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Chapter 20: Problem 89

Calculate the solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}\left(K_{\text {sp }}=8.9 \times 10^{-12}\right.\) ) in an aqueous solution buffered at \(\mathrm{pH}=9.42\).

Short Answer

Expert verified
The solubility of Mg(OH)鈧� in the buffered solution with pH 9.42 is approximately \(9.92 \times 10^{-5}\) mol/L.

Step by step solution

01

Write the dissociation equation

Mg(OH)2 (s) <-> Mg2+ (aq) + 2OH- (aq) #Step 2: Convert pH to pOH and find the concentration of OH- ions#
02

Calculate pOH and concentration of OH- ions

Given, pH = 9.42 pOH = 14 - pH pOH = 14 - 9.42 = 4.58 Now, convert pOH to the concentration of hydroxide ions (OH-) using the formula: [OH-] = 10^(-pOH) [OH-] = 10^(-4.58) #Step 3: Set up the Ksp expression for Mg(OH)2 and substitute the known values#
03

Solve for the solubility using Ksp expression

Ksp(Mg(OH)2) = [Mg2+][OH-]^2 Given, Ksp = 8.9 脳 10^(-12) Let x be the solubility of Mg(OH)2 in the buffered solution. Then, after dissociation, [Mg2+] = x and [OH-] = 2x + 10^(-4.58), as there will be twice the number of OH- ions produced from the dissociation compared to the initial amount of OH- ions in the buffered solution. Substitute these values into the Ksp expression: 8.9 脳 10^(-12) = (x)(2x + 10^(-4.58))^2 #Step 4: Solve the equation for x (solubility)#
04

Solve for solubility, x

Since Ksp is a very low value, we can assume that the value of x is small compared to the initial concentration of OH- ions (10^(-4.58)) and approximate the equation as follows: 8.9 脳 10^(-12) = (x)(2x)^2 x = 鈭�(8.9 脳 10^(-12) / 4) x 鈮� 9.92 脳 10^(-5) mol/L Therefore, the solubility of Mg(OH)2 in the buffered solution with pH 9.42 is approximately 9.92 脳 10^(-5) mol/L.

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Key Concepts

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

Chemical Equilibrium
When reactions occur, they tend to move towards a state where the rate of the forward reaction equals the rate of the reverse reaction, a condition known as chemical equilibrium. At this point, the concentrations of reactants and products remain constant over time. This is evident in the dissolution of compounds like Mg(OH)2 where the solid compound is in constant exchange with its ions in solution. Understanding this dynamic balance is crucial for predicting how changing conditions, such as pH, can affect the solubility of substances.
Ksp (Solubility Product Constant)
The solubility product constant (Ksp) is an essential value in chemistry that gives us insight into the solubility of sparingly soluble compounds. The Ksp is determined at a particular temperature and measures the product of the molar concentrations of the ions in a saturated solution, each raised to the power of their stoichiometric coefficients. For instance, for Mg(OH)2, the Ksp expression is \[Ksp = [Mg^{2+}][OH^-]^2\] where the concentrations are those at equilibrium. Understanding and calculating Ksp values helps determine the extent to which a compound will dissociate in solution, which is particularly useful when dealing with ionic compounds in various chemical environments.
pH and pOH Conversions
The pH and pOH scales are logarithmic measures of the acidity and basicity of a solution, respectively. The pH scale ranges from 0 to 14, with 7 being neutral. pH and pOH are inversely related and always add up to 14. The relation is given by the formula: \[pOH = 14 - pH\] Knowing the pH of a solution, one can quickly determine its pOH, which is essential for understanding the concentration of hydroxide ions present ([OH^-]). The concentration of hydroxide ions can be calculated from the pOH using the formula \[\text{[OH^-]} = 10^{-\text{pOH}}\] This calculation is a fundamental aspect of solving for the solubility of compounds within solutions of a specific pH, as seen with Mg(OH)2 in a buffered solution.
Dissociation of Compounds
The process by which a compound separates into its ions in solution is known as dissociation. For ionic compounds such as Mg(OH)2, this involves the solid compound splitting into Mg2+ cations and OH- anions in water. The degree to which a compound dissociates is influenced by factors such as the Ksp and the pH of the solution. When we calculate the solubility under different pH conditions, we consider the dissociation equilibrium and how it shifts in response to changes in ion concentrations, guided by Le Ch芒telier's Principle.

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Most popular questions from this chapter

Captain Kirk has set a trap for the Klingons who are threatening an innocent planet. He has sent small groups of fighter rockets to sites that are invisible to Klingon radar and put a decoy in the open. He calls this the "fishhook" strategy. Mr. Spock has sent a coded message to the chemists on the fighters to tell the ships what to do next. The outline of the message is Fill in the blanks of the message using the following clues. (1) Symbol of the halogen whose hydride has the second highest boiling point in the series of HX compounds that are hydrogen halides. (2) Symbol of the halogen that is the only hydrogen halide, \(\mathrm{HX}\), that is a weak acid in aqueous solution. (3) Symbol of the element whose existence on the sun was known before its existence on earth was discovered. (4) The Group \(5 \mathrm{~A}\) element in Table \(20.13\) that should have the most metallic character. (5) Symbol of the Group 6 A element that, like selenium, is a semiconductor. (6) Symbol for the element known in rhombic and monoclinic forms. (7) Symbol for the element that exists as diatomic molecules in a yellow-green gas when not combined with another element. (8) Symbol for the most abundant element in and near the earth's crust. (9) Symbol for the element that seems to give some protection against cancer when a diet rich in this element is consumed. (10) Symbol for the smallest noble gas that forms compounds with fluorine having the general formula \(\mathrm{AF}_{2}\) and \(\mathrm{AF}_{4}\) (reverse the symbol and split the letters as shown). (11) Symbol for the toxic element that, like phosphorus and antimony, forms tetrameric molecules when uncombined with other elements (split the letters of the symbol as shown). (12) Symbol for the element that occurs as an inert component of air but is a very prominent part of fertilizers and explosives.

Thallium and indium form \(+1\) and \(+3\) oxidation states when in compounds. Predict the formulas of the possible compounds between thallium and oxygen and between indium and chlorine. Name the compounds.

Write equations describing the reactions of Sn with each of the following: \(\mathrm{Cl}_{2}, \mathrm{O}_{2}\), and \(\mathrm{HCl}\).

Use bond energies (see Table 8.4) to show that the preferred products for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{3}\) are \(\mathrm{NO}_{2}\) and \(\mathrm{NO}\) rather than \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}\). (The \(\mathrm{N}-\mathrm{O}\) single bond energy is 201 \(\mathrm{kJ} / \mathrm{mol}\).) (Hint: Consider the reaction kinetics.)

Many structures of phosphorus-containing compounds are drawn with some \(\mathrm{P}=\mathrm{O}\) bonds. These bonds are not the typical \(\pi\) bonds we've considered, which involve the overlap of two \(p\) orbitals. Instead, they result from the overlap of a \(d\) orbital on the phosphorus atom with a \(p\) orbital on oxygen. This type of \(\pi\) bonding is sometimes used as an explanation for why \(\mathrm{H}_{3} \mathrm{PO}_{3}\) has the first structure below rather than the second: Draw a picture showing how a \(d\) orbital and a \(p\) orbital overlap to form a \(\pi\) bond.
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