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Chapter 20: Problem 85

A saturated solution of \(\mathrm{Sr}(\mathrm{OH})_{2}(a q)\) at \(25^{\circ} \mathrm{C}\) has a measured \(\mathrm{pH}\) of \(13.50 .\) Estimate the solubility of \(\mathrm{Sr}(\mathrm{OH})_{2}(s)\) in water at \(25^{\circ} \mathrm{C}\) in grams per 100 milliliters of solution.

Short Answer

Expert verified
The solubility of Sr(OH)鈧� in water is approximately 1.92 g/100 mL.

Step by step solution

01

Find the concentration of OH鈦� ions

The given pH of the solution is 13.50. We first determine the pOH by using the relation:\[pOH = 14 - ext{pH} = 14 - 13.50 = 0.50\]Then, we find the concentration of hydroxide ions, \([ ext{OH}^-]\), using:\[[ ext{OH}^-] = 10^{- ext{pOH}} = 10^{-0.50} \approx 0.3162 ext{ M}\]
02

Relate Hydroxide Ion Concentration to Sr(OH)鈧� Solubility

In a saturated solution of \( ext{Sr(OH)}_2\), the dissociation is:\[ ext{Sr(OH)}_2 (s) \rightleftharpoons \text{Sr}^{2+} (aq) + 2 ext{OH}^- (aq)\]Let \(s\) be the molar solubility of \( ext{Sr(OH)}_2\), which means:\[[ ext{OH}^-] = 2s\]Therefore, solving for \(s\):\[2s = 0.3162 \implies s = \frac{0.3162}{2} = 0.1581 \text{ M}\]
03

Convert Molar Solubility to Mass Solubility

First, calculate the molar mass of \( ext{Sr(OH)}_2\):- Atomic mass of Sr = 87.62 g/mol- Atomic mass of O = 16.00 g/mol- Atomic mass of H = 1.01 g/molThe molar mass is:\[87.62 + (2 \times (16.00 + 1.01)) = 121.64 ext{ g/mol}\]Now, convert the molar solubility to grams per liter:\[0.1581 ext{ mol/L} \times 121.64 ext{ g/mol} = 19.2225 ext{ g/L}\]
04

Convert Solubility to Grams per 100 mL

To find the solubility in grams per 100 mL, divide the solubility in grams per liter by 10:\[\frac{19.2225 ext{ g/L}}{10} = 1.92225 ext{ g/100 mL}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Saturated Solution
In chemistry, a saturated solution is one that contains the maximum amount of solute that can dissolve in a particular solvent at a specific temperature. This means that any additional solute added will not dissolve because equilibrium has been reached. Saturated solutions are important in solubility calculations as they provide a way to gauge how much solute a solution can contain before it becomes unable to dissolve more.

When dealing with substances like \(\mathrm{Sr}(\mathrm{OH})_{2}\), the process involves solute particles dissolving and then recombining out of the saturated solution. This dynamic equilibrium makes understanding saturated solutions fundamental for solubility analysis.
  • Maximum capacity of solute in solvent
  • Equilibrium between dissolved ions and solid particles
  • Temperature dependence on solubility
Understanding the concept of saturation is crucial for solving problems related to the solubility of different compounds in water and other solvents.
pH and pOH Relationship
The pH and pOH relationship is a fundamental concept in chemistry that ties together the acidity and basicity of a solution. The pH scale measures the concentration of hydrogen ions \((\mathrm{H}^+)\) while pOH measures the concentration of hydroxide ions \((\mathrm{OH}^-)\). The two are interrelated by the equation:

\[\text{pH} + \text{pOH} = 14\]

