Question

Will a precipitate of \(\mathrm{Mg}(\mathrm{OH})_{2}\) form when \(25.0 \mathrm{mL}\) of \(0.010 \mathrm{M} \mathrm{NaOH}\) is combined with \(75.0 \mathrm{mL}\) of a \(0.10 \mathrm{M}\) solution of magnesium chloride?

Short answer

Yes, a precipitate of \( \text{Mg(OH)}_2 \) will form because \( Q > K_{sp} \).

Step-by-step solution

Calculate Moles of OH-

First, calculate the moles of hydroxide ions (OH⁻) present in the solution. Use the formula: \( ext{moles} = ext{volume (L)} \times ext{molarity (M)} \).For NaOH: \[ 0.025 ext{ L} \times 0.010 ext{ M} = 0.00025 ext{ moles of OH⁻} \]

Calculate the Total Volume of the Solution

Determine the total volume of the combined solutions.\[ 25.0 ext{ mL (from NaOH)} + 75.0 ext{ mL (from MgCl}_2 ext{)} = 100.0 ext{ mL} = 0.100 ext{ L} \]

Calculate Concentration of OH- in Mixed Solution

Calculate the concentration of OH⁻ in the combined solution:\[ \text{Concentration of } ext{OH}^- = \frac{0.00025 ext{ moles}}{0.100 ext{ L}} = 0.0025 ext{ M} \]

Calculate Moles of Mg2+

Calculate the moles of Mg²⁺ present using the magnesium chloride solution:\[ 0.075 ext{ L} \times 0.10 ext{ M} = 0.0075 ext{ moles of Mg}^{2+} \]

Calculate Concentration of Mg2+ in Mixed Solution

Calculate the concentration of Mg²⁺ in the combined solution:\[ \text{Concentration of } ext{Mg}^{2+} = \frac{0.0075 ext{ moles}}{0.100 ext{ L}} = 0.075 ext{ M} \]

Write the Solubility Product Expression

For Mg(OH)₂, the solubility product constant (Ksp) is:\[ K_{sp} = [ ext{Mg}^{2+}][ ext{OH}^-]^2 \]

Calculate the Reaction Quotient, Q

Plug in the concentrations into the expression to determine the reaction quotient, Q:\[ Q = [ ext{Mg}^{2+}][ ext{OH}^-]^2 = (0.075)(0.0025)^2 = 4.6875 \times 10^{-7} \]

Compare Q with Ksp

The solubility product constant for Mg(OH)₂ is approximately \( K_{sp} = 5.6 \times 10^{-12} \)Compare Q to \( K_{sp} \): If \( Q > K_{sp} \), a precipitate will formSince \( 4.6875 \times 10^{-7} > 5.6 \times 10^{-12} \), a precipitate will form.

Key concepts

Solubility Product Constant (Ksp)

The solubility product constant, often abbreviated as \( K_{sp} \), is a measure used in chemistry to describe the solubility of a sparingly soluble compound. In a saturated solution, there is an equilibrium between the undissolved solid and its dissolved ions. The general formula to represent this equilibrium is: \[ K_{sp} = [\text{Cation}]^a [\text{Anion}]^b \] where the exponents \( a \) and \( b \) correspond to the stoichiometric coefficients from the dissociation equation of the compound. For instance, in

• \( \text{Mg}(\text{OH})_2 \) dissociating to \( \text{Mg}^{2+} \) and \( \text{OH}^- \), the solubility product expression becomes \( K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \).

This expression indicates that any increase in the concentration of magnesium ions or hydroxide ions can shift the equilibrium, potentially leading to precipitation if it exceeds the value of \( K_{sp} \). The \( K_{sp} \) value for each compound is specific and can be found in chemical data tables. Knowing this value is crucial in predicting whether a precipitate will form under certain conditions.

Reaction Quotient (Q)

The reaction quotient, denoted as \( Q \), serves as an immediate snapshot of the state of a reaction at a given moment. It is used to determine whether a system is at equilibrium, and if not, the direction in which the system will proceed to reach equilibrium. The expression for \( Q \) is derived similarly to \( K_{sp} \), but instead of using equilibrium concentrations, \( Q \) uses the concentrations present at any given point:

• \( Q = [\text{Mg}^{2+}][\text{OH}^-]^2 \)

By calculating \( Q \), we can compare it to \( K_{sp} \):

• If \( Q < K_{sp} \), the solution is unsaturated, hence no precipitation occurs.
• If \( Q = K_{sp} \), the system is at equilibrium, and the solution is saturated.
• If \( Q > K_{sp} \), the solution is supersaturated, leading to precipitation.

In the given example, the calculation of \( Q \) exceeded the \( K_{sp} \) for \( \text{Mg}(\text{OH})_2 \), indicating that the solution was supersaturated and a precipitate would form.

Molarity Calculations

Molarity, represented as \( M \), is a crucial concept used to express the concentration of a solution. Molarity is defined as the number of moles of solute per liter of solution. The formula is given by \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]Calculating molarity involves a few key steps:

• Determine moles of solute: Use the initial concentration and volume to find the moles. For example, in the solution with \(\text{NaOH}\), \(0.025\,\text{L} \times 0.010\,\text{M} = 0.00025\,\text{moles of OH}^-\).
• Total solution volume: Add volumes of combined solutions to find the total, in this case \( 0.100\,\text{L} \).
• Calculate new concentration: Divide the moles of ions by the total volume for new molarity.

Molarity calculations are essential for predicting the outcome of reactions, such as precipitation reactions, by allowing us to compute \( Q \) and compare it with \( K_{sp} \). Understanding these basics of molarity helps in better comprehension of how concentrations affect chemical equilibrium and the potential for forming precipitates.