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Chapter 16: Problem 120

The \(K_{\mathrm{sp}}\) for \(Q,\) a slightly soluble ionic compound composed of \(\mathrm{M}_{2}^{2+}\) and \(\mathrm{X}^{-}\) ions, is \(4.5 \times 10^{-29} .\) The electron configuration of \(\mathrm{M}^{+}\) is $[\mathrm{Xe}] 6 s^{1} 4 f^{4} 5 d^{10} .\( The \)\mathrm{X}^{-}$ anion has 54 electrons. What is the molar solubility of \(Q\) in a solution of \(\mathrm{NaX}\) prepared by dissolving \(1.98 \mathrm{g}\) \(\mathrm{NaX}\) in \(150 .\) mL solution?

Short Answer

Expert verified
The molar solubility of Q (TbI2) in the NaI solution is approximately \(5.79 \times 10^{-28} M\).

Step by step solution

01

Find the ions composing the ionic compound Q.

To determine what elements correspond to the M^2+ and X- ions, we'll look at their electron configurations. The element with an electron configuration of [Xe] 6s1 4f4 5d10 when it has lost 2 electrons will correspond to the M2+ ion. The X- ion has 54 electrons, so it has one more electron than its corresponding neutral element; we need to find which element has 53 protons. We can use the periodic table to determine that the neutral element with 53 protons is Iodine (I), so X- ion will be I-. The M^+ ion corresponds to Tb^+ with an atomic number of 65 and its 2+ charged ion as Tb^2+. Now we know that Q is composed of Tb^2+ and I- ions, the formula of Q can be written as TbI2.
02

Write the balanced dissolution equation of Q and setup the Ksp expression.

The balanced chemical equation for the dissolution of Q in water is: \( TbI_2(s) \rightleftharpoons Tb^{2+}(aq) + 2I^-(aq) \) The Ksp expression for this dissolution process can be written as: \( K_{sp} = [Tb^{2+}][I^-]^2 \) The given Ksp value is \(4.5 \times 10^{-29}\).
03

Find the concentration of I- ions from NaX solution.

We're given 1.98 g of NaI is dissolved in a 150 mL solution. To find the concentration of I- ions, we need to convert grams of NaI into moles and then divide by the volume of the solution in liters. First, calculate the moles of NaI: \(\frac{1.98~g~NaI}{1~mol~NaI/149.89~g~NaI} = 0.0132~mol~NaI\) Now, calculate the concentration of I- ions: \(\frac{0.0132~mol~I^-}{0.150~L} = 0.088~M\)
04

Calculate the molar solubility of Q.

To find the molar solubility of Q, let x be the concentration of Tb^2+ ions in the solution. Since there are two I- ions for every Tb^2+ ion, the concentration of I- ions will be (0.088+2x). Now we can plug these concentrations into the Ksp expression and solve for x: \( K_{sp} = [Tb^{2+}][I^-]^2 \) \( 4.5 \times 10^{-29} = x(0.088 + 2x)^2 \) We can assume that x is much smaller than 0.088, which simplifies the equation: \( 4.5 \times 10^{-29} = x(0.088)^2 \) Now, solve for x: \( x = \frac{4.5 \times 10^{-29}}{(0.088)^2} \) \( x = 5.79 \times 10^{-28} \) So, the molar solubility of Q (TbI2) in the NaI solution is approximately \(5.79 \times 10^{-28} M\).

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Most popular questions from this chapter

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 \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}.\)

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

Calculate the final concentrations of $\mathrm{K}^{+}(a q), \mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q),\( \)\mathrm{Ba}^{2+}(a q),$ and \(\operatorname{Br}^{-}(a q)\) in a solution prepared by adding 0.100 \(\mathrm{L}\) of \(0.200M\) \(\mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) to 0.150 \(\mathrm{L}\) of \(0.250 M\) \(\mathrm{BaBr}_{2}\) . (For $\mathrm{BaC}_{2} \mathrm{O}_{4}, K_{\mathrm{sp}}=2.3 \times 10^{-8} . )$

The copper(I) ion forms a chloride salt that has $K_{\mathrm{sp}}= 1.2 \times 10^{-6} .\( Copper (I) also forms a complex ion with \)\mathrm{Cl}^{-} :$ $$\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \qquad K=8.7 \times 10^{4}$$ a. Calculate the solubility of copper(I) chloride in pure water. (Ignore \(\mathrm{CuCl}_{2}^{-}\) formation for part a.) b. Calculate the solubility of copper(I) chloride in 0.10\(M\) \(\mathrm{NaCl}\).

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? $\left(K_{\mathrm{sp}} \text { for } \mathrm{CaF}_{2}=4.0 \times 10^{-11}\right)$
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