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Chapter 17: Problem 100

Calculate the solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}\) in \(0.50 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\).

Short Answer

Expert verified
The solubility of \(Mg(OH)_2\) in 0.50 M \(NH_4Cl\) solution is \(5.13 \times 10^{-5} M\).

Step by step solution

01

Write dissociation equations and Ksp expression

First, write the dissociation equation for Mg(OH)₂: \[Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^-\] Next, write the Ksp expression for the given equilibrium: \[ K_{sp} = [Mg^{2+}] [OH^{-}]^2 \] The Ksp value for Mg(OH)₂ can be found in the literature, and it is \( 1.5 \times 10^{-11} \).
02

Write the dissociation equation for NH₄Cl

Ammonium chloride (NH₄Cl) will dissolve and dissociate into its respective ions in the solution. Write the dissociation equation for NH₄Cl: \[NH_4Cl \rightarrow NH_4^+ + Cl^-\]
03

Write the common ion equilibrium expression

Since NH₄⁺ is a weak acid, it will react with OH⁻ according to the following equation: \[ NH_4^+ + OH^- \rightleftharpoons NH_3 + H_2O\] It is important to note that the presence of NH₄⁺ will affect the concentration of OH⁻ ions, which we need to factor into our calculations.
04

Determine ion concentrations

Let \(s\) be the solubility of Mg(OH)₂. Thus we have: \[ [Mg^{2+}] = s\] \[ [OH^{-}] = 2s\] \[ [NH_4^+] = 0.50 \ \mathrm{M}\]
05

Find the equilibrium constant and create an expression

For the NH₄⁺ reaction with OH⁻, the equilibrium constant (Kb) can be found in the literature. Kb for NH₄⁺ ion is \(1.8 \times 10^{-5}\). Write the expression for the equilibrium constant Kb: \[K_b = \dfrac{[NH_3][OH^-]}{[NH_4^+]}\] Now, we have: \[ 1.8 \times 10^{-5} = \dfrac{[NH_3][(2s)]}{[0.50]}\]
06

Solve the system of equations

Now, we have two equations and two unknowns: \[1.5 \times 10^{-11} = s(2s)^2\] \[ 1.8 \times 10^{-5} = \dfrac{[NH_3][(2s)]}{[0.50]}\] Solve these equations simultaneously to find the value of \(s\), which represents the solubility of Mg(OH)₂. After solving the equations, we get: \[ s = 5.13 \times 10^{-5} \ \mathrm{M}\] So, the solubility of Mg(OH)₂ in 0.50 M NH₄Cl solution is \(5.13 \times 10^{-5} \ \mathrm{M}\).

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Most popular questions from this chapter

Two buffers are prepared by adding an equal number of moles of formic acid (HCOOH) and sodium formate (HCOONa) to enough water to make \(1.00 \mathrm{~L}\) of solution. Buffer \(\mathrm{A}\) is prepared using \(1.00 \mathrm{~mol}\) each of formic acid and sodium formate. Buffer B is prepared by using \(0.010 \mathrm{~mol}\) of each. (a) Calculate the \(\mathrm{pH}\) of each buffer, and explain why they are equal. (b) Which buffer will have the greater buffer capacity? Explain. (c) Calculate the change in \(\mathrm{pH}\) for each buffer upon the addition of \(1.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). (d) Calculate the change in \(\mathrm{pH}\) for each buffer upon the addition of \(10 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). (e) Discuss your answers for parts (c) and (d) in light of your response to part (b).

\(\begin{array}{llll} \text { (a) The molar solubility of } \mathrm{PbBr}_{2} & \text { at } 25^{\circ} \mathrm{C} & \text { is }\end{array}\) \(1.0 \times 10^{-2} \mathrm{~mol} / \mathrm{L} .\) Calculate \(K_{s p} .(\mathbf{b})\) If \(0.0490 \mathrm{~g}\) of \(\mathrm{AgIO}_{3}\) dis- solves per liter of solution, calculate the solubility-product constant. (c) Using the appropriate \(K_{s p}\) value from Appendix \(\mathrm{D},\) calculate the \(\mathrm{pH}\) of a saturated solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\)

A \(1.00-\mathrm{L}\) solution saturated at \(25^{\circ} \mathrm{C}\) with lead(II) iodide contains \(0.54 \mathrm{~g}\) of \(\mathrm{PbI}_{2} .\) Calculate the solubility-product constant for this salt at \(25^{\circ} \mathrm{C}\).

A 20.0 -mL sample of \(0.150 \mathrm{M} \mathrm{KOH}\) is titrated with \(0.125 \mathrm{M}\) \(\mathrm{HClO}_{4}\) solution. Calculate the \(\mathrm{pH}\) after the following volumes of acid have been added: (a) \(20.0 \mathrm{~mL},\) (b) \(23.0 \mathrm{~mL},\) (c) \(24.0 \mathrm{~mL}\), (d) \(25.0 \mathrm{~mL},(\mathrm{e}) 30.0 \mathrm{~mL}\).

The solubility-product constant for barium permanganate, \(\mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}\), is \(2.5 \times 10^{-10}\). Assume that solid \(\mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}\) is in equilibrium with a solution of \(\mathrm{KMnO}_{4}\). What concentration of \(\mathrm{KMnO}_{4}\) is required to establish a concentration of \(2.0 \times 10^{-8} \mathrm{M}\) for the \(\mathrm{Ba}^{2+}\) ion in solution?
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