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In qualitative analysis \(\mathrm{NH}_{4} \mathrm{Cl}\) is added before \(\mathrm{NH}_{4} \mathrm{OH}\) (a) to increase \(\left[\mathrm{OH}^{-}\right]\)conc (b) for making \(\mathrm{HCl}\) (c) to decrease \(\left[\mathrm{OH}^{-}\right]\)conc (d) statement is wrong

Short Answer

Expert verified
(c) to decrease \( \left[\mathrm{OH}^{-} \right] \) conc.

Step by step solution

01

Interpretation of the Question

The question asks us why \( \mathrm{NH}_{4} \mathrm{Cl} \) is added before \( \mathrm{NH}_{4} \mathrm{OH} \) in qualitative analysis. We are given four statements and must determine which is true.
02

Understanding the Role of \( \mathrm{NH}_{4} \mathrm{Cl} \)

\( \mathrm{NH}_{4} \mathrm{Cl} \) dissociates into \( \mathrm{NH}_{4}^{+} \) and \( \mathrm{Cl}^{-} \) ions in solution. The presence of \( \mathrm{NH}_{4}^{+} \) ions can influence the concentration of \( \mathrm{OH}^{-} \) ions.
03

Analyzing the Reaction Environment

