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Among different types of salts have nearly same solubility product constant, \(K_{\mathrm{sn}}\) but much smaller than one, the most soluble salt is that which (a) produces maximum number of ions (b) produces minimum number of ions (c) produces high charge on ions (d) produces low charges on ions

Short Answer

Expert verified
The most soluble salt among those with nearly the same solubility product constant and much smaller than one, is the one that (a) produces maximum number of ions.

Step by step solution

01

Identify the Key Concept

Understand that solubility product, denoted as Ksp, is a constant for a given solid at a specific temperature that relates the extent to which it can dissolve in water to form ions. A smaller Ksp suggests that the salt is less soluble. For salts with similar Ksp values, the one producing more ions will have a higher actual solubility because each dissolution event leads to more solute particles in solution.
02

Consider the Formation of Ions

Contemplate that when a salt dissolves, it dissociates into its constituent ions. The actual number of ions produced from the dissolution of each formula unit should be considered. A higher number of ions increases the molarity of the solution for a given mass of salt, indicating greater solubility.
03

Apply the Concept to the Options

Evaluate the provided options and deduce that the most soluble salt will be the one that dissociates to produce the maximum number of ions, leading to a higher concentration of ions in the solution from the same amount of solid.
04

Find the Correct Answer

Conclude that the most soluble salt, assuming similar Ksp values and the salts having a Ksp much smaller than one, is the salt that produces the maximum number of ions upon dissolution.

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Key Concepts

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

Dissolution of Salts
When we talk about the dissolution of salts, we're referring to the process by which a salt breaks apart into its constituent ions in water. The solubility product constant, or Ksp, comes into play here as a crucial factor. It quantifies the degree to which a salt can dissolve. But let's dive in further. During dissolution, a solid salt will disintegrate into cations and anions, its two types of ions, each surrounded by water molecules. These molecules stabilize the ions through a process called hydration, which is vital for the dissolution of salts. The Ksp value gives us insight into this balance between the solid and the ions in solution. Generally, higher solubility corresponds to more ions in the solution, which is why salts that release more ions upon dissolution are generally more soluble. Thus, option (a), which produces the maximum number of ions, identifies the most soluble salt among those with similar Ksp values.

When considering the dissolution process, it's essential to think about the initial solid and its interactions with the water molecules. This interaction drives the separation of the salt into ions, and the balance between the forces holding the salt together and the interactions with water determines how readily a salt will dissolve. To enhance student learning, we can illustrate this with a simple experiment. Dissolve different types of salt in water and observe how the size and charge of ions affect how readily they dissolve.
Ion Formation
Now let's zero in on ion formation. Essentially, we're looking at how salt, when placed in water, separates into positively and negatively charged particles. But it's not just about splitting; it's the specifics that matter—like how many and what kind of ions emerge. The nature and number of ions that form directly influence the solubility of the salt. For instance, when common table salt (NaCl) dissolves, it forms two ions—one Na+ and one Cl-. In contrast, a salt like CaF2 forms three ions— one Ca2+ and two F- ions. More ions typically lead to a higher ion concentration, boosting solubility.

Now, when salts have comparable Ksp values, the ones that split into a greater number of ions when dissolving do, indeed, tend to be the most soluble, which is the essence of Step 2 in our solution steps. To drive this point home for students, we can visualize the process of dissolution as a series of steps in which each salt molecule is surrounded and gradually pulled apart by water molecules, resulting in a mixture that's uniform at the molecular level—better known as a solution.
Solubility of Salts in Water
Last but not least, we explore the solubility of salts in water. This is essentially the measure of how much of a particular salt can dissolve in water before reaching saturation—where no more salt can dissolve. Let's get technical: Solubility can be expressed in grams per 100 milliliters of water, but in the world of chemistry, we often use molarity—the number of moles of solute (the salt, in this case) per liter of solution. The interesting thing is, for salts with similar low Ksp values, solubility isn't solely about how much can fit into the water. It's also about how many individual ions one formula unit of the salt generates. More ions mean a greater total number of solute particles, which equals higher solubility. More ions also mean a higher osmotic pressure, which can affect the colligative properties of the solution.

As we inferred in Step 3 of our solution, the most soluble salt will consequently be the one that produces the most ions for a given mass. This concept elevates our understanding of the direct relationship between the number of ions produced upon the dissolution of a salt and its solubility in water. When teaching students, it is valuable to have them practice by calculating the molarity of ions in solution for various salts to see this relationship in action. Ultimately, option (a) from our original exercise is correct because a higher number of ions typically indicates a more soluble salt.

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Most popular questions from this chapter

What is the pH of a \(0.50 \mathrm{M}\) aqueous \(\mathrm{NaCN}\) solution? \(\mathrm{p} K_{\mathrm{b}}\) of \(\mathrm{CN}^{-}\) is \(4.70 .\) \((\log 2=0.3)\) (a) \(3.0\) (b) \(11.0\) (c) \(4.7\) (d) \(9.3\)

What is the solubility of \(\mathrm{MnS}\) in pure water, assuming hydrolysis of \(\mathrm{S}^{2-}\) ions? \(K_{\mathrm{sp}}\) of \(\mathrm{MnS}=2.5 \times 10^{-10}, K_{\mathrm{al}}=1 \times 10^{-7}\) and \(K_{\mathrm{a} 2}=1 \times 10^{-14}\) for \(\mathrm{H}_{2} \mathrm{~S} .\left(0.63^{3}=0.25\right)\) (a) \(6.3 \times 10^{-4} \mathrm{M}\) (b) \(2.5 \times 10^{-4} \mathrm{M}\) (c) \(6.3 \times 10^{-3} \mathrm{M}\) (d) \(1.58 \times 10^{-5} \mathrm{M}\)

The hydronium ion concentration in an aqueous solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is \(2.0 \times 10^{-4} \mathrm{M}\) at \(25^{\circ} \mathrm{C}\). The hydroxide ion concentration in the solution is (a) 0 (b) \(2.0 \times 10^{-4} \mathrm{M}\) (c) \(5 \times 10^{3} \mathrm{M}\) (d) \(5 \times 10^{-11} \mathrm{M}\)

What is the aqueous ammonia concentration of a solution prepared by dissolving \(0.15\) mole of \(\mathrm{NH}_{4}^{+} \mathrm{CH}_{3} \mathrm{COO}^{-}\) in 1 L of water? Given: \(K_{\text {a }}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) \(=1.8 \times 10^{-5} ; K_{\mathrm{b}}\left(\mathrm{NH}_{4} \mathrm{OH}\right)=1.8 \times 10^{-5}\) (a) \(8.3 \times 10^{-4} \mathrm{M}\) (b) \(0.15 \mathrm{M}\) (c) \(5.52 \times 10^{-3} \mathrm{M}\) (d) \(3.8 \times 10^{-4} \mathrm{M}\)

How many times solubility of \(\mathrm{CaF}_{2}\) is decreased in \(4 \times 10^{-3} \mathrm{M}-\mathrm{KF}(\mathrm{aq})\) solution as compared to pure water at \(25^{\circ} \mathrm{C}\). Given: \(K_{\text {sp }}\left(\mathrm{CaF}_{2}\right)=3.2 \times 10^{-11}\) (a) 50 (b) 100 (c) 500 (d) 1000

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