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Chapter 16: 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, \(K_{sp}\), for the ionic compound \(M_{2}X_{3}\) is approximately \(1.24 \times 10^{-43}\).

Step by step solution

01

Convert solubility from g/L to mol/L

First, we need to convert the given solubility, which is expressed in g/L, to a molar solubility expressed in mol/L. To do this, we will divide the given solubility in g/L by the given molar mass of the compound. \(Molar\: solubility = \frac{Solubility\: (g/L)}{Molar\: mass\: (g/mol)}\)
02

Calculate ion concentration in the saturated solution

We will use the molar solubility to find the concentration of ions in a saturated solution of M₂X₃. Each M₂X₃ unit dissociates into 2 M⁺ ions and 3 X⁻ ions. So, if 's' is the molar solubility, the resulting concentrations will be: MM: 2s XX: 3s
03

Calculate Ksp

Now that we have the ion concentrations in the saturated solution, we can calculate the Ksp using the following expression: \(K_{sp} = [M^+]^2 [X^-]^3\) Plugging in the ion concentrations (2s for M⁺ and 3s for X⁻) into the formula and simplifying: \(K_{sp} = (2s)^2 (3s)^3\) Now we can substitute the molar solubility from Step 1 and compute the Ksp.
04

Substitute the molar solubility and simplify

Substitute the molar solubility found in Step 1 into the Ksp expression and simplify: \( K_{sp} = (2 \times (\frac{3.60 \times 10^{-7}}{288}))^2 (3 \times (\frac{3.60 \times 10^{-7}}{288}))^3 \) After substituting and solving for Ksp: \(K_{sp} \approx 1.24 \times 10^{-43}\) The solubility product constant, Ksp, for the ionic compound M₂X₃ is approximately 1.24 x 10⁻⁴³.

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Most popular questions from this chapter

Write equations for the stepwise formation of each of the following complex ions. a. \(\mathrm{Ni}(\mathrm{CN})_{4}^{2-}\) b. \(\mathrm{V}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}{ }^{3-}\)

A solution is prepared by mixing \(75.0 \mathrm{~mL}\) of \(0.020 \mathrm{M} \mathrm{BaCl}_{2}\) and \(125 \mathrm{~mL}\) of \(0.040 \mathrm{M} \mathrm{K}_{2} \mathrm{SO}_{4}\). What are the concentrations of barium and sulfate ions in this solution? Assume only \(\mathrm{SO}_{4}^{2-}\) ions (no \(\left.\mathrm{HSO}_{4}^{-}\right)\) are present.

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

Consider \(1.0 \mathrm{~L}\) of an aqueous solution that contains \(0.10 \mathrm{M}\) sulfuric acid to which \(0.30\) mol barium nitrate is added. Assuming no change in volume of the solution, determine the \(\mathrm{pH}\), the concentration of barium ions in the final solution, and the mass of solid formed.

a. Using the \(K_{\mathrm{sp}}\) value for \(\mathrm{Cu}(\mathrm{OH})_{2}\left(1.6 \times 10^{-19}\right)\) and the overall formation constant for \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\left(1.0 \times 10^{13}\right)\), calculate the value for the equilibrium constant for the following reaction: \(\mathrm{Cu}(\mathrm{OH})_{2}(s)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)\) b. Use the value of the equilibrium constant you calculated in part a to calculate the solubility (in \(\mathrm{mol} / \mathrm{L}\) ) of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0 \mathrm{M}\) \(\mathrm{NH}_{3} .\) In \(5.0 \mathrm{M} \mathrm{NH}_{3}\) the concentration of \(\mathrm{OH}^{-}\) is \(0.0095 \mathrm{M} .\)
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