Question

What mass of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) must be added to \(1.0 \mathrm{L}\) of a \(1.0-M \mathrm{HF}\) 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

The mass of Ca(NO₃)₂ to be added to 1.0 L of 1.0 M HF solution to begin precipitation of CaF₂ can be calculated as follows: 1. Calculate the F⁻ concentration from HF dissociation: [F⁻] = √(Ka * [HF]) 2. Calculate the required Ca²⁺ concentration for precipitation: [Ca²⁺] = Ksp / [F⁻]² 3. Calculate the moles of Ca(NO₃)₂: moles of Ca(NO₃)₂ = [Ca²⁺] * volume of solution 4. Calculate the mass of Ca(NO₃)₂: mass of Ca(NO₃)₂ = moles of Ca(NO₃)₂ * molar mass of Ca(NO₃)₂ Using the given values of Ksp and Ka, and following the steps above, we can calculate the required mass of Ca(NO₃)₂ to be added to the solution.

Step-by-step solution

Calculate the F⁻ concentration from the equilibrium of HF dissociation

The dissociation of HF can be represented as: HF ⇌ H⁺ + F⁻ The Ka expression for HF is given by: Ka = [H⁺][F⁻] / [HF] We know the concentration of HF ([HF]) = 1.0 M and Ka = 7.2 x 10⁻⁴. Since the dissociation of HF is very small, we assume that the concentration of HF remains almost unchanged, and [H⁺] ≈ [F⁻]. So, Ka = [F⁻]² / [HF] Now, we can solve for the F⁻ concentration: [F⁻]² = Ka * [HF] [F⁻] = √(Ka * [HF])

Calculate the required Ca²⁺ concentration for precipitation

The dissolution of CaF₂ can be represented as: CaF₂(s) ⇌ Ca²⁺ + 2F⁻ The Ksp expression for CaF₂ is given by: Ksp = [Ca²⁺][F⁻]² We know the Ksp = 4.0 x 10⁻¹¹ and [F⁻] from the previous step. Now, we can solve for the Ca²⁺ concentration: [Ca²⁺] = Ksp / [F⁻]²

Calculate the moles of Ca(NO₃)₂

From the previous step, we have found the required concentration of Ca²⁺ ions to precipitate CaF₂. Since the volume of the solution is 1.0 L and we assume no volume change on addition, the number of moles of Ca(NO₃)₂ to be added can be calculated as: moles of Ca(NO₃)₂ = [Ca²⁺] * volume of solution

Calculate the mass of Ca(NO₃)₂

Now that we have the number of moles of Ca(NO₃)₂, we can calculate the mass of Ca(NO₃)₂ to be added using its molar mass. The molar mass of Ca(NO₃)₂ = 164.10 g/mol. Thus, mass of Ca(NO₃)₂ = moles of Ca(NO₃)₂ * molar mass of Ca(NO₃)₂ The calculated mass of Ca(NO₃)₂ is the required mass to be added to the 1.0 L of 1.0 M HF solution to begin precipitation of CaF₂.

Key concepts

Solubility Product Constant (Ksp)

Understanding the solubility product constant, abbreviated as Ksp, is pivotal in exploring the solubility of sparingly soluble compounds. Essentially, Ksp provides an insight into the extent to which a compound can dissolve in water, defining its solubility equilibrium.

For a general salt represented as AB, which dissociates into A+ and B- ions in solution, the Ksp is expressed as follows:
\[ K_{sp} = [A^+][B^-] \]
It's the product of the concentrations of the ions, each raised to the power of their stoichiometric coefficients in the balanced equation for dissolution. The smaller the Ksp value, the less soluble the compound is in solution.

When solving exercises involving Ksp, we often work under the premise that the concentration of these ions comes from a solid's dissociation at equilibrium. For example, calcium fluoride (CaF2) in our exercise dissociates as follows:
\[ \text{CaF}_{2}(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^-(aq) \]
The Ksp expression for this reaction is: \[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \]
To begin precipitation, or the process of forming a solid from a solution, the product of the ion concentrations must exceed the Ksp value. Knowing the value of Ksp allows us to calculate the saturation point of the solution, beyond which additional solid will precipitate out. This concept was applied in the exercise to determine the mass of calcium nitrate needed to initiate precipitation.

Acid Dissociation Constant (Ka)

The acid dissociation constant, Ka, is an incredibly significant figure representing the strength of an acid in solution. It quantifies the acid's tendency to donate a proton to the surrounding water molecules, producing hydronium (H3O+) or hydrogen (H+) ions and its conjugate base.

For a simple acid HA in water, the dissociation can be given by:
\[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \]
The Ka expression for this equilibrium is:
\[ K_{a} = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]
The higher the value of Ka, the stronger the acid, as it indicates a higher concentration of dissociated ions at equilibrium. In the context of our exercise, we use the Ka value for hydrofluoric acid (HF) to calculate the concentration of fluoride ions (F-) in solution, which directly impacts the solubility equilibrium of CaF2 and therefore the precipitation process.

It is critical to grasp that the acid's dissociation reaction is dynamic and dependent on the concentrations of all species at equilibrium. Substantial variation in these concentrations can shift the equilibrium, altering the Ksp and the solubility of a salt tied to the anion of the acid, such as in the calcium fluoride example.

Stoichiometry

Stoichiometry is the foundation of chemical reactions that deals with the quantitative relationships between reactants and products. It draws from the principle of conservation of mass, where the numbers of atoms of each element must be the same before and after a chemical reaction.

A balanced chemical equation is the starting point for stoichiometric calculations. It informs us of the proportions in which substances react or form. For instance, the stoichiometry of calcium nitrate reacting with fluoride ions is grounded on the balanced equation representing the dissolution of the precipitate, CaF2. The molar ratios of reactants and products are used to convert between masses, moles, and numbers of particles.

In practical applications, such as the exercise provided, stoichiometry allows us to calculate the precise mass of a reactant required to start a reaction—in this case, to precipitate CaF2 from a solution containing fluoride ions. By understanding the stoichiometric relationships and using the molar mass of the reactants, we can solve for the mass of calcium nitrate necessary to reach the precipitation threshold, blending the concepts of both solubility equilibrium and acid-base chemistry.