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Chapter 7: Problem 123

A \((\mathrm{OH})_{2}\) is a partially soluble substance. Its \(\mathrm{Ksp}\) value is \(4 \times 10^{-12}\), which of the following statement is correct? (1) The solubility is unaffected by \(\mathrm{pH}\) of the medium (2) Its solubility has been decreased in a buffered medium at \(\mathrm{pH}\) at \(1 \mathrm{~L}\) (3) Its solubility has been increased in a buffered medium having \(\mathrm{pH}\) at 9 (4) Its saturated solution has \(\mathrm{pH}\) is equal to \(10.3\) (a) \(1,2,3\) (b) 3 and 4 (c) 2 and 3 (d) 2,3 and 4

Short Answer

Expert verified
The correct option is (d) 2, 3, and 4.

Step by step solution

01

Understanding Solubility and Ksp

The solubility product (Ksp) is the equilibrium constant for a solid substance dissolving in an aqueous solution. For a compound A(OH)_2, which dissociates as \[ A(OH)_2(s) \rightleftharpoons A^{2+} (aq) + 2 OH^- (aq) \] the solubility product expression is given by \( K_{sp} = [A^{2+}][OH^-]^2 \).
02

Effect of pH on Solubility

The solubility of A(OH)_2 is affected by the pH of the solution. An increase in OH^- concentration (higher pH) causes the equilibrium to shift left, reducing solubility, while a decrease in OH^- concentration (lower pH) shifts equilibrium right, increasing solubility.
03

Evaluating Statements

Let's evaluate each statement: 1. Solubility is affected by pH because changes in OH^- concentration affect equilibrium.2. In a buffered solution with high H+ (low pH), solubility of the base decreases due to common ion effect.3. Increased pH (more OH^-) initially suggests decreased solubility, however, low base solubility can mean improved solubility due to less competition for OH^-.4. Calculating pH: At saturation, \( 2[OH^-] = s \), \( pH = 14 - pOH = 10.3 \) implies \( [OH^-] = 5 imes 10^{-11} \), hence plausible.
04

Conclusion

Based on the analysis, statements 1 is false as pH affects solubility. Statements 2 and 3 are true, indicating decreased solubility at lower pH and possible increased solubility at higher pH under specific conditions. Statement 4 is true as calculations align. Therefore, the correct options are 2, 3, and 4.

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Key Concepts

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

Effect of pH on Solubility
The **pH of a solution** plays a significant role in determining the solubility of substances, especially for compounds involving ions like hydroxide ( OH^- ). For the compound A(OH)_2 , which is a sparingly soluble hydroxide, the pH affects the equilibrium of its dissolution reaction.

When the pH of a solution is increased, meaning the solution becomes more basic, the concentration of hydroxide ions ( OH^- ) rises. This increase causes the dissolution equilibrium to shift to the left, according to Le Chatelier's Principle. As a result, the solubility decreases because the hydroxide ions already present in the solution discourage the compound from dissolving further.

Conversely, when the solution becomes more acidic (lower pH), the concentration of hydroxide ions decreases. In this scenario, the equilibrium shifts to the right, favoring the dissolution of A(OH)_2 to restore balance. Thus, the solubility of the compound increases as there is less competition from OH^- ions, making more room for the compound to dissociate. Therefore, understanding pH impact is crucial for predicting solubility changes in variable pH environments.
Common Ion Effect
The common ion effect is a concept in solubility that describes how the presence of a common ion in a solution can influence the solubility of a salt.

For example, if we consider the dissolution equilibrium of A(OH)_2 , which dissociates into A^{2+} and OH^- ions, introducing another source of OH^- ions like a different base or a buffer can significantly affect its solubility.

