quotient (Q) using the concentrations of Mg2+ and OH- ions: \[Q = [Mg^{2+}][OH^-]^2\] \[Q = (2.0 \times 10^{-4})(1.0 \times 10^{-4})^2\] \[Q = 2.0 \times 10^{-12}\]

Compare Q to Ksp

Compare the reaction quotient (Q) to the Ksp of Mg(OH)2. The Ksp for Mg(OH)2 is approximately \(1.5 \times 10^{-11} \). If Q > Ksp, a precipitate will form; if Q < Ksp, no precipitate will form: \[Q = 2.0 \times 10^{-12} < Ksp = 1.5 \times 10^{-11}\] Since Q is less than Ksp, no precipitate will form upon mixing the two solutions.

Key concepts

Understanding Reaction Quotient (Q)

The reaction quotient, denoted as \( Q \), helps us understand how far a reaction has progressed towards equilibrium. It is similar to the equilibrium constant but applies to reactions that have not yet reached equilibrium.

For a reaction \( aA + bB \leftrightarrow cC + dD \), the expression for \( Q \) is given by: \[ Q = \frac{{[C]^c[D]^d}}{{[A]^a[B]^b}} \] Where \( [A] \), \( [B] \), \( [C] \), and \( [D] \) are the molar concentrations of the reactants and products.

To determine if a reaction will proceed towards forming more products or reactants, we compare \( Q \) to the equilibrium constant \( K_{sp} \). This comparison can reveal several states:

• If \( Q < K_{sp} \), the reaction will move in the forward direction to reach equilibrium.
• If \( Q > K_{sp} \), the reaction will move in the reverse direction.
• If \( Q = K_{sp} \), the system is already at equilibrium.

Understanding \( Q \) is vital in predicting if a precipitate will form in a solution.

Exploring Solubility Product Constant (Ksp)

The solubility product constant, represented as \( K_{sp} \), quantifies the solubility of a sparingly soluble salt. It represents the equilibrium between the ions in a solution and the undissolved salt.

For a salt \( AB \) that dissociates into \( A^+ \) and \( B^- \), its \( K_{sp} \) expression is given by \[ K_{sp} = [A^+][B^-] \]. This value tells us the maximum product of the concentrations of the individual ions that can exist in a solution before the salt starts to precipitate.

The \( K_{sp} \) value is unique to each compound and influences the extent of solubility. A small \( K_{sp} \) indicates low solubility (i.e., high tendency to precipitate), while a larger \( K_{sp} \) suggests higher solubility.

When comparing \( Q \) and \( K_{sp} \), if \( Q \) exceeds \( K_{sp} \), the solution becomes supersaturated, leading to precipitation. Understanding \( K_{sp} \) alongside \( Q \) is essential to predict precipitants in mixed solutions.

The Role of Chemical Equilibrium

Chemical equilibrium describes a state in which the rate of the forward reaction equals the rate of the reverse reaction. In this state, the concentrations of reactants and products remain constant over time, although both reactions continue to occur.

Consider the general reaction: \[ aA + bB \leftrightarrow cC + dD \] At equilibrium, the concentration ratio of products to reactants is constant and expressed by the equilibrium constant \( K \).

Reactions may reach equilibrium after an initial period of favoring either the forward or reverse reaction, but once equilibrium is achieved, changes in concentration, pressure, or temperature can disturb it. When these disturbances occur, the system adjusts to restore equilibrium, a concept described by Le Châtelier's Principle.

Understanding equilibrium is not only theoretical; it's practical in deducing whether products or reactants will dominate in a chemical setup, such as predicting precipitation in solutions.

Solution Concentration Calculations

Calculating concentration is essential in predicting and analyzing chemical reactions. Concentration is typically expressed in molarity (\( M \)), which refers to the number of moles of solute per liter of solution.

When dealing with reactions in solutions, especially when mixing two different solutions, determining the resulting concentrations is crucial. Here's how to calculate it:

• Determine initial concentrations using the formula \( M = \frac{n}{V} \), where \( n \) is moles of solute and \( V \) is volume in liters.
• Account for any dilution or concentration changes resulting from mixing by using the principle \( C_1V_1 = C_2V_2 \).
• Calculate the concentrations in the combined volume, which is the sum of the two volumes mixed.

In the context of a precipitation reaction, these calculations help assess whether the product of the ion concentrations exceeds the \( K_{sp} \), indicating precipitate formation. Thus, concentration calculations are a practical tool for understanding and predicting chemical behavior.