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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 the presence of 0.50 M NH₄Cl is approximately \(1.31 \times 10^{-4} \mathrm{M}\).

Step by step solution

01

Write down the dissolution reaction

When Mg(OH)₂ dissolves in water, it forms Mg²⁺ and OH⁻ ions. The dissolution reaction can be written as: \[ \mathrm{Mg(OH)}_{2}(s) \rightleftharpoons \mathrm{Mg}^{2+}(aq) + 2\mathrm{OH}^{-}(aq) \]
02

Determine the solubility product constant (Ksp) of Mg(OH)₂

The solubility product constant, Ksp, is a measure of the solubility of a slightly soluble salt in water. This value is temperature-dependent and can be found in reference tables. For magnesium hydroxide (Mg(OH)₂) at 25°C, Ksp is \(1.5 \times 10^{-11}\).
03

Find the initial concentration of Cl⁻ ions

Given that NH₄Cl is a strong electrolyte, it will completely dissociate in water. So, the initial concentration of Cl⁻ ions is the same as the concentration of NH₄Cl, which is 0.50 M.
04

Set up an ICE table and the Ksp expression

Using an Initial-Change-Equilibrium (ICE) table, we can relate the concentrations of the ions involved in the reaction to the solubility (s) of Mg(OH)₂: ``` Mg(OH)₂ (s) ↔ Mg²⁺ (aq) + 2OH⁻ (aq) Initial s 0 0 Change -s +s +2s Equilibrium 0 s 2s ``` Now, we can write the Ksp expression for the dissolution reaction: \[K_{sp} = [\mathrm{Mg^{2+}}][\mathrm{OH^-}]^2\] Plug in the solubility of Mg²⁺ and OH⁻ ions from the ICE table: \[K_{sp} = (s)(2s)^2\]
05

Use the common ion effect

As NH₄Cl is present in the solution, its Cl⁻ ions will combine with Mg²⁺ ions to form MgCl₂, reducing the concentration of Mg²⁺ ions. On the other hand, some of the OH⁻ ions will combine with NH₄⁺ ions to form NH₄OH, reducing the concentration of OH⁻ ions. Since the concentration of Mg²⁺ ions decreases, the solubility of Mg(OH)₂ will also decrease.
06

Solve for solubility (s)

We are given the Ksp of Mg(OH)₂ and the ICE table, so we can solve for solubility (s). Substitute the Ksp value we found in "Step 2" into the Ksp expression: \[(1.5 \times 10^{-11}) = (s)(2s)^2\] To solve for s, simplify the above equation: \[(1.5 \times 10^{-11}) = 4s^3\] Now, divide \(1.5 \times 10^{-11}\) by 4: \[s^3 = \frac{1.5 \times 10^{-11}}{4}\] Then, find the cube root of the result to obtain s: \[s = \sqrt[3]{\frac{1.5 \times 10^{-11}}{4}}\] Calculate the final value of s: \[s \approx 1.31 \times 10^{-4} \mathrm{M}\]
07

Find the solubility of Mg(OH)₂ in the presence of NH₄Cl

The solubility of Mg(OH)₂ in the presence of NH₄Cl is approximated by s, which is \(1.31 \times 10^{-4} \mathrm{M}\).

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Most popular questions from this chapter

Excess \(\mathrm{Ca}(\mathrm{OH})_{2}\) is shaken with water to produce a saturated solution. The solution is filtered, and a \(50.00-\mathrm{mL}\) sample titrated with \(\mathrm{HCl}\) requires \(11.23 \mathrm{~mL}\) of \(0.0983 \mathrm{M}\) \(\mathrm{HCl}\) to reach the end point. Calculate \(K_{s p}\) for \(\mathrm{Ca}(\mathrm{OH})_{2}\) Compare your result with that in Appendix D. Do you think the solution was kept at \(25^{\circ} \mathrm{C}\) ?

A \(30.0-\mathrm{mL}\) 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) \(30.0 \mathrm{~mL}\), (b) \(35.0 \mathrm{~mL}\), (c) \(36.0 \mathrm{~mL}\), (d) \(37.0 \mathrm{~mL}\), (e) \(40.0 \mathrm{~mL}\).

How many milliliters of \(0.0850 \mathrm{M} \mathrm{NaOH}\) are required to titrate each of the following solutions to the equivalence point: (a) \(40.0 \mathrm{~mL}\) of \(0.0900 \mathrm{M} \mathrm{HNO}_{3}\), (b) \(35.0 \mathrm{~mL}\) of \(0.0850 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\), (c) \(50.0 \mathrm{~mL}\) of a solution that con- tains \(1.85 \mathrm{~g}\) of \(\mathrm{HCl}\) per liter?

(a) Calculate the \(\mathrm{pH}\) of a buffer that is \(0.105 \mathrm{M}\) in \(\mathrm{NaHCO}_{3}\) and \(0.125 \mathrm{M}\) in \(\mathrm{Na}_{2} \mathrm{CO}_{3} .\) (b) Calculate the \(\mathrm{pH}\) of a solution formed by mixing \(65 \mathrm{~mL}\) of \(0.20 \mathrm{M}\) \(\mathrm{NaHCO}_{3}\) with \(75 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\)

(a) Explain the difference between solubility and solubility-product constant. (b) Write the expression for the solubility-product constant for each of the following ionic compounds: \(\mathrm{MnCO}_{3}, \mathrm{Hg}(\mathrm{OH})_{2}\), and \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).

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