Impact of Buffers:
  • A buffered solution typically maintains a particular pH by providing a constant source of hydrogen ions ( H^+ ) or hydroxide ions ( OH^- ).
  • When a common ion is added, such as OH^- , the increased concentration of OH^- shifts the solubility equilibrium to the left, decreasing the solubility of the compound due to the common ion effect.
  • This occurs because the equilibrium favors the formation of the solid compound A(OH)_2 when the product ions are already present in the solution, inhibiting further dissolution.
By understanding the common ion effect, you can explain why adding substances that release similar ions into the solution can lead to reduced solubility of certain compounds.
Equilibrium in Solubility Reactions
In solubility reactions, an equilibrium is established between a solid and its dissolved ions once the solution becomes saturated. For the solubility of a substance like A(OH)_2 , this is an example of a dynamic equilibrium where both dissolution and precipitation occur at equal rates.

Importance of Equilibrium:
  • The solubility product constant ( K_{sp} ) quantifies this equilibrium, representing the maximum product of the ion concentrations at saturation.
  • If the ion concentrations exceed the K_{sp} , the solution is supersaturated, leading to precipitation until equilibrium is restored.
  • Conversely, if the ion concentrations are below K_{sp} , more of the solid can dissolve until equilibrium is reached.
Understanding equilibrium in solubility reactions helps predict how changes in conditions, like pH or ion concentrations, can shift the balance of this dynamic system, altering the solubility and stability of different compounds.

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Most popular questions from this chapter

The equilibrium between water and its vapour, in an open vessel: (a) Can be achieved (b) Depends upon pressure (c) Cannot be achieved (d) Depends upon temperature

What is the correct sequence of active masses in increasing order in gaseous mixture, containing one gram per litre of each of the following: 1\. \(\mathrm{NH}_{3}\) 2\. \(\mathrm{N}_{2}\) 3\. \(\mathrm{H}_{2}\) 4\. \(\mathrm{O}_{2}\) Select the correct answer using the codes given below: (a) \(3,1,4,2\) (b) \(3,4,2,1\) (c) \(2,1,4,3\) (d) \(4,2,1,3\)

Phosphorous pentachloride dissociates as follows, in a closed reaction vessel. \(\mathrm{PCI}_{5}(\mathrm{~g}) \longrightarrow \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\). If total pressure at equilibrium of the reaction mixture is \(\mathrm{P}\) and degree of dissociation of \(\mathrm{PC} 1_{5}\) is \(x\), the partial pressure of \(\mathrm{PCl}_{3}\) will be: (a) \(\left(\frac{x}{(x+1)}\right) \mathrm{P}\) (b) \(\left(\frac{2 x}{(x-1)}\right) \mathrm{P}\) (c) \(\left(\frac{x}{(x-1)}\right) \mathrm{P}\) (d) \(\left(\frac{x}{(1-x)}\right) \mathrm{P}\)

Consider an endothermic reaction \(\mathrm{X} \longrightarrow \mathrm{Y}\) with the activation energies \(E_{b}\) and \(E_{f}\) for the backward and forward reactions, respectively. In general: (a) \(\mathrm{E}_{\mathrm{b}}<\mathrm{E}_{\mathrm{f}}\) (b) \(\mathrm{E}_{\mathrm{b}}>\mathrm{E}_{\mathrm{f}}\) (c) \(\mathrm{E}_{\mathrm{b}}=\mathrm{E}_{\mathrm{f}}\) (d) There is no definite relation between \(\mathrm{E}_{\mathrm{b}}\) and \(\mathrm{E}_{\mathrm{f}}\)

\(1.25\) moles of NOCl were placed in a \(2.50 \mathrm{~L}\) reaction chamberat \(427^{\circ} \mathrm{C}\). After equilibrium was reached, 1.10 molesofNOClremained. Calculatetheequilibrium constant \(\mathrm{K}_{\mathrm{c}}\) for the reaction, \(2 \mathrm{NOC} 1(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}):\) (a) \(1.6 \times 10^{-3}\) (b) \(5.6 \times 10^{-4}\) (c) \(2.6 \times 10^{-3}\) (d) \(4.6 \times 10^{-4}\)
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