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Chapter 18: Problem 9

A solution is saturated with magnesium palmitate \(\left[\mathrm{Mg}\left(\mathrm{C}_{16} \mathrm{H}_{31} \mathrm{O}_{2}\right)_{2}, \text { a component of bathtub ring }\right] \mathrm{at}\) \(50^{\circ} \mathrm{C} .\) How many milligrams of magnesium palmitate will precipitate from \(965 \mathrm{mL}\) of this solution when it is cooled to \(25^{\circ} \mathrm{C} ?\) For \(\mathrm{Mg}\left(\mathrm{C}_{16} \mathrm{H}_{31} \mathrm{O}_{2}\right)_{2},\) \(K_{\mathrm{sp}}=4.8 \times 10^{-12}\) at \(50^{\circ} \mathrm{C}\) and \(3.3 \times 10^{-12}\) at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
To find the amount of magnesium palmitate that will precipitate, subtract the molar solubility at \(25^{\circ}C\) from the molar solubility at \(50^{\circ}C\) and multiply by the volume of the solution in liters. Then multiply by the molar mass of magnesium palmitate and convert to milligrams. Please plug in the values into the formula mentioned in Step 3 for obtaining the exact figure.

Step by step solution

01

Find the Solubility at given Temperatures

Using the formula for solubility product \(K_{sp} = [Mg^{2+}][A^{2-}]^2\), where A is the palmitate ion, we can find the molar solubility of the salt at the given temperatures by taking the square root of each \(K_{sp}\). Let's denote the molar solubility at \(50^{\circ}C\) as \(x\) and the molar solubility at \(25^{\circ}C\) as \(y\). So, \(x = \sqrt{4.8 \times 10^{-12}}\) and \(y = \sqrt{3.3 \times 10^{-12}}\).
02

Calculate the Difference in Solubility

The difference in molar solubility tells us how much of the solute will precipitate out when the solution is cooled from \(50^{\circ}C\) to \(25^{\circ}C\). This difference, \(d\), is found by subtraction: \(d = x - y\).
03

Determine the Mass in milligrams

A solution's concentration is its molarity, which is the number of moles per liter of solution. Because we have 965 mL of solution, we need to convert this to liters to find the mass of the precipitate. The number of moles of precipitate is \(d \times 0.965\) L. Then, using the molecular weight of magnesium palmitate (i.e., 591.27 g/mol), we calculate the mass of the precipitate in grams and then convert it to milligrams: \(Precipitate (mg) = d \times 0.965 \times 591.27 \times 10^3\).

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Most popular questions from this chapter

Both \(\mathrm{Mg}^{2+}\) and \(\mathrm{Cu}^{2+}\) are present in the same aqueous solution. Which of the following reagents would work best in separating these ions, precipitating one and leaving the other in solution: \(\mathrm{NaOH}(\mathrm{aq}), \mathrm{HCl}(\mathrm{aq})\), \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq}),\) or \(\mathrm{NH}_{3}(\mathrm{aq}) ?\) Explain your choice.

Reaction (18.10), described in the Integrative Example, is called a carbonate transposition. In such a reaction, anions of a slightly soluble compound (for example, hydroxides and sulfates) are obtained in a sufficient concentration in aqueous solution that they can be identified by qualitative analysis tests. Suppose that \(3 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) is used and that an anion concentration of \(0.050 \mathrm{M}\) is sufficient for its detection. Predict whether carbonate transposition will be effective for detecting (a) \(\mathrm{SO}_{4}^{2-}\) from \(\mathrm{BaSO}_{4}(\mathrm{s}) ;\) (b) \(\mathrm{Cl}^{-}\) from \(\mathrm{AgCl}(\mathrm{s}) ;(\mathrm{c}) \mathrm{F}^{-}\) from \(\mathrm{MgF}_{2}(\mathrm{s})\).

Show that in qualitative cation analysis group \(1,\) if you obtain \(1.00 \mathrm{mL}\) of saturated \(\mathrm{PbCl}_{2}(\mathrm{aq})\) at \(25^{\circ} \mathrm{C}\), sufficient \(\mathrm{Pb}^{2+}\) should be present to produce a precipitate of \(\mathrm{PbCrO}_{4}(\mathrm{s}) .\) Assume that you use \(1 \mathrm{drop}\) \((0.05 \mathrm{mL})\) of \(1.0 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) for the test.

What is the solubility of \(\mathrm{MnS}\), in grams per liter, in a buffer solution that is \(0.100 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}-0.500 \mathrm{M}\) \(\mathrm{NaCH}_{3} \mathrm{COO} ?\) For \(\mathrm{MnS}, K_{\mathrm{spa}}=3 \times 10^{7}\).

The solubility of \(\mathrm{CdCO}_{3}(\mathrm{s})\) in \(1.00 \mathrm{M} \mathrm{KI}(\mathrm{aq})\) is \(1.2 \times 10^{-3} \mathrm{mol} / \mathrm{L} .\) Given that \(K_{\mathrm{sp}}\) of \(\mathrm{CdCO}_{3}\) is \(5.2 \times 10^{-12},\) what is \(K_{\mathrm{f}}\) for \(\left[\mathrm{CdI}_{4}\right]^{2-} ?\)
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