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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 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 by step solution

01

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: \[HF(aq) \rightleftharpoons H^+(aq) + F^-(aq)\]
02

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: \[4.0 \times 10^{-11} = [x][2x]^2\]
03

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: \[7.2 \times 10^{-4} = \frac{x \times x}{1.0 - x}\]
04

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.

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Most popular questions from this chapter

Tooth enamel is composed of the mineral hydroxyapatite. The \(K_{\mathrm{sp}}\) of hydroxyapatite, $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH},\( is \)6.8 \times 10^{-37}$ . Calculate the solubility of hydroxyapatite in pure water in moles per liter. How is the solubility of hydroxyapatite affected by adding acid? When hydroxyapatite is treated with fluoride, the mineral fluorapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F}\) , forms. The \(K_{\mathrm{sp}}\) of this substance is \(1 \times 10^{-60}\) . Calculate the solubility of fluorapatite in water. How do these calculations provide a rationale for the fluoridation of drinking water?

Write balanced equations for the dissolution reactions and the corresponding solubility product expressions for each of the following solids a. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) b. \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) c. \(\mathrm{BaF}_{2}\)

Which of the following will affect the total amount of solute that can dissolve in a given amount of solvent? a. The solution is stirred. b. The solute is ground to fine particles before dissolving. c. The temperature changes

Calculate the equilibrium concentrations of $\mathrm{NH}_{3}, \mathrm{Cu}^{2+}\( \)\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2+}, \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}^{2+}, \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}^{2+},$ and \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) in a solution prepared by mixing \(500.0 \mathrm{mL}\) of \(3.00M\) \(\mathrm{NH}_{3}\) with $500.0 \mathrm{mL}\( of \)2.00 \times 10^{-3} \mathrm{M} \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}$ . The step wise equilibria are $$\mathrm{Cu}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{CuNH}_{3}^{2+}(a q) \quad K_{1}=1.86 \times 10^{4}$$ $$\mathrm{CuNH}_{3}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}^{2+}(a q) \quad K_{2}=3.88 \times 10^{3}$$ $$\mathrm{Cu}\left(\mathrm{NH}_{2}\right)_{2}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{2}\right)_{3}^{2+}(a q) \quad K_{3}=1.00 \times 10^{3}$$ $$\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \quad K_{4}=1.55 \times 10^{2}$$

Solubility is an equilibrium position, whereas \(K_{\mathrm{sp}}\) is an equilibrium constant. Explain the difference.
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