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The correct increasing order of solubility of the following substances in \(\mathrm{g} / 100 \mathrm{ml}\) is \(\mathrm{PbSO}_{4}\left(K_{\mathrm{sp}}=2 \times 10^{-9}\right), \mathrm{ZnS}\left(K_{\mathrm{sp}}=1\right.\) \(\left.\times 10^{-22}\right)\), AgBr \(\left(K_{\text {sp }}=4 \times 10^{-13}\right), \mathrm{CuCO}_{3}\) \(\left(K_{\mathrm{sp}}=1 \times 10^{-8}\right) .\) (Atomic masses: \(\mathrm{Pb}=208\), \(\mathrm{Zn}=65, \mathrm{Ag}=108, \mathrm{Br}=80, \mathrm{Cu}=63)\) (a) \(\mathrm{PbSO}_{4}<\mathrm{ZnS}<\mathrm{AgBr}<\mathrm{CuCO}_{3}\) (b) \(\mathrm{PbSO}_{4}<\mathrm{CuCO}_{3}<\mathrm{AgBr}<\mathrm{ZnS}\) (c) \(\mathrm{ZnS}<\mathrm{AgBr}<\mathrm{CuCO}_{3}<\mathrm{PbSO}_{4}\) (d) \(\mathrm{ZnS}<\mathrm{AgBr}<\mathrm{PbSO}_{4}<\mathrm{CuCO}_{3}\)

Short Answer

Expert verified
(c) ZnS < AgBr < CuCO3 < PbSO4

Step by step solution

01

Understand the Solubility Product

The solubility product, or Ksp, is a measure of the solubility of a compound. A lower Ksp value indicates a compound is less soluble in water, while a higher Ksp value indicates greater solubility. Since the Ksp values are given, we can compare them directly to arrange the solubilities of the given substances.
02

Arrange Substances by Ksp Values

Start by listing the compounds and their Ksp values: PbSO4 (Ksp = 2 x 10^{-9}), ZnS (Ksp = 1 x 10^{-22}), AgBr (Ksp = 4 x 10^{-13}), CuCO3 (Ksp = 1 x 10^{-8}). Next, arrange these compounds in order of increasing Ksp values which correlates to increasing solubility.
03

Determine the Correct Order

Arrange the substances in ascending order based on their Ksp values: ZnS (Ksp = 1 x 10^{-22}) < AgBr (Ksp = 4 x 10^{-13}) < PbSO4 (Ksp = 2 x 10^{-9}) < CuCO3 (Ksp = 1 x 10^{-8}). This order reflects their relative solubilities in water from least to most soluble.

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Key Concepts

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

Ksp
When diving into the world of inorganic chemistry, one crucial concept is the Solubility Product Constant, commonly known as Ksp. Ksp helps us understand just how much of a particular solid can dissolve in water to form a saturated solution. It's an equilibrium constant specific to the dissolution of sparingly soluble salts. The rule of thumb here is straightforward: the higher the Ksp value, the more soluble the compound is in water. This constant is a true lifesaver when predicting whether a precipitate will form in a reaction.

For example, if you're given several compounds and their Ksp values, to determine their solubility, you simply need to compare these Ksp values. Remember, Ksp values are product concentrations at equilibrium, so they provide a direct comparison – larger Ksp, greater solubility. This is exactly what we did with the compounds PbSO4, ZnS, AgBr, and CuCO3, ranking them from the least soluble based on the lowest Ksp value to the most soluble with the highest Ksp value.
Solution Equilibrium
You might be wondering what happens at a microscopic level when a salt starts dissolving in water. This is where solution equilibrium comes into play. It's a state where the rate at which the solid dissolves equals the rate at which the dissolved ions recombine to form the solid. At this point, the concentrations of dissolved ions are constant over time, which defines a saturated solution.

Understanding this will illuminate why Ksp is so significant. It's calculated when the solution is saturated, at equilibrium. Maintaining this balance ensures that no more of the solid can dissolve unless some external condition changes, such as temperature or pressure. In the exercise with the four compounds, the various Ksp values implied the different points at which each compound reached its own solution equilibrium in water.
Inorganic Chemistry
The branch of chemistry concerning the properties and behavior of inorganic compounds is fittingly called inorganic chemistry. This area is incredibly broad, covering compounds without carbon-hydrogen bonds, exceptions being carbonates, bicarbonates, and a few others. Inorganic compounds, such as those in our exercise – PbSO4, ZnS, AgBr, and CuCO3 – have myriad applications, from industrial catalysts to medicines.

