Problem 119 - Chemistry

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 \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(s).\)

Step-by-Step Solution

Answer

To determine the mass of $\mathrm{Ca(NO_3)_2}$ that needs to be added to a 1.0 L of 1.0 M $\mathrm{HF}$ solution to begin precipitation of $\mathrm{CaF_2}$, we can follow these steps: 1. Write the chemical reactions for CaF₂ dissolution and HF ionization. 2. Use the $K_{sp}$ value for CaF₂ to find the maximum concentration of F⁻ that causes precipitation. 3. Calculate the concentration of F⁻ ions in the 1.0 M $\mathrm{HF}$ solution using the given $K_{a}$ value. 4. Determine the additional mass of F⁻ needed to begin precipitation and calculate the mass of $\mathrm{Ca(NO_3)_2}$ required. Upon finding the concentration of F⁻ needed for precipitation and the concentration of F⁻ in the $\mathrm{HF}$ solution, we can calculate the mass of Ca(NO₃)₂: \[Mass \ of \ Ca(NO_{3})_{2} = \frac{Mass \ of \ F⁻ \ needed}{2}\] Substitute the obtained values from steps 2 and 3 and calculate the mass of $\mathrm{Ca(NO_3)_2}$ needed.

Step 1: Write the Chemical Reactions

First, we need to write down the chemical reactions for the dissolution of CaF₂ and the ionization of HF. For CaF₂ dissolution: \[CaF_{2}(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)\] For HF ionization: \[HF(aq) \rightleftharpoons H^+(aq) + F^-(aq)\]

Step 2: Calculate Maximum Concentration of F⁻ to Begin Precipitation

To find the concentration of F⁻ ions that will cause precipitation of CaF₂, we will utilize the Ksp value given: \[K_{sp} = [Ca^{2+}][F^-]^2\] Since the precipitation will just begin, the concentration of Ca²⁺ and F⁻ ions will be equal at the point of saturation, so: \[K_{sp} = [x][2x]^2\] where x is the concentration of Ca²⁺ ions, and 2x the concentration of F⁻ ions. Now, substitute the given Ksp value: \[4.0 \times 10^{-11} = [x][2x]^2\]

Step 3: Calculate the concentration of F⁻ ions in 1.0 M HF

To calculate the concentration of F⁻ ions in 1.0 M HF, we will use the Ka value for HF: \[K_{a} = \frac{[H^{+}][F^{-}]}{[HF]}\] At equilibrium, [HF] = 1.0 - x, [H⁺] = x, and [F⁻] = x. Substitute the given Ka value, and solve for x: \[7.2 \times 10^{-4} = \frac{x \times x}{1.0 - x}\]

Step 4: Determine the mass of Ca(NO₃)₂ that needs to be added

Now that we have the concentration of F⁻ ions needed to begin precipitation and the concentration of F⁻ ions in the 1.0 M HF solution, we can calculate the additional mass of F⁻ needed: \[Mass \ of \ F⁻ \ needed = (2\times[Ca(NO_{3})_{2}])\times \left[(2 \times \ Concentration \ of \ F⁻ \ needed) - (Concentration \ of \ F⁻ \ in \ HF \ solution)\right]\] Then, we can calculate the mass of Ca(NO₃)₂ that needs to be added: \[Mass \ of \ Ca(NO_{3})_{2} = \frac{Mass \ of \ F⁻ \ needed}{2}\] Calculate the mass of Ca(NO₃)₂ by substituting the values obtained from the previous steps, and that will be the answer to our problem.

Key Concepts

Ksp (Solubility Product Constant)

The Solubility Product Constant, abbreviated as \( K_{sp} \), is key when discussing the equilibrium of sparingly soluble compounds. It represents the maximum amount of a solid that can dissolve in a solution before the onset of precipitation. This constant is especially useful in predicting whether a precipitate will form under certain conditions.

For instance, consider \( CaF_2 \), which dissociates in water as follows:

• \( CaF_{2}(s) \rightleftharpoons Ca^{2+}(aq) + 2F^{-}(aq) \)

The \( K_{sp} \) expression for this process is \([Ca^{2+}][F^{-}]^2\), where \([Ca^{2+}]\) is the concentration of calcium ions and \([F^{-}]\) the fluoride ions in solution at equilibrium.

In simple terms, \( K_{sp} \) can help you find out exactly how much of the solute can dissolve before the solution becomes saturated and begins to precipitate the excess substance out of the solution.

Ka (Acid Dissociation Constant)

The Acid Dissociation Constant, known as \( K_a \), is crucial for understanding how acids behave in solutions. It measures the strength of an acid in a solvent, typically water. The larger the \( K_a \), the stronger the acid and the more it dissociates into its ions in solution.

For example, hydrofluoric acid (\( HF \)) dissociates as follows:

• \( HF(aq) \rightleftharpoons H^{+}(aq) + F^{-}(aq) \)

The \( K_a \) expression for this reaction is \( \frac{[H^{+}][F^{-}]}{[HF]} \), where \([H^{+}]\) and \([F^{-}]\) are the concentrations of hydrogen and fluoride ions, and \([HF]\) is the concentration of the undissociated acid still in solution.

Using \( K_a \), you can determine the extent of acid ionization and thus predict the acidic behavior in equilibrium conditions. This helps in calculating the concentration of ions in solutions, which can influence other reactions, such as precipitation.

Precipitation Reactions

Precipitation reactions occur when solutions containing soluble salts are mixed, leading to the formation of an insoluble solid, or precipitate. These reactions are fundamental in achieving equilibrium in certain systems.

A precipitate forms when the product of the ion concentrations of the compound in a solution exceeds its \( K_{sp} \). This critical point marks where the dissolved ions recombine into a solid lattice.

• Example Reaction: Mixing \( Ca^{2+} \) ions with \( F^{-} \) ions will lead to the precipitation of \( CaF_2 \).

Precipitation helps in removing ions from solutions and is used in processes like water purification and qualitative analysis in chemistry labs. By manipulating ion concentrations, chemists control when and how substances form precipitates.

Ionic Equations

Ionic equations are an invaluable tool in understanding chemical reactions, especially those happening in aqueous solutions. They provide a clear picture by showing only the species that change during the reaction process.

There are two main types of ionic equations:

• Complete Ionic Equations: Show all of the ions present in the reaction mixture, spotlighting the decomposition into ions.
• Net Ionic Equations: Focus exclusively on the ions that undergo changes, omitting spectator ions that do not participate in the reaction.

For instance, in the reaction forming \( CaF_2 \):- The complete ionic equation captures all ions in play.- The net ionic equation, however, boils down to only change-causing ions.

Understanding ionic equations allows for a more in-depth insight into how ions interact, precipitate, or remain unreacted in the solution, aiding in many practical chemical applications.