Question

The precipitate of \(\mathrm{CaF}_{2}\left(\mathrm{Ksp}=1.7 \times 10^{-10}\right)\) is obtained when equal volumes of the following are mixed: (a) \(10^{-2} \mathrm{M} \mathrm{Ca}^{2+}+10^{-3} \mathrm{MF}^{-}\) (b) \(10^{-3} \mathrm{M} \mathrm{Ca}^{2+}+10^{-5} \mathrm{MF}^{-}\) (c) \(10^{-4} \mathrm{M} \mathrm{Ca}^{2+}+10^{-4} \mathrm{MF}^{-}\) (d) \(10^{-5} \mathrm{M} \mathrm{Ca}^{2+}+10^{-3} \mathrm{MF}^{-}\)

Short answer

Precipitate forms in (a) since its IP exceeds the Ksp.

Step-by-step solution

Calculate Ionic Product for Each Combination

For each pair provided, the ionic product (IP) of \( \mathrm{CaF}_2 \) is given by \( \text{IP} = [\mathrm{Ca}^{2+}][\mathrm{F}^-]^2 \). We need to use the final concentrations after mixing, which means concentrations will reduce by half due to equal volume mixing.(a) Initial: \([\mathrm{Ca}^{2+}] = 10^{-2} \ \mathrm{M}, [\mathrm{F}^-] = 10^{-3} \ \mathrm{M} \), after mixing \([\mathrm{Ca}^{2+}] = 5 \times 10^{-3} \ \mathrm{M}, [\mathrm{F}^-] = 5 \times 10^{-4} \ \mathrm{M} \), so \( \text{IP} = 5 \times 10^{-3} \times (5 \times 10^{-4})^2 = 1.25 \times 10^{-9} \).(b) Initial: \([\mathrm{Ca}^{2+}] = 10^{-3} \ \mathrm{M}, [\mathrm{F}^-] = 10^{-5} \ \mathrm{M} \), after mixing \([\mathrm{Ca}^{2+}] = 5 \times 10^{-4} \ \mathrm{M}, [\mathrm{F}^-] = 5 \times 10^{-6} \ \mathrm{M} \), so \( \text{IP} = 5 \times 10^{-4} \times (5 \times 10^{-6})^2 = 1.25 \times 10^{-15} \).(c) Initial: \([\mathrm{Ca}^{2+}] = 10^{-4} \ \mathrm{M}, [\mathrm{F}^-] = 10^{-4} \ \mathrm{M} \), after mixing \([\mathrm{Ca}^{2+}] = 5 \times 10^{-5} \ \mathrm{M}, [\mathrm{F}^-] = 5 \times 10^{-5} \ \mathrm{M} \), so \( \text{IP} = 5 \times 10^{-5} \times (5 \times 10^{-5})^2 = 1.25 \times 10^{-13} \).(d) Initial: \([\mathrm{Ca}^{2+}] = 10^{-5} \ \mathrm{M}, [\mathrm{F}^-] = 10^{-3} \ \mathrm{M} \), after mixing \([\mathrm{Ca}^{2+}] = 5 \times 10^{-6} \ \mathrm{M}, [\mathrm{F}^-] = 5 \times 10^{-4} \ \mathrm{M} \), so \( \text{IP} = 5 \times 10^{-6} \times (5 \times 10^{-4})^2 = 1.25 \times 10^{-13} \).

Key concepts

Ionic Product

When we talk about the ionic product (IP) of a solution, we're essentially discussing the product of the concentrations of the ions in solution. This is especially important when dealing with substances that can form precipitates. For example, in the case of calcium fluoride (\[\text{CaF}_2\]), the ionic product is calculated using the equation:\[\text{IP} = [\text{Ca}^{2+}][\text{F}^-]^2\]This formula shows us that the ionic product depends on the concentration of the calcium ions \([\text{Ca}^{2+}]\) and the square of the concentration of fluoride ions \([\text{F}^-]^2\).This is because fluoride ions are present as two ions per formula unit of calcium fluoride. When solutions are mixed, as in our example, each of their initial concentrations drops because they are diluted to half their initial concentration. To find out if a precipitate will form when mixing solutions, we compare the calculated ionic product to the known solubility product of the substance.

Precipitation Reactions

A precipitation reaction occurs when ions in a solution combine to form an insoluble compound. If this compound is less soluble than its ionic precursors, it will precipitate out of the solution. How does this relate to the ionic product? When the ionic product of a solution exceeds the known solubility product (Ksp) of the compound, a precipitate forms. This is because the concentrations of ions in the solution are too high for the compound to remain dissolved.

• If the IP is greater than the Ksp, a precipitate forms.
• If the IP is less than the Ksp, no precipitate forms.
• If the IP equals the Ksp, the system is at equilibrium.

Understanding these concepts helps predict whether certain reactions will occur and what products might form. By checking these inequalities, one can decide if a combination of ions will result in precipitation.

Calcium Fluoride K_sp

The solubility product (\(K_{sp}\)) is a constant for a given substance at a particular temperature. For calcium fluoride (\(\text{CaF}_2\)), the \(K_{sp}\) is given as \(1.7 \times 10^{-10}\). This value gives us crucial information: it represents the threshold below which calcium fluoride will remain dissolved in solution. When calculating whether a precipitate will form, as in the example problems:

• Calculate the ionic product for the given solutions.
• Compare this with the \(K_{sp}\) value of calcium fluoride.

For example, if a solution has \(\text{IP} = 1.25 \times 10^{-9}\), since this is below \(K_{sp}\), \(\text{CaF}_2\) will not precipitate. However, if \(\text{IP}\) were higher than \(K_{sp}\), you would expect precipitation. Thus, the concept of \(K_{sp}\) is crucial in determining solubility and the conditions under which precipitation happens.