Question

What mass of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) must be added to \(1.0 \mathrm{~L}\) of a \(1.0 \mathrm{M} \mathrm{HF}\) solution to begin precipitation of \(\mathrm{CaF}_{2}(s) ?\) For \(\mathrm{CaF}_{2}, K_{\mathrm{sp}}=4.0 \times\) \(10^{-11}\) and \(K_{\mathrm{a}}\) for \(\mathrm{HF}=7.2 \times 10^{-4}\). Assume no volume change on addition of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(s)\).

Short answer

To begin the precipitation of CaF₂ in 1.0 L of 1.0 M HF solution, we first determine the F⁻ ion concentration in the HF solution, which is approximately 0.0268 M. Then, we find the necessary Ca²⁺ concentration to start precipitation using the given Ksp value, which is approximately \(5.6 \times 10^{-6}\) M. Finally, we calculate the mass of Ca(NO₃)₂ needed to achieve that Ca²⁺ concentration, resulting in approximately \(9.2 \times 10^{-4}\) g of Ca(NO₃)₂.

Step-by-step solution

Write the solubility product expression for CaF₂

The solubility product expression for CaF₂ can be written as the concentration of Ca²⁺ ions multiplied by the square of the concentration of F⁻ ions: \[ K_{sp} = [Ca^{2+}][F^-]^2 \] We are given that \(K_{sp} = 4.0 \times 10^{-11}\).

Write the dissociation expression for HF

The equilibrium expression for the dissociation of HF is given by: \[ K_a = \frac{[H^+][F^-]}{[HF]} \] We are given that \(K_a = 7.2 \times 10^{-4}\).

Determine the F⁻ concentration from the HF dissociation

From the dissociation of HF, we have the equilibrium concentrations: - For F⁻ => \(x\) - For H⁺ => \(x\), since for every F⁻ ion produced, a corresponding H⁺ ion is also produced - For undissociated HF => \(1.0 - x\), since there's initially 1.0 M of HF We can now write the \(K_a\) expression in terms of x: \[ 7.2 \times 10^{-4} = \frac{x^2}{1-x} \] Since the \(K_a\) is relatively small we can assume \(x \ll 1\), so we can write: \[ 7.2 \times 10^{-4} \approx x^2 \] Solving for x: \[ x = \sqrt{7.2 \times 10^{-4}} \approx 0.0268 \,M \] So, the concentration of F⁻ ions in the solution due to the dissociation of HF is approximately 0.0268 M.

Determine the concentration of Ca²⁺ needed to start precipitation

From the solubility product expression, we can determine the concentration of Ca²⁺ that is needed to start precipitation: \[ 4.0 \times 10^{-11} = [Ca^{2+}](0.0268)^2 \] Solving for [Ca²⁺]: \[ [Ca^{2+}] \approx \frac{4.0 \times 10^{-11}}{(0.0268)^2} \approx 5.6 \times 10^{-6} \,M \]

Calculate the mass of Ca(NO₃)₂ needed

In 1.0 L of solution, we need the concentration of Ca²⁺ to be 5.6 × 10⁻⁶ M. So the amount of Ca²⁺ (in mol) required is: \[ 5.6 \times 10^{-6}\, mol \times 1.0\, L = 5.6 \times 10^{-6}\, mol \] Since the molar mass of Ca(NO₃)₂ = 164 g/mol (40 g/mol for Ca, 14 g/mol each for the two N, and 48 g/mol for the six O), we can calculate the mass of Ca(NO₃)₂ needed: \[ 5.6 \times 10^{-6}\, mol \times 164\, g/mol \approx 9.2 \times 10^{-4}\, g \] Therefore, approximately 9.2 × 10⁻⁴ g of Ca(NO₃)₂ must be added to the 1.0 L of 1.0 M HF solution to begin precipitation of CaF₂.

Key concepts

Calcium Fluoride Precipitation

Precipitation occurs when a solid forms out of a solution. This often happens when the concentrations of the involved ions exceed their solubility product, an expression symbolized as \( K_{sp} \). In our exercise, the focus is on calcium fluoride \( \text{CaF}_2 \), which precipitates out of the solution when the product of the calcium and fluoride ion concentrations exceeds \( 4.0 \times 10^{-11} \). This value represents the maximum product of ion concentrations that can exist in a solution without forming a solid.

• For \( \text{CaF}_2 \), the \( K_{sp} \) is particularly low, indicating low solubility in water.
• Precipitation begins specifically when the solution cannot hold any more dissolved ions without forming a solid.

Understanding this process is crucial because it explains why certain substances appear to magically form solids in liquid solutions—a key concept in many chemistry applications.

Equilibrium Expressions

Equilibrium expressions are algebraic representations used to describe the concentrations of reactants and products in a chemical reaction at equilibrium. For calcium fluoride \( \text{CaF}_2 \), the solubility product expression is \( K_{sp} = [Ca^{2+}][F^-]^2 \). This expression helps us understand how the concentrations of calcium and fluoride ions relate to the solubility of \( \text{CaF}_2 \).
For weak acids like hydrogen fluoride (HF), the corresponding equilibrium expression is \( K_a = \frac{[H^+][F^-]}{[HF]} \). The constant \( K_a \) describes the acidity and dissociation extent of HF.

• The smaller the \( K_a \) value, the weaker the acid and the less it dissociates.
• In our scenario, knowing \( K_a \) enables us to calculate the fluoride concentration from HF dissociation.

These expressions are foundational in chemistry, helping predict reaction outcomes under varied conditions.

Dissociation of Weak Acids

Dissociation refers to the process by which molecules split into smaller particles, such as ions. Weak acids, like hydrofluoric acid (HF), do not completely dissociate in solution. Instead, they reach a state of dynamic equilibrium where both the intact acid and dissociated ions exist.
In our exercise, HF dissociates only partially forming hydrogen ions \( [H^+] \) and fluoride ions \( [F^-] \), as governed by:- \( K_a = \frac{[H^+][F^-]}{[HF]} \)

• The small \( K_a \) value, \( 7.2 \times 10^{-4} \), suggests limited dissociation.
• The strategy involves estimating ion concentrations assuming that very little HF dissociates (i.e., \( x \ll 1 \)).
• This assumption simplifies calculations while still yielding accurate fluoride ion concentrations.

Understanding weak acid dissociation is crucial for solving equilibrium problems and for applications involving buffers and pH adjustments.

Molar Mass Calculations

Molar mass calculations are vital for converting between moles and grams. The molar mass of a compound is the sum of the atomic masses of its constituent elements. For \( \text{Ca(NO}_3\text{)}_2 \), the calculation is straightforward: Add the atomic masses of calcium, nitrogen, and oxygen.
- Calcium (Ca) = 40 g/mol- Nitrogen (N) = 14 g/mol- Oxygen (O) = 16 g/mol Combining these gives us:- \( 40 + 2(14) + 6(16) = 164 \) g/molWith this value, you can convert the required moles of \( \text{Ca(NO}_3\text{)}_2 \) into grams needed to start precipitation. If the calculation yields \( 5.6 \times 10^{-6} \) moles of calcium are necessary, multiplying this by 164 g/mol gives the mass required.
Accurate molar mass calculations ensure you add the correct amount of substance to achieve the desired chemical outcome.