Question

Will precipitation of \(\mathrm{MgF}_{2}(\mathrm{s})\) occur if a \(22.5 \mathrm{mg}\) sample of \(\mathrm{MgCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) is added to \(325 \mathrm{mL}\) of \(0.035 \mathrm{M} \mathrm{KF}\) ?

Short answer

Yes, precipitation of \( MgF_{2}(s) \) will occur.

Step-by-step solution

Calculate mass to mole

Find the number of moles of \( Mg^{2+} \) ions in the 22.5 mg of \( MgCl_{2} \cdot 6H_{2}O \). This can be done using the molar mass of \( MgCl_{2} \cdot 6H_{2}O \) which is approximately 203.31 g/mol. So, the moles of \( Mg^{2+} \) = \( \frac{0.0225 g}{203.31 g/mol} = 1.11 \times 10^{-4} mol \).

Calculate concentration of ions

Determine the concentrations of \( Mg^{2+} \) and \( F^- \) ions. For \( Mg^{2+} \), divide the moles by the volume in liters which is \( \frac{1.11 \times 10^{-4} mol}{0.325 L} = 0.342 M \). For \( F^- \), it is originally given as 0.035 M.

Calculate ion product

Calculate the ion product of \( Mg^{2+} \) and \( F^− \) which is \( [Mg^{2+}] [F^−]^2 = (0.342 M) (0.035 M)^2 = 1.23 \times 10^{-4} \).

Compare ion product to solubility product

The ion product is large than the \( K_{sp} \) ( \( K_{sp} = 6.4 \times 10^{-9} \) ). Therefore, precipitation of \( MgF_{2}(s) \) will occur.

Key concepts

Molar Mass Calculation

Molar mass calculation is a fundamental aspect of chemistry that involves determining the mass of one mole of a substance. It is expressed in grams per mole ((g/mol)). This concept is crucial when converting the mass of a substance to the number of moles, which is required for stoichiometric calculations in chemical reactions.

In the context of the exercise, calculating the molar mass of MgCl_2 6H2Ois the first step toward determining whether MgF_2(s) will precipitate when mixed with KF. To calculate molar mass, you add together the atomic masses of each element in the compound, multiplied by the number of atoms of that element present in the formula. For example, MgCl_2 6H2O contains magnesium (Mg), chlorine (Cl), hydrogen (H), and oxygen (O). Using the periodic table, you can find the atomic masses and calculate the molar mass as follows:

• Mg: 24.305 g/mol
• Cl: 35.45 g/mol x 2
• H: 1.008 g/mol x 12
• O: 15.999 g/mol x 6

Summing up the masses gives the compound's molar mass, which you then use to convert the given sample mass into moles.

Solubility Product Constant

The solubility product constant (K_{sp}) is a special kind of equilibrium constant that applies to the dissolution of sparingly soluble compounds. It provides a quantitative measure of the solubility of a compound in a solvent under standardized conditions. The K_{sp} is determined by multiplying the concentrations of the ions, each raised to the power of its coefficient in the dissociation equation.

For instance, MgF_2(s)dissolves to produce one Mg^{2+}ion and two F^-ions, so its K_{sp}expression would be written as K_{sp} = [Mg^{2+}][F^-]^2. When comparing the calculated ion product of Mg^{2+}and F^-ions to the tabulated K_{sp}of MgF_2, if the ion product is greater, it indicates that the solution is supersaturated, and precipitation is expected to occur, as in the exercise.

Mole Conversion

Mole conversion is the process of converting between mass, number of particles, and volume of a gas at standard temperature and pressure (STP), using the mole as the intermediary. The mole is a basic unit in chemistry that provides a bridge between the atomic or molecular scale and the macroscopic scale. It is defined as the amount of substance that contains as many entities (atoms, ions, molecules) as there are atoms in 12 grams of carbon-12.

In our exercise, mole conversion was used to find the number of moles of Mg^{2+} from a known mass of MgCl_2 6H2O. The relationship used is: number of moles = mass of substance (g) / molar mass (g/mol). This step is critical because only the number of moles allows us to understand the stoichiometry of the reactants in the solution and predict the formation of the precipitate.

Ion Concentration Calculation

Ion concentration calculation involves finding the molarity (M), which is a measure of the concentration of a solute in a solution, or how many moles of a given solute are present in one liter of the solution. Molarity is calculated by dividing the number of moles of the solute by the volume of the solution in liters (moles/L).

In the provided exercise, the concentration of Mg^{2+} ions is found after determining the moles of Mg^{2+} present in the given volume of solution. For the F^- ions, the concentration was already provided. These concentrations are essential for calculating the ion product, as both the number of moles and the volume of the solution are needed to determine whether the ion product exceeds the K_{sp}, leading to precipitation of MgF_2(s). Understanding ion concentration calculations is vital for any solution chemistry problem involving reactions in solution.