Question

Which one among the following pairs of ions can not be separated by \(\mathrm{H}_{2} \mathrm{~S}\) in dilute hydrochloric acid? (a) \(\mathrm{Bi}^{3+}, \mathrm{Sn}^{4+}\) (b) \(\mathrm{Al}^{3+}, \mathrm{Hg}^{2+}\) (c) \(\mathrm{Zn}^{2+}, \mathrm{Cu}^{2+}\) (d) \(\mathrm{Ni}^{2+}, \mathrm{Cu}^{2+}\)

Short answer

The pair (Ni^{2+}, Cu^{2+}) cannot be separated by  extrm{H}_2 extrm{S} in dilute HCl.

Step-by-step solution

Understanding the Precipitation Process

In qualitative inorganic analysis, hydrogen sulfide ( extrm{H}_2 extrm{S}) is often used to precipitate metal ions as sulfides. However, the ability of these ions to form sulfides depends on the acidity of the solution. In dilute hydrochloric acid ( extrm{HCl}), the concentration of  extrm{H}^+ ions influences which metal ions will precipitate as sulfides.

Evaluating Ion Separations in Dilute HCl

In a dilute  extrm{HCl} solution, only certain metal ions will form insoluble sulfides with  extrm{H}_2 extrm{S} gas. Generally,  extrm{Cu}^{2+},  extrm{Bi}^{3+},  extrm{Sn}^{4+}, and  extrm{Hg}^{2+} form insoluble sulfides under these conditions, whereas ions like  extrm{Al}^{3+},  extrm{Zn}^{2+}, and  extrm{Ni}^{2+} do not form sulfides as readily because they require more alkaline conditions.

Determine the Non-Separable Ion Pair

Under dilute  extrm{HCl} conditions,  extrm{Cu}^{2+} forms an insoluble sulfide, while  extrm{Ni}^{2+} does not fully precipitate. Therefore, the pair ( extrm{Ni}^{2+}, extrm{Cu}^{2+}) cannot be effectively separated using  extrm{H}_2 extrm{S} in dilute  extrm{HCl}. Other ion pairs such as ( extrm{Bi}^{3+}, extrm{Sn}^{4+}) and ( extrm{Zn}^{2+}, extrm{Cu}^{2+}) will form distinct sulfides that allow for separation.

Key concepts

Metal Ion Precipitation

Metal ion precipitation is a fundamental concept in qualitative inorganic analysis. Precipitation involves the formation of an insoluble solid from the reaction of ions in a solution. This process is often utilized to separate or identify metal ions based on their distinct chemical properties. By using specific reagents, such as hydrogen sulfide (H\(_2\)S), chemists can precipitate metal ions as insoluble sulfides. Not all metal ions precipitate equally.
Some ions, like copper (Cu\(^{2+}\)), bismuth (Bi\(^{3+}\)), or mercury (Hg\(^{2+}\)), readily form insoluble sulfides. This characteristic allows them to be separated from ions that don't precipitate as easily under similar conditions. The selective precipitation of sulfide ions plays a crucial role in analyzing complex mixtures of metals.

Hydrogen Sulfide Precipitation

Hydrogen sulfide precipitation is a technique used to separate metal ions by forming specific sulfide compounds. In a chemical reaction involving hydrogen sulfide (H\(_2\)S) gas and a solution containing metal ions, sulfur combines with the metals to create insoluble sulfides. These sulfides then precipitate out of the solution, allowing for distinct separation and analysis.
Different metal ions require varying conditions to precipitate as sulfides. Metal ions such as Cu\(^{2+}\) and Bi\(^{3+}\) react readily with H\(_2\)S to form insoluble sulfides, while others, like Zn\(^{2+}\) and Ni\(^{2+}\), remain in solution. Understanding these reactions helps chemists identify and separate ions in mixed solutions.

Acidity Influence on Precipitation

The acidity of a solution heavily influences precipitation reactions. In qualitative analysis, the concentration of hydrogen ions (H\(^+\)) affects the solubility of metal sulfides. A more acidic environment, like that created by dilute hydrochloric acid (HCl), suppresses the formation of sulfides for some metal ions, while promoting it for others.
In general, metal ions that form stable complexes with water or require a higher pH to precipitate will not do so in acidic conditions. For instance, in a dilute HCl solution, Cu\(^{2+}\) readily forms an insoluble sulfide with H\(_2\)S, while Ni\(^{2+}\) does not precipitate effectively. This allows chemists to manipulate acidity and achieve selective precipitation for separation and analysis.

Separation of Metal Ions

Separating metal ions from each other is a critical step in qualitative inorganic analysis. By taking advantage of the different chemical properties of ions, chemists use various methods to isolate specific ions from a mixture. Precipitation with reagents such as H\(_2\)S enables the formation of distinct compounds that are insoluble in the solution.
In practice, the separation process can involve several steps, each requiring precision in controlling conditions such as pH and reagent concentration. Metal ions like Bi\(^{3+}\) and Sn\(^{4+}\) can be efficiently separated because they form different sulfides easily. However, some pairs, such as Ni\(^{2+}\) and Cu\(^{2+}\), pose challenges due to their similar responses under given conditions.

Chemical Separations in Solution

Chemical separations in solution rely on chemical reactions that manipulate ionic solubility and reactivity. This process is essential in separating and identifying elements in complex solutions. By altering solution conditions, such as acidity, temperature, and reagent concentrations, chemists can selectively precipitate or keep ions in solution.
Understanding the solubility principles and reactivity of ions assists in designing procedures to achieve successful separation. In the context of hydrogen sulfide, the ability of some ions to precipitate in dilute HCl while others do not is a quintessential example of using chemical separations strategically. These techniques form the backbone of many analytical methods in chemistry, enabling accurate identification of components within a solution.