Studying their solubility is a key aspect of inorganic chemistry. It allows chemists to predict outcomes in reactions, synthesize new materials, and even understand environmental impacts. The notion of predicting the solubility of a substance by comparing Ksp values is a fundamental part of this field and one that demonstrates the intrinsic link between theoretical principles and practical applications.

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Most popular questions from this chapter

An aqueous solution is prepared by dissolving \(0.1\) mole \(\mathrm{H}_{2} \mathrm{CO}_{3}\) in sufficient water to get \(100 \mathrm{ml}\) solution at \(25^{\circ} \mathrm{C}\). For \(\mathrm{H}_{2} \mathrm{CO}_{3}, \quad K_{\mathrm{a} 1}=4.0 \times 10^{-6}\) and \(K_{\mathrm{a} 2}=5.0 \times 10^{-11} .\) The only incorrect equilibrium concentration is (a) \(\left[\mathrm{H}^{+}\right]=6.32 \times 10^{-4} \mathrm{M}\) (b) \(\left[\mathrm{HCO}_{3}\right]=2 \times 10^{-3} \mathrm{M}\) (c) \(\left[\mathrm{CO}_{3}^{2-}\right]=5 \times 10^{-11} \mathrm{M}\) (d) \(\left[\mathrm{OH}^{-}\right]=5 \times 10^{-12} \mathrm{M}\)

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

In an attempted determination of the solubility product constant of \(\mathrm{Tl}_{2} \mathrm{~S}\), the solubility of this compound in pure \(\mathrm{CO}_{2}\) free water was determined as \(2.0 \times 10^{-6} \mathrm{M}\). Assume that the dissolved sulphide hydrolyses almost completely to \(\mathrm{HS}^{-}\) and that the further hydrolysis to \(\mathrm{H}_{2} \mathrm{~S}\) can be neglected, what is the computed \(K_{\mathrm{sp}}\) ? For \(\mathrm{H}_{2} \mathrm{~S}, K_{\mathrm{al}}=1.4 \times 10^{-7}, K_{\mathrm{a} 2}=1.0 \times 10^{-14}\) (a) \(6.4 \times 10^{-23}\) (b) \(1.6 \times 10^{-23}\) (c) \(3.2 \times 10^{-17}\) (d) \(3.2 \times 10^{-24}\)

Following five solutions of \(\mathrm{KOH}\) were prepared as: first, \(0.1\) mole in \(1 \mathrm{~L} ;\) second. \(0.2\) mole in \(2 \mathrm{~L}\); third, \(0.3\) mole in \(3 \mathrm{~L}\); fourth, \(0.4\) mole in \(4 \mathrm{~L} ;\) fifth, \(0.5\) mole in \(5 \mathrm{~L}\). The \(\mathrm{pH}\) of resultant solution, when all these solutions are mixed, is (a) 2 (b) 1 (c) 13 (d) 7

Small amount of freshly precipitated \(\begin{array}{llll}\text { magnesium } & \text { hydroxides } & \text { are } & \text { stirred }\end{array}\) vigorously in a buffer solution containing \(0.25 \mathrm{M}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(0.05 \mathrm{M}\) of \(\mathrm{NH}_{4} \mathrm{OH}\). \(\left[\mathrm{Mg}^{2}\right]\) in the resulting solution is \(\left(K_{\mathrm{b}}\right.\) for \(\mathrm{NH}_{4} \mathrm{OH}=2.0 \times 10^{-5}\) and \(K_{\mathrm{sp}}\) of \(\mathrm{Mg}(\mathrm{OH})_{2}\) \(\left.=8.0 \times 10^{-12}\right)\) (a) \(4 \times 10^{-6} \mathrm{M}\) (b) \(2 \times 10^{-6} \mathrm{M}\) (c) \(0.5 \mathrm{M}\) (d) \(2.0 \mathrm{M}\)

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