Problem 55 - Chemistry

Question

A solution is prepared by mixing 100.0 \(\mathrm{mL}\) of \(1.0 \times 10^{-2} \mathrm{M}\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and 100.0 \(\mathrm{mL}\) of \(1.0 \times 10^{-3} \mathrm{M} \mathrm{NaF} .\) Will \(\mathrm{PbF}_{2}(s)\) \(\left(K_{\mathrm{sp}}=4 \times 10^{-8}\right)\) precipitate?

Step-by-Step Solution

Answer

In the case of mixing 100.0 mL of \(1.0 \times 10^{-2} M\) Pb(NO₃)₂ and 100.0 mL of \(1.0 \times 10^{-3} M\) NaF, we find the concentrations of Pb²⁺ and F⁻ ions after mixing to be \(5.0 \times 10^{-3} M\) and \(5.0 \times 10^{-4} M\), respectively. Then, we calculate the ion product Q as \(Q = (5.0 \times 10^{-3})(5.0 \times 10^{-4})^2 = 12.5 \times 10^{-11}\). Comparing Q (\(12.5 \times 10^{-11}\)) to the solubility product constant Ksp of PbF₂ (\(4 \times 10^{-8}\)), we find that Q > Ksp, meaning a precipitate of PbF₂ will form.

Step 1: Calculate the moles of each ion in the initial solutions

First, let's find the moles of lead (II) ions and fluoride ions in the initial solutions: Moles of Pb²⁺ ions: \( (100.0\,mL) * (1.0 \times 10^{-2}\,M) \) Moles of F⁻ ions: \( (100.0\,mL) * (1.0 \times 10^{-3}\,M) \) (Keep in mind that the volume should be converted to liters when calculating moles.)

Step 2: Calculate the concentrations of the ions after mixing

After the solutions are mixed, the total volume of the solution becomes 200.0 mL. Now, we can calculate the new concentration of each ion as follows: Concentration of Pb²⁺ ions: \( \frac{(100.0\,mL) * (1.0 \times 10^{-2}\,M)}{(200.0\,mL)} \) Concentration of F⁻ ions: \( \frac{(100.0\,mL) * (1.0 \times 10^{-3}\,M)}{(200.0\,mL)} \)

Step 3: Calculate the ion product Q

Now we can find the ion product (Q) of the solution by multiplying the concentrations of the ions: \( Q = [\mathrm{Pb}^{2+}] [\mathrm{F^-}]^2 \) Make sure to square the concentration of F⁻ ions.

Step 4: Compare Q to Ksp

Finally, compare the ion product Q calculated in the previous step to the solubility product constant (Ksp) of PbF2: - If Q > Ksp, a precipitate will form. - If Q < Ksp, there will be no precipitate. - If Q = Ksp, the solution is saturated and on the edge of precipitation. Depending on the result, it will be determined if PbF2 will precipitate or not.

Key Concepts

Solution Concentration

Solution concentration indicates the amount of solute present in a given volume of solvent. When preparing a solution, it's crucial to accurately calculate this concentration.

• Molarity (M) is a common measurement for solution concentration. It's defined as the number of moles of solute per liter of solution.
• In the exercise given, we had two solutions: one with lead(II) nitrate, \(\mathrm{Pb(NO}_{3})_{2}\), and the other with sodium fluoride, \(\mathrm{NaF}\).
• The concentration of the lead(II) nitrate solution was \(1.0 \times 10^{-2} \,\mathrm{M}\), and for sodium fluoride, it was \(1.0 \times 10^{-3} \, \mathrm{M}\).

When these solutions were mixed, the total volume doubled, which means we needed to recalculate the concentration of each ion. The formula \(\text{Concentration (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}\) helps us re-calculate concentration after mixing.

Solubility Product Constant

The solubility product constant, \(K_{sp}\), helps us determine how soluble a compound is in a solution. Specifically, it relates to equilibrium in slightly soluble salts.

• Each ionic compound dissolves in water to a certain extent, reaching a balance between the dissolved ions and the undissolved solid.
• The \(K_{sp}\) value is a specific constant for a given compound at a given temperature, representing the product of the concentrations of its ions raised to the power of their stoichiometric coefficients.

In our example, the solubility product constant for \(\mathrm{PbF}_{2}\) is \(4 \times 10^{-8}\). This value provides a threshold. If the product of the ion concentrations in the solution exceeds this value, \(\mathrm{PbF}_{2}\) will commence precipitation.

Ion Product

The ion product, \(Q\), is essential when deciding whether a precipitate will form in a solution. It measures the present concentrations of the ions in the mixed solution.

• If \(Q\), the ion product, exceeds the solubility product constant, \(K_{sp}\), the solution is supersaturated, and precipitation occurs.
• If \(Q\) is less than \(K_{sp}\), the solution lacks enough ion concentration to form a precipitate.
• When \(Q\) equals \(K_{sp}\), the solution is saturated but no precipitate forms.

In the textbook's problem, you calculate \(Q\) by finding the concentrations of \([\mathrm{Pb}^{2+}]\) and \([\mathrm{F}^{-}]\) ions after mixing and then using the formula \(Q = [\mathrm{Pb}^{2+}][\mathrm{F}^{-}]^2\). This helps determine the state of saturation in the solution.

Chemical Equilibrium

Chemical equilibrium is a state where the forward and reverse reactions occur at the same rate, leading to no net change in the concentration of reactants and products.

• In the context of solubility, equilibrium is reached when the rate at which a solid dissolves equals the rate at which it precipitates.
• This concept is crucial to understanding both the \(K_{sp}\) and the ion product.

For the solution in our example, equilibrium concepts help decide the ions' behavior in water. When mixing the solutions, if the concentration of ions surpasses the threshold defined by \(K_{sp}\), the forward reaction (precipitation) overtakes the reverse (dissolution), and a solid forms. Understanding chemical equilibrium allows precise predictions about precipitation during mixing solutions.