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Chapter 16: Problem 73

Compare the molar solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}\) in water and in a solution buffered at a pH of 9.0 .

Short Answer

Expert verified
The molar solubility of Mg(OH)2 in water is approximately \(1.64 \times 10^{-4} M\) and in a solution buffered at a pH of 9.0 is approximately \(0.12 M\). This shows that the molar solubility increases significantly in the buffer solution.

Step by step solution

01

Determine the \(K_{sp}\) of Mg(OH)2

We know that \(\mathrm{Mg}(\mathrm{OH})_{2}\) dissociates into ions as follows: \(\mathrm{Mg}(\mathrm{OH})_{2} \rightarrow \mathrm{Mg}^{2+} + 2\mathrm{OH}^{-}\). In this case, it's known that the solubility product constant, \(K_{sp}\), of \(\mathrm{Mg(OH)2}\) is \(1.2 \times 10^{-11}\).
02

Calculate molar solubility in water

Let the molar solubility of \(\mathrm{Mg(OH)2}\) in water be s. Then, we can write the expression for \(K_{sp}\) as follows: \(K_{sp} = [\mathrm{Mg}^{2+}] [(\mathrm{OH}^-)]^2 = (s)((2s)^2) = 4s^3\). Solving for s, we find that the molar solubility of Mg(OH)2 in water is approximately \(1.64 \times 10^{-4} M\).
03

Calculate molar solubility in a buffered solution

The pH of the buffer gives us the concentration of OH- ions, which is \(10^{-pOH}\) or \(10^{-14+pH}\). Given that the pH of the solution is 9.0, the pOH is 5.0. Therefore, the concentration of OH- ions is \(10^{-5}\). Now, we can write the expression for \(K_{sp}\) again, but this time in terms of the buffered solution: \(K_{sp} = [\mathrm{Mg}^{2+}] [(\mathrm{OH}^-)]^2 = ([s_{buffered}]) ((10^{-5})^2)\) = 1.2 \times 10^{-11}\). Solving for \(s_{buffered}\), we find that the molar solubility of Mg(OH)2 in the buffer is approximately \(0.12 M\).
04

Compare the molar solubilities

We see that the molar solubility of Mg(OH)2 in the buffer solution is much larger than that in pure water, which demonstrates the effect of the buffer on the solubility of the compound.

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Most popular questions from this chapter

The iodide impurity in a \(4.50-\mathrm{g}\) sample of a metal nitrate is precipitated as silver iodide. If \(5.54 \mathrm{~mL}\) of \(0.186 M\) AgNO \(_{3}\) solution is needed for the precipitation, calculate the mass percent of iodide in the sample.

A \(10.0-\mathrm{mL}\) solution of \(0.300 \mathrm{M} \mathrm{NH}_{3}\) is titrated with a \(0.100 M \mathrm{HCl}\) solution. Calculate the \(\mathrm{pH}\) after the following additions of the \(\mathrm{HCl}\) solution: (a) \(0.0 \mathrm{~mL},\) (b) \(10.0 \mathrm{~mL}\) (c) \(20.0 \mathrm{~mL}\) (d) \(30.0 \mathrm{~mL}\) (e) \(40.0 \mathrm{~mL}\)

Write the solubility product expression for the ionic compound \(\mathrm{A}_{x} \mathrm{~B}_{y}\)

The molar mass of a certain metal carbonate, \(\mathrm{MCO}_{3}\), can be determined by adding an excess of \(\mathrm{HCl}\) acid to react with all the carbonate and then "back titrating" the remaining acid with a \(\mathrm{NaOH}\) solution. (a) Write equations for these reactions. (b) In a certain experiment, \(18.68 \mathrm{~mL}\) of \(5.653 \mathrm{M} \mathrm{HCl}\) were added to a \(3.542-\mathrm{g}\) sample of \(\mathrm{MCO}_{3}\). The excess HCl required \(12.06 \mathrm{~mL}\) of \(1.789 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the molar mass of the carbonate and identify \(\mathrm{M}\).

Which of the following solutions can act as a buffer. (a) \(\mathrm{KCl} / \mathrm{HCl}\), (b) \(\mathrm{KHSO}_{4} / \mathrm{H}_{2} \mathrm{SO}_{4}\) (c) \(\mathrm{Na}_{2} \mathrm{HPO}_{4} /\) \(\mathrm{NaH}_{2} \mathrm{PO}_{4}\) (d) \(\mathrm{KNO}_{2} / \mathrm{HNO}_{2} ?\)
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