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Chapter 5: Problem 138

If \(\mathrm{S}\) and \(\mathrm{Ksp}\) be the solubility and solubility product respectively, then a. for \(\mathrm{AgCl}: \mathrm{S}=\sqrt{\mathrm{K}} \mathrm{sp}\) and for \(\mathrm{Al}(\mathrm{OH})_{3}: \mathrm{S}=\) b. for \(\mathrm{CaSO}_{4}: \mathrm{S}=\sqrt{\mathrm{K} s p}\) and for \(\mathrm{KI}_{3}: \mathrm{S}=\sqrt{\mathrm{K} \operatorname{sp}}\) c. for \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}: \mathrm{S}={ }^{5} \sqrt{(\mathrm{Ksp} / 108)}\) and for \(\mathrm{Agl}\) : \(\mathrm{S}=\sqrt{\mathrm{Ksp}}\) d. for \(\mathrm{Bi}_{2} \mathrm{~S}_{3}\) and for \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) both : \(\mathrm{S}={ }^{5} \sqrt{(\mathrm{Ksp} /}\) 108)

Short Answer

Expert verified
a. S = \( \sqrt[4]{\frac{\text{Ksp}}{27}} \) for Al(OH)3; c. S = \( \sqrt[5]{\frac{\text{Ksp}}{108}} \) for Ca3(PO4)2, repeated for both Bi2S3 and Ca3(PO4)2.

Step by step solution

01

Understanding Solubility and Solubility Product

The solubility product (Ksp) is an equilibrium constant for the dissolution of a sparingly soluble compound. The solubility (S) is the concentration of a solute that can dissolve in a solvent to form a saturated solution. For some ionic compounds, there is a direct formula relating S and Ksp.
02

Solubility of AgCl

For AgCl, the dissolution is represented by: \[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \] The expression for Ksp is: \[ \text{Ksp} = [\text{Ag}^+][\text{Cl}^-] = S^2 \] Thus, the solubility \( S = \sqrt{\text{Ksp}} \) for AgCl.
03

Solubility of Al(OH)3

For Al(OH)3, the dissolution is represented by: \[ \text{Al(OH)}_3 \rightleftharpoons \text{Al}^{3+} + 3\text{OH}^- \] The expression for Ksp is: \[ \text{Ksp} = [\text{Al}^{3+}][\text{OH}^-]^3 = S \cdot (3S)^3 = 27S^4 \] Therefore, the solubility equation is: \[ S = \sqrt[4]{\frac{\text{Ksp}}{27}} \]
04

Solubility of CaSO4

For CaSO4, the dissolution is: \[ \text{CaSO}_4 \rightleftharpoons \text{Ca}^{2+} + \text{SO}_4^{2-} \] The expression for Ksp is: \[ \text{Ksp} = [\text{Ca}^{2+}][\text{SO}_4^{2-}] = S^2 \] Thus, \( S = \sqrt{\text{Ksp}} \) for CaSO4.
05

Solubility of KI3

For KI3, the dissolution is: \[ \text{KI}_3 \rightleftharpoons \text{K}^+ + \text{I}_3^- \] The expression for Ksp is: \[ \text{Ksp} = [\text{K}^+][\text{I}_3^-] = S^2 \] Thus, \( S = \sqrt{\text{Ksp}} \) for KI3.
06

Solubility of Ca3(PO4)2

For \( \text{Ca}_3(\text{PO}_4)_2 \), the dissolution is: \[ \text{Ca}_3(\text{PO}_4)_2 \rightleftharpoons 3\text{Ca}^{2+} + 2\text{PO}_4^{3-} \] The expression for Ksp is: \[ \text{Ksp} = [\text{Ca}^{2+}]^3[\text{PO}_4^{3-}]^2 = (3S)^3(2S)^2 = 108S^5 \] Therefore, \( S = \sqrt[5]{\frac{\text{Ksp}}{108}} \).
07

Solubility of AgI

For AgI, the dissolution is: \[ \text{AgI} \rightleftharpoons \text{Ag}^+ + \text{I}^- \] The expression for Ksp is: \[ \text{Ksp} = [\text{Ag}^+][\text{I}^-] = S^2 \] Thus, \( S = \sqrt{\text{Ksp}} \).
08

Solubility of Bi2S3 and Review for Ca3(PO4)2

For Bi2S3, the dissolution is: \[ \text{Bi}_2\text{S}_3 \rightleftharpoons 2\text{Bi}^{3+} + 3\text{S}^{2-} \] The expression for Ksp is: \[ \text{Ksp} = [\text{Bi}^{3+}]^2[\text{S}^{2-}]^3 = (2S)^2(3S)^3 = 108S^5 \] Thus, \( S = \sqrt[5]{\frac{\text{Ksp}}{108}} \) is again derived for Bi2S3. This reconfirms the formula previously found for Ca3(PO4)2.

