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Chapter 16: Problem 119

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

Expert verified
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 CaF鈧� dissolution and HF ionization. 2. Use the $K_{sp}$ value for CaF鈧� to find the maximum concentration of F鈦� that causes precipitation. 3. Calculate the concentration of F鈦� ions in the 1.0 M $\mathrm{HF}$ solution using the given $K_{a}$ value. 4. Determine the additional mass of F鈦� needed to begin precipitation and calculate the mass of $\mathrm{Ca(NO_3)_2}$ required. Upon finding the concentration of F鈦� needed for precipitation and the concentration of F鈦� in the $\mathrm{HF}$ solution, we can calculate the mass of Ca(NO鈧�)鈧�: \[Mass \ of \ Ca(NO_{3})_{2} = \frac{Mass \ of \ F鈦� \ needed}{2}\] Substitute the obtained values from steps 2 and 3 and calculate the mass of $\mathrm{Ca(NO_3)_2}$ needed.

Step by step solution

01

Write the Chemical Reactions

First, we need to write down the chemical reactions for the dissolution of CaF鈧� and the ionization of HF. For CaF鈧� dissolution: \[CaF_{2}(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)\] For HF ionization: \[HF(aq) \rightleftharpoons H^+(aq) + F^-(aq)\]
02

Calculate Maximum Concentration of F鈦� to Begin Precipitation

To find the concentration of F鈦� ions that will cause precipitation of CaF鈧�, we will utilize the Ksp value given: \[K_{sp} = [Ca^{2+}][F^-]^2\] Since the precipitation will just begin, the concentration of Ca虏鈦� and F鈦� ions will be equal at the point of saturation, so: \[K_{sp} = [x][2x]^2\] where x is the concentration of Ca虏鈦� ions, and 2x the concentration of F鈦� ions. Now, substitute the given Ksp value: \[4.0 \times 10^{-11} = [x][2x]^2\]
03

Calculate the concentration of F鈦� ions in 1.0 M HF

To calculate the concentration of F鈦� 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鈦篯 = x, and [F鈦籡 = x. Substitute the given Ka value, and solve for x: \[7.2 \times 10^{-4} = \frac{x \times x}{1.0 - x}\]
04

Determine the mass of Ca(NO鈧�)鈧� that needs to be added

Now that we have the concentration of F鈦� ions needed to begin precipitation and the concentration of F鈦� ions in the 1.0 M HF solution, we can calculate the additional mass of F鈦� needed: \[Mass \ of \ F鈦� \ needed = (2\times[Ca(NO_{3})_{2}])\times \left[(2 \times \ Concentration \ of \ F鈦� \ needed) - (Concentration \ of \ F鈦� \ in \ HF \ solution)\right]\] Then, we can calculate the mass of Ca(NO鈧�)鈧� that needs to be added: \[Mass \ of \ Ca(NO_{3})_{2} = \frac{Mass \ of \ F鈦� \ needed}{2}\] Calculate the mass of Ca(NO鈧�)鈧� by substituting the values obtained from the previous steps, and that will be the answer to our problem.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

In the presence of \(\mathrm{CN}^{-}, \mathrm{Fe}^{3+}\) forms the complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-} .\) The equilibrium concentrations of \(\mathrm{Fe}^{3+}\) and \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) are \(8.5 \times 10^{-40} \mathrm{M}\) and \(1.5 \times 10^{-3} M,\) respectively, in a \(0.11-M\) KCN solution. Calculate the value for the overall formation constant of \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) . $$\mathrm{Fe}^{3+}(a q)+6 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q) \qquad K_{\text { overall }}=?$$

The solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) in a \(0.10-M \mathrm{KIO}_{3}\) solution is \(2.6 \times 10^{-11} \mathrm{mol} / \mathrm{L}\) . Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}.\)

Order the following solids (a鈥揹) from least soluble to most soluble. Ignore any potential reactions of the ions with water. a. \(\mathrm{AgCl} \quad K_{s p}=1.6 \times 10^{-10}\) b. \(\mathrm{Ag}_{2} \mathrm{S} \quad K_{\mathrm{sp}}=1.6 \times 10^{-49}\) c. \(\mathrm{CaF}_{2} \quad K_{\mathrm{sp}}=4.0 \times 10^{-11}\) d. \(\mathrm{CuS} \quad K_{\mathrm{sp}}=8.5 \times 10^{-45}\)

The U.S. Public Health Service recommends the fluoridation of water as a means for preventing tooth decay. The recommended concentration is 1 \(\mathrm{mg} \mathrm{F}^{-}\) per liter. The presence of calcium ions in hard water can precipitate the added fluoride. What is the maximum molarity of calcium ions in hard water if the fluoride concentration is at the USPHS recommended level? \(\left(K_{\mathrm{sp}} \text { for } \mathrm{CaF}_{2}=4.0 \times 10^{-11}\right)\)

Calculate the molar solubility of \(\mathrm{Co}(\mathrm{OH})_{3}, K_{\mathrm{sp}}=2.5 \times 10^{-43}.\)
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