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Which of the following compounds, when added to water, will increase the solubility of CdS: (a) \(\mathrm{LiNO}_{3},\) (b) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) (c) \(\mathrm{KCN}\), (d) \(\mathrm{NaClO}_{3}\) ?

Short Answer

Expert verified
\(\mathrm{KCN}\) (option c) increases the solubility of CdS by forming a soluble complex.

Step by step solution

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01

Understanding the Problem

The question asks which compound, when added to water, will increase the solubility of cadmium sulfide (CdS). CdS is a sparingly soluble compound, and various additives can impact its solubility through common ion effects, complex formation, or changes in ionic strength.
02

Analyzing Each Option

We will analyze each compound to determine how it affects the solubility of CdS: (a) \(\mathrm{LiNO}_{3}\) - Contains Li\(^+\) and NO\(_3^-\), neither forming a strong complex with Cd\(^{2+}\) nor common ions with CdS.(b) \(\mathrm{Na}_2 \mathrm{SO}_4\) - Contains SO\(_4^{2-}\), which can decrease solubility due to common ion effect with sulfide.(c) \(\mathrm{KCN}\) - Contains CN\(^-\), which can form a soluble complex with Cd\(^{2+}\), increasing solubility.(d) \(\mathrm{NaClO}_3\) - Contains ClO\(_3^-\), with no expected interaction influencing solubility of CdS.
03

Selecting the Correct Compound

From the analysis, \(\mathrm{KCN}\) (option c) forms a complex ion with Cd\(^{2+}\) (\(\mathrm{Cd(CN)}_4^{2-}\)), which increases the solubility of CdS in water. Complexing agents like CN\(^-\) are known to enhance solubility by binding with metal ions and increasing their concentration in solution.