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Key Concepts

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

Ionic Equilibrium
Understanding ionic equilibrium is essential in chemistry as it describes the balance in a solution where reversible reactions occur with electrolytes. When an ionic compound dissolves in water, it separates into ions. The balance between these ions, the undissolved solid, and the water is what we call ionic equilibrium.
In a saturated solution, a dynamic state persists where the rate of dissolution equals the rate of precipitation. At this point, the concentration of ions in the solution remains constant, illustrating ionic equilibrium. Recognizing this equilibrium is key to solving solubility-related problems and understanding how different compounds behave in solution.
  • This concept helps explain why certain compounds, like AgCl or Al(OH)鈧�, exhibit predictable solubility patterns.
  • Understanding the equilibrium allows chemists to calculate and predict the behavior of ions in various situations, such as in reactions and precipitation.
Recognizing the nuances of ionic equilibrium can greatly aid in understanding solubility product calculations and formulating Ksp expressions.
Solubility Calculations
Solubility calculations are essential for determining how much of a solute can dissolve in water to form a saturated solution. These calculations are linked to the solubility product (Ksp), which is specific for each compound. Knowing how to calculate solubility lets us understand the maximum concentration of ions that can be in a solution before a precipitate forms.
For simple ionic compounds like AgCl, the relationship between solubility (S) and Ksp is straightforward: For AgCl, where one mole of the salt gives one mole each of \[ \text{Ag}^+ \text{ and } \text{Cl}^- \]the equation is \[ S = \sqrt{\text{Ksp}} \]
For more complex dissociations, solubility may require different calculations. Take Al(OH)鈧�, where the dissolution into one Al鲁鈦� and three OH鈦� ions means the equation becomes more involved:
\[ \text{Ksp} = 27S^4 \] leading to \[ S = \sqrt[4]{\frac{\text{Ksp}}{27}} \]
  • This variety in calculations emphasizes the importance of understanding the chemical formula and the dissociation context of each ionic compound.
Grasping solubility calculations allows one to predict ion concentrations and determine conditions for precipitate formation.
Ksp Expressions
The concept of the solubility product constant, Ksp, is fundamental to understanding the solubility of sparingly soluble salts. The Ksp value provides insights into how much of the compound can dissolve under given conditions.
Writing a Ksp expression involves identifying the ions produced when a compound dissolves in water and expressing the product of their concentrations raised to the power of their coefficients from the balanced equation.
For example, when we dissolve AgCl:
\[ \text{Ksp} = [\text{Ag}^+][\text{Cl}^-] \] each concentration is equal to S, leading to \[ \text{Ksp} = S^2 \] For Ca鈧�(PO鈧�)鈧�, which dissolves into three \text{Ca}^{2+} and two \text{PO鈧剗^{3-} ions, the expression becomes a bit more complex:
\[ \text{Ksp} = (3S)^3 (2S)^2 = 108S^5 \]
  • These expressions are crucial for calculating solubility and predicting when a precipitate will form.
  • Understanding how to derive these expressions allows us to better predict and manipulate chemical reactions involving salts.
In-depth knowledge of Ksp expressions is indispensable for anyone studying chemistry as it underpins many practical applications in fields like pharmacology, environmental science, and material sciences.

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Most popular questions from this chapter

Which of the following statement is/are correct for a reversible reaction? a. At a given temperature both \(Q\) and \(K\) vary with the progress of the reaction. b. When \(Q>K\), the reaction proceeds in backward direction before coming to stand still. c. Reaction quotient (Q) is the ratio of the product or arbitrary molar concentrations of the products to those of the reactants. d. \(Q\) may be \(<>=K\).

For the reaction $$ \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \leftrightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) $$ at a given temperature, the equilibrium amount of \(\mathrm{CO}_{2}(\mathrm{~g})\) can be increased by a. adding a suitable catalyst b. adding an inert gas c. decreasing the volume of the container d. increasing the amount of \(\mathrm{CO}(\mathrm{g})\)

Which of the following is/are correct? a. The equilibrium constant does not depend upon pressure b. When pressure is applied on ice \(\rightleftharpoons\) water equilibrium more water will be formed c. The equilibrium constant increases when a catalyst is introduced d. Changes with temperature

Match the following: Column I Column II A. \(0.5 \mathrm{M} \mathrm{NO}_{3}+0.1 \mathrm{M} \quad\) (p) 7 \(\mathrm{NH}_{4} \mathrm{OH}\) B. \(0.1 \mathrm{M} \mathrm{KCl}+0.1 \mathrm{M} \quad\) (q) greater then 7 \(\mathrm{KNO}_{3}\) C. \(0.2 \mathrm{M} \mathrm{NaOH}+0.5 \mathrm{M}(\mathrm{r})\) between 1 to 7 \(\mathrm{HCOOH}\) D. \(0.1 \mathrm{NH}_{4} \mathrm{Cl}+0.1 \mathrm{M}(\mathrm{s}) 0.7\) \(\mathrm{KOH}\)

For the decomposition of \(\mathrm{PCl}_{5}(\mathrm{~g})\) in a closed vessel, the degree of dissociation is \(\alpha\) at total pressure (P). \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) ; \mathrm{K}_{\mathrm{p}}\) Which among the following relations is correct? a. \(\left.\alpha=\sqrt{[}\left(\mathrm{K}_{\mathrm{p}}+\mathrm{P}\right) / \mathrm{K}_{\mathrm{p}}\right]\) b. \(\left.\alpha=\sqrt{[} \mathrm{K}_{\mathrm{p}} /\left(\mathrm{K}_{\mathrm{p}}+\mathrm{P}\right)\right]\) c. \(\alpha=1 / \sqrt{\left(K_{p}+P\right)}\) d. \(\alpha=\sqrt{\left(K_{p}+P\right)}\)
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