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(a) True or false: "solubility" and "solubility-product con. stant" are the same number for a given compound. (b) Write the expression for the solubility- product constant for each of the following ionic compounds: MnCO3,Hg(OH)2, and Cu3(PO4)2.

Short Answer

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(a) False. Solubility refers to the maximum concentration of a solute that can be dissolved in a solvent, while the solubility-product constant (Ksp) is an equilibrium constant representing the relationship between dissolved and undissolved species. (b) Solubility-Product Constant Expressions: 1. MnCO3: Ksp=[Mn2+][CO32] 2. Hg(OH)2: Ksp=[Hg2+][OH]2 3. Cu3(PO4)2: Ksp=[Cu2+]3[PO43]2

Step by step solution

01

Answer to Part (a) - Solubility versus Solubility-Product Constant

Solubility refers to the maximum concentration of a solute that can be dissolved in a solvent to form a saturated solution at a given temperature. It is expressed in terms of moles per liter (M) or grams per liter (g/L). On the other hand, the solubility-product constant, often denoted by Ksp, is a numerical value that represents the equilibrium between the dissolved ions and the undissolved solid in a saturated solution. In short, solubility refers to the maximum concentration of a solute that can be dissolved, while the solubility-product constant is an equilibrium constant representing the relationship between dissolved and undissolved species. So, the statement "solubility" and "solubility-product constant" are the same number for a given compound is False.
02

Answer to Part (b) - Writing Solubility-Product Constant Expressions

For a given ionic compound AbCd, which dissociates into a cations A^+ and d anions B^-, the balanced dissociation equation and solubility-product constant expression would be: AbCd <-> aA^+ + bB^- Ksp = [A^+]^a[B^-]^b 1. For MnCO3: Dissociation Equation: MnCO3(s)Mn2+(aq)+CO32(aq) Solubility-Product Constant Expression: Ksp=[Mn2+][CO32] 2. For Hg(OH)2: Dissociation Equation: Hg(OH)2(s)Hg2+(aq)+2OH(aq) Solubility-Product Constant Expression: Ksp=[Hg2+][OH]2 3. For Cu3(PO4)2: Dissociation Equation: Cu3(PO4)2(s)3Cu2+(aq)+2PO43(aq) Solubility-Product Constant Expression: Ksp=[Cu2+]3[PO43]2

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Key Concepts

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

Understanding Solubility
When it comes to understanding how substances dissolve in liquids, the term solubility is central. Solubility is the maximum amount of a solute that can dissolve in a given solvent at a specific temperature and pressure, resulting in a saturated solution. It can be influenced by various factors, including temperature, pressure, and the nature of the solvent and solute.

Imagine solubility as the 'limit' to how much sugar you can dissolve in your tea; once you've reached that limit, additional sugar won't dissolve and will settle at the bottom. Solubility is typically measured in units like moles per liter (M) or grams per liter (g/L), showing how much solute the solvent can hold before becoming saturated.

This concept is key in fields such as chemistry, environmental science, and pharmaceuticals because it determines the extent to which substances mix and react with one another in solutions.
Equilibrium Constant (Ksp) Basics
In solubility context, the equilibrium constant, specifically known as the solubility-product constant (Ksp), plays a vital role. It quantifies the relationship between the dissolved ions and solid particles in a saturated solution at equilibrium. Unlike solubility, which is expressed as a concentration, the Ksp value is unitless, serving as a mathematical representation of the product of the ion concentrations, each raised to the power of their respective coefficients in a balanced chemical equation.

The Ksp value gives us insight into the solute's tendency to dissolve in a solvent. A larger Ksp indicates greater solubility, while a lower Ksp suggests lower solubility. This value is particularly useful for predicting whether a precipitate will form when two solutions are mixed. In essence, knowing the Ksp assists chemists in anticipating the behavior of ionic compounds in various conditions.
Ionic Compounds & Solubility
An ionic compound consists of positively charged ions (cations) and negatively charged ions (anions) held together by electrostatic forces known as ionic bonds. Common examples include table salt (NaCl) and Epsom salt (MgSO4). In water, many of these compounds dissociate into their ions, a process influenced by the compound's solubility.

Factors such as lattice energy— the energy required to separate the ions in the solid—and hydration energy—the energy release when ions interact with water—play crucial roles in the solubility of ionic compounds. The balance between these energies helps determine if an ionic compound will be highly soluble or sparingly soluble in water.

Understanding the solubility of ionic compounds is important in areas such as medicine, for drug formulation, and in environmental science, to predict the movement and impact of ions in water systems.
The Dissociation Equation Connection
A dissociation equation represents the process in which a solid ionic compound separates into its component ions in a solvent. The equation is written as a reversible reaction, indicating that the dissociation is an equilibrium process, where the rate of the solid dissolving into ions is equal to the rate of ions coming together to form the solid.

For example, when salt dissolves in water, the dissociation equation is written as:
NaCl(s) ↔ Na+(aq) + Cl-(aq)This equation illustrates that solid sodium chloride (NaCl) dissociates into sodium ions (Na+) and chloride ions (Cl-) in water (aq denotes aqueous or dissolved in water).

The dissociation equation lays the groundwork for calculating the solubility-product constant (Ksp), which requires balancing the equation and using the concentrations of the resulting ions. It is the cornerstone of understanding the solubility behavior of an ionic compound and is widely used in predicting the outcomes of chemical reactions that involve precipitation.

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Most popular questions from this chapter

Tooth enamel is composed of hydroxyapatite, whose simplest formula is Cas(PO4)3OH, and whose corresponding Ksp=6.8×1027. As discussed in the "Chemistry and Life" box on page 755, fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, Ca5(PO4)3 F whose Kp=1.0×1040, (a) Write the expression for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds.

An unknown solid is entirely soluble in water. On addition of dilute HCl, a precipitate forms. After the precipitate is filtered off, the pH is adjusted to about 1 and H2 S is bubbled in; a precipitate again forms. After filtering off this precipitate, the pH is adjusted to 8 and H2 S is again added; no precipitate forms. No precipitate forms upon addition of (NH4)2HPO4. (See Figure 7.23.) The remaining solution shows a yellow color in a flame test (see Figure 7.22). Based on these observations, which of the following compounds might be present, which are definitely present, and which are definitely absent: CdS, Pb(NO3)2,HgO3,ZnSO4,Cd(NO3)2, and Na2SO4 ?

Calculate the pH at the equivalence point in titrating 0.100M solutions of each of the following with 0.080MNaOH : (a) hydrobromic acid (HBr), (b) chlorous acid (HClO2), (c) benzoic acid ( C4H3COOH).

A 35.0-mL sample of 0.150M acetic acid (CH3COOH) is titrated with 0.150MNaOH solution. Calculate the pH after the following volumes of base have been added: (a) 0 mL, (b) 17.5 mL (c) 34.5 mL (d) 35.0 mL (e) 35.5 mL (f) 50.0 mL..

(a) If the molar solubility of CaF2 at 35C is 1.24×103 mol/L, what is Ksp  at this temperature? (b) It is found that 1.1×102 gSrF2 dissolves per 100 mL of aqueous solution at 25C. Calculate the solubility product for SrF2. (c) The Kpp  of Ba(IO3)2 at 25C is 6.0×1010. What is the molar solubility of Ba(IO3)2 ?

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