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Chapter 15: Problem 24

The solubility of the ionic compound \(\mathrm{M}_{2} \mathrm{X}_{3},\) having a molar mass of \(288 \mathrm{g} / \mathrm{mol},\) is \(3.60 \times 10^{-7} \mathrm{g} / \mathrm{L} .\) Calculate the \(K_{\mathrm{sp}}\) of the compound.

Short Answer

Expert verified
The solubility product constant (Ksp) of the compound M2X3 is approximately \(2.58 \times 10^{-32}\).

Step by step solution

01

Write the dissolution equation for the compound M2X3

The equation representing the dissolution of M2X3 in water is: \(M_2X_3 (s) \rightleftharpoons 2M^{a+}(aq) + 3X^{b-}(aq)\) where "a" and "b" are the charges of M and X, respectively.
02

Find the molar solubility

We are given the solubility of the ionic compound M2X3 in grams per liter, and we need to find the molar solubility, or the mol/L solubility. Molar solubility = \( \frac{Solubility (g/L)}{Molar Mass (g/mol)} \) Given the solubility = \(3.60 \times 10^{-7}\,g/L\) and molar mass = \(288\,g/mol\), Molar solubility = \(\frac{3.60 \times 10^{-7}\,g/L}{288\,g/mol} = 1.25 \times 10^{-9}\,mol/L\)
03

Set up expressions for the concentrations of the ions

Using the stoichiometry from the dissolution equation in Step 1, we can write the expressions for the equilibrium concentrations of M+ and X- ions in terms of the molar solubility. For the M+ ion, its molar solubility is 2 times that of M2X3: [M+] = \(2 \times (1.25 \times 10^{-9}\,mol/L) = 2.50 \times 10^{-9}\,mol/L\) For the X- ion, its molar solubility is 3 times that of M2X3: [X-] = \(3 \times (1.25 \times 10^{-9}\,mol/L) = 3.75 \times 10^{-9}\,mol/L\)
04

Calculate the Ksp

Now, we can use the expressions for the concentrations of the ions to calculate the Ksp of the compound: \(K_{sp} = [M^{a+}]^{2}[X^{b-}]^{3}\) Substituting the values for [M+] and [X-]: \(K_{sp} = (2.50 \times 10^{-9} \,mol/L)^{2}(3.75 \times 10^{-9} \,mol/L)^{3}\) \(K_{sp} = 2.578 \times 10^{-32}\) So, the solubility product constant (Ksp) of the compound M2X3 is approximately \(2.58 \times 10^{-32}\).

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Most popular questions from this chapter

Consider a solution made by mixing \(500.0 \mathrm{mL}\) of \(4.0\) \(M\) \(\mathrm{NH}_{3}\) and \(500.0 \mathrm{mL}\) of \(0.40\) \(M\) \(\mathrm{AgNO}_{3} . \mathrm{Ag}^{+}\) reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{AgNH}_{3}^{+}\) and \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}:\) $$\begin{aligned} \mathrm{Ag}^{+}(a q)+\mathrm{NH}_{3}(a q) & \rightleftharpoons \mathrm{AgNH}_{3}^{+}(a q) & K_{1} &=2.1 \times 10^{3} \\ \mathrm{AgNH}_{3}^{+}(a q)+\mathrm{NH}_{3}(a q) & \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) & K_{2} &=8.2 \times 10^{3} \end{aligned}$$ Determine the concentration of all species in solution.

Use the following data to calculate the \(K_{\mathrm{sp}}\) value for each solid. a. The solubility of \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is \(6.2 \times 10^{-12} \mathrm{mol} / \mathrm{L}\). b. The solubility of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) is \(7.4 \times 10^{-2} \mathrm{mol} / \mathrm{L}\).

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

\(\mathrm{Ag}_{2} \mathrm{S}(s)\) has a larger molar solubility than CuS even though \(\mathrm{Ag}_{2} \mathrm{S}\) has the smaller \(K_{\mathrm{sp}}\) value. Explain how this is possible.

A solution contains 0.018 molel each of \(\mathrm{I}^{-}, \mathrm{Br}^{-},\) and \(\mathrm{Cl}^{-}\). When the solution is mixed with \(200 . \mathrm{mL}\) of \(0.24\) \(M\) \(\mathrm{AgNO}_{3}\), what mass of \(\mathrm{AgCl}(s)\) precipitates out, and what is \(\left[\mathrm{Ag}^{+}\right] ?\) Assume no volume change. $$\begin{aligned} \operatorname{AgI}: K_{\mathrm{sp}} &=1.5 \times 10^{-16} \\ \operatorname{AgBr}: K_{\mathrm{sp}} &=5.0 \times 10^{-13} \\ \mathrm{AgCl}: K_{\mathrm{sp}} &=1.6 \times 10^{-10} \end{aligned}$$
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