Key Concepts

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

cadmium sulfide
Cadmium sulfide (CdS) is an important compound in the field of chemistry and materials science. It is known as a sparingly soluble, yellow crystalline substance that is often used as a pigment and a semiconductor. In water, CdS has very limited solubility, which means it doesn’t easily dissolve, forming a solid precipitate. This characteristic poses a challenge in certain industries where removal or use of soluble cadmium is preferred. Understanding ways to increase the solubility of compounds like CdS is important for diverse applications such as waste treatment and manufacturing. Cadmium sulfide dissolves in water through a reaction forming cadmium ions \[ \text{CdS (s)} \rightleftharpoons \text{Cd}^{2+} (aq) + \text{S}^{2-} (aq) \]. This equilibrium describes the very low concentration of dissolved ions from CdS, contributing to its characterization as a sparingly soluble salt. Its solubility can be modified by changing certain environmental factors or by adding specific substances to the solution.
complex ion formation
Complex ion formation refers to the process where metal ions form a complex with other ions or molecules, often increasing their solubility in a solution. In the context of cadmium sulfide, adding a compound like KCN results in the formation of a complex ion with cadmium, \[ \text{Cd}^{2+} + 4 \text{CN}^- \rightleftharpoons \text{Cd(CN)}_4^{2-} \]. This complex is more soluble in water than cadmium ions alone, which explains why adding potassium cyanide (KCN) increases the solubility of CdS. Complex ion formation is an important reaction in chemistry since it can fundamentally alter the properties of a solution, making otherwise insoluble compounds dissolve.
  • It stabilizes metal ions in solutions.
  • It alters conductivity and reactivity.
  • It influences the separation and purification of metals in industrial processes.
By utilizing complex ions, chemists can manipulate reactions and processes for desired outcomes, enhancing solubility where necessary.
common ion effect
The common ion effect occurs when the solubility of a compound decreases due to the presence of a common ion in the solution. In the case of cadmium sulfide, an example is the addition of sodium sulfate (Na extsubscript{2}SO extsubscript{4}). Sodium sulfate dissociates into sodium ions and sulfate ions, the latter being the common ion here:\[ \text{Na}_2\text{SO}_4 \rightarrow 2\text{Na}^+ + \text{SO}_4^{2-} \].The sulfate ion is present in both Na extsubscript{2}SO extsubscript{4} and CdS because sulfide ions ( ext{S}^{2-}) and sulfate ions ( ext{SO}_4^{2-}) can be perceived as related in common ion contexts. This decreases the dissociation of CdS due to the increase in nearby anion concentration, thus reducing its solubility. In general, the common ion effect acts to suppress the dissociation of weak electrolytes. Understanding this effect helps in predicting and controlling the solubility of salts in solutions, also finding applications in buffers and precipitation processes.
ionic strength
Ionic strength is a measure of the total concentration of ions in a solution. It affects how different ions interact with each other and their solubility. Increasing the ionic strength of a solution can, in some cases, make sparingly soluble salts more soluble or cause precipitation. This is because it influences the activity coefficients of the ions in solution. For cadmium sulfide, adding neutral salts like LiNO extsubscript{3} doesn't result in chemical complexes or introduce common ions. However, it still affects ionic strength. With more ions present, interaction among ions intensifies, sometimes increasing solubility by effectively screening the attractive forces between oppositely charged ions.
  • High ionic strength impacts reaction rates.
  • It modifies solubility and stability of molecules.
  • Used in manipulating industrial chemical processes.
Thus, understanding ionic strength is essential for chemists in designing solutions where control over solubility and reaction pathways is crucial.
soluble complexes
Soluble complexes are chemical species in solution that result in the increased solubility of otherwise insoluble compounds. In the exercise involving cadmium sulfide, the formation of a soluble complex with CN extsuperscript{-} ions enhances the solubility of CdS. When complex ions form, they encapsulate the central metal ion (Cd extsuperscript{2+}) with ligands like the cyanide ion, hence stabilizing it in solution. This stabilization effectively increases the cadmium ion concentration that can remain dissolved, making it a successful move to resolve solubility issues for industrial and environmental processes.
  • They help in metal recovery processes.
  • Lead to greater solubility for precipitation reactions.
  • Used to selectively dissolve specific metal ions.
Soluble complexes are valuable tools in chemical industries and environmental engineering, providing solutions for dissolving and removing specific ions from waste or raw materials.

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Most popular questions from this chapter

The \(\mathrm{p} K_{\mathrm{a}}\) of phenolphthalein is \(9.10 .\) Over what \(\mathrm{pH}\) range does this indicator change from 95 percent HIn to 95 percent \(\mathrm{In}^{-} ?\)

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A student mixes \(50.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) with \(86.4 \mathrm{~mL}\) of \(0.494 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} .\) Calculate the mass of \(\mathrm{BaSO}_{4}\) formed and the \(\mathrm{pH}\) of the mixed solution.

Barium is a toxic substance that can seriously impair heart function. For an \(X\) ray of the gastrointestinal tract, a patient drinks an aqueous suspension of \(20 \mathrm{~g} \mathrm{BaSO}_{4}\). If this substance were to equilibrate with the \(5.0 \mathrm{~L}\) of the blood in the patient's body, what would be \(\left[\mathrm{Ba}^{2+}\right] ?\) For a good estimate, we may assume that the \(K_{\mathrm{sp}}\) of \(\mathrm{BaSO}_{4}\) at body temperature is the same as at \(25^{\circ} \mathrm{C}\). Why is \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) not chosen for this procedure?

\(\mathrm{CaSO}_{4}\left(K_{\mathrm{sp}}=2.4 \times 10^{-5}\right)\) has a larger \(K_{\mathrm{sp}}\) value than that of \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\left(K_{\mathrm{sp}}=1.4 \times 10^{-5}\right)\). Does it necessarily follow that \(\mathrm{CaSO}_{4}\) also has greater solubility \((\mathrm{g} / \mathrm{L}) ?\) Explain.

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