This equation allows you to calculate one value if the other is known, providing crucial insight into the nature of the solution.
  • At pH 7, the solution is neutral
  • pH less than 7 indicates acidity
  • pH greater than 7 indicates basicity
In our original exercise, we used the given pH of 13.50 to find the pOH, which was crucial for determining the concentration of \(\mathrm{OH}^-\) ions. This step is essential for linking the pH and pOH scales to calculate the solubility of compounds.
Molar Solubility Calculation
Molar solubility is the number of moles of solute that can be dissolved in a liter of solution, particularly when the solution becomes saturated. Calculating molar solubility is vital for understanding reactions and equilibrium in solutions.
In the original problem, the dissociation of \(\mathrm{Sr(OH)}_2\) was used to determine molar solubility. This compound dissociates into one \(\mathrm{Sr}^{2+}\) ion and two \(\mathrm{OH}^-\) ions. From the concentration of hydroxide ions calculated from the pOH, we used the relation \([\mathrm{OH}^-] = 2s\) to solve for \(s\), the molar solubility.
  • Utilizes concentration of ions at equilibrium
  • Relates initial dissociation of compounds to solubility
  • Makes use of stoichiometry from balanced chemical equations
Understanding how to calculate molar solubility is essential for determining the concentrations of substances in saturated solutions, thereby predicting their behavior in diverse chemical scenarios.
Conversion of Units in Chemistry
Conversions are critical in chemistry to ensure all components of a calculation are expressed in coherent units, allowing the final results to be meaningful and standardized.
In the initial problem, we converted molar solubility from moles per liter (mol/L) to grams per liter (g/L), taking into account the molar mass of \(\mathrm{Sr(OH)}_2\). Following this, the solubility was further converted to grams per 100 milliliters (g/100 mL), the commonly used metric for expressing solubility in solutions.
  • Ensures consistency and accuracy in calculations
  • Facilitates comparison between different measurements
  • Useful in laboratory settings and real-world applications
Understanding how to perform these unit conversions accurately helps avoid errors and misinterpretations, significantly aiding in scientific communication and understanding.

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Most popular questions from this chapter

Various aluminum salts, such as \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(s)\) and \(\mathrm{KAl}\left(\mathrm{SO}_{4}\right)_{2}(s)\), are used as additives to increase the acidity of soils for "acid-loving" plants such as azaleas and tomatoes. Explain how these salts increase soil acidity.

Sodium propanoate, \(\mathrm{NaCH}_{3} \mathrm{CH}_{2} \mathrm{COO}(s)\), is used as a food preservative. Calculate the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of a \(0.20-\mathrm{M}\) solution of \(\mathrm{NaCH}_{3} \mathrm{CH}_{2} \mathrm{COO}(a q)\), taking \(K_{\mathrm{a}}=1.4 \times 10^{-5} \mathrm{M}\) for propanoic acid.

The autoprotonation constant for the solvent ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)\), is \(8 \times 10^{-20} \mathrm{M}^{2}\) at \(25^{\circ} \mathrm{C}\). The autoprotonation equilibrium equation is \(2 \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l) \leftrightharpoons\) $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}_{2}^{+}(a l c)+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}(a l c) $$ where (alc) denotes a solute in alcohol solution. (a) We define a \(\mathrm{pH}\) scale in alcohol by the equation $$ \mathrm{pH}=-\log \left(\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}_{2}^{*}\right] / \mathrm{M}\right) $$ Calculate the \(\mathrm{pH}\) of a neutral alcohol solution at \(25^{\circ} \mathrm{C} .\) (b) Calculate the \(\mathrm{pH}\) of a \(0.010\) -M solution of sodium ethoxide, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-} \mathrm{Na}^{+}(a l c)\), in alcohol at \(25^{\circ} \mathrm{C}\). Assume that \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-} \mathrm{Na}^{+}(a l c)\) is completely dis- sociated in alcohol.

A saturated solution of the strong base \(\mathrm{Mg}(\mathrm{OH})_{2}(a q)\) at \(25^{\circ} \mathrm{C}\) has a \(\mathrm{pH}\) of \(10.52 .\) Estimate the solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}(s)\) in water at \(25^{\circ} \mathrm{C}\) in grams per 100 milliliters.

Autoprotonation occurs in solvents other than water. For liquid ammonia, the autoprotonation equilibrium is described by the equation $$ 2 \mathrm{NH}_{3}(l) \leftrightharpoons \mathrm{NH}_{\mathrm{d}}^{+}(\mathrm{amm})+\mathrm{NH}_{2}^{-}(\mathrm{amm}) $$ with \(K=1 \times 10^{-30} \mathrm{M}^{2}\) at \(-50^{\circ} \mathrm{C}\), where \((a m m)\) denotes a solute in liquid ammonia. Estimate the concentration of the ammonium ion in liquid ammonia at \(-50^{\circ} \mathrm{C}\). How many molecules of ammonia are dissociated per mole of ammonia? Take the density of liquid ammonia as \(0.77 \mathrm{~g} \cdot \mathrm{mL}^{-1}\) at \(-50^{\circ} \mathrm{C}\).
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