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Chapter 17: Problem 94

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)鈧� in the presence of 0.50 M NH鈧凜l is approximately \(1.31 \times 10^{-4} \mathrm{M}\).

Step by step solution

01

Write down the dissolution reaction

When Mg(OH)鈧� dissolves in water, it forms Mg虏鈦� and OH鈦� 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)鈧�

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)鈧�) at 25掳C, Ksp is \(1.5 \times 10^{-11}\).
03

Find the initial concentration of Cl鈦� ions

Given that NH鈧凜l is a strong electrolyte, it will completely dissociate in water. So, the initial concentration of Cl鈦� ions is the same as the concentration of NH鈧凜l, 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)鈧�: ``` Mg(OH)鈧� (s) 鈫� Mg虏鈦� (aq) + 2OH鈦� (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 Mg虏鈦� and OH鈦� ions from the ICE table: \[K_{sp} = (s)(2s)^2\]
05

Use the common ion effect

As NH鈧凜l is present in the solution, its Cl鈦� ions will combine with Mg虏鈦� ions to form MgCl鈧�, reducing the concentration of Mg虏鈦� ions. On the other hand, some of the OH鈦� ions will combine with NH鈧勨伜 ions to form NH鈧凮H, reducing the concentration of OH鈦� ions. Since the concentration of Mg虏鈦� ions decreases, the solubility of Mg(OH)鈧� will also decrease.
06

Solve for solubility (s)

We are given the Ksp of Mg(OH)鈧� 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)鈧� in the presence of NH鈧凜l

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

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Most popular questions from this chapter

From the value of \(K_{f}\) listed in Table \(17.1\), calculate the concentration of Curt in \(1.0 \mathrm{~L}\) of a solution that contains a total of \(1 \times 10^{-3} \mathrm{~mol}\) of copper(II) ion and that is \(0.10 \mathrm{M}\) in \(\mathrm{NH}_{3}\).

Suggest how the cations in each of the following solution mixtures can be separated: (a) \(\mathrm{Na}^{+}\) and \(\mathrm{Cd}^{2+}\), (b) \(\mathrm{Cu}^{2+}\) and \(\mathrm{Mg}^{2+}\), (c) \(\mathrm{Pb}^{2+}\) and \(\mathrm{Al}^{3+}\), (d) \(\mathrm{Ag}^{+}\) and \(\mathrm{Hg}^{2+}\).

(a) What is the common-ion effect? (b) Give an example of a salt that can decrease the ionization of \(\mathrm{HNO}_{2}\) in solution.

How many milliliters of \(0.105 \mathrm{M} \mathrm{HCl}\) are needed to titrate each of the following solutions to the equivalence point: (a) \(45.0 \mathrm{~mL}\) of \(0.0950 \mathrm{M} \mathrm{NaOH}\), (b) \(22.5 \mathrm{~mL}\) of \(0.118 \mathrm{M} \mathrm{NH}_{3}\) (c) \(125.0 \mathrm{~mL}\) of a solution that contains \(1.35 \mathrm{~g}\) of \(\mathrm{NaOH}\) per liter?

Use information from Appendix \(\mathrm{D}\) to calculate the \(\mathrm{pH}\) of (a) a solution that is \(0.150 \mathrm{M}\) in sodium formate \((\mathrm{HCOONa})\) and \(0.200 \mathrm{M}\) in formic acid \((\mathrm{HCOOH})\) (b) a solution that is \(0.210 \mathrm{M}\) in pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\) and \(0.350 \mathrm{M}\) in pyridinium chloride \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHCl}\right) ;\) (c) a solution that is made by combining \(125 \mathrm{~mL}\) of \(0.050 \mathrm{M}\) hydrofluoric acid with \(50.0 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) sodium fluoride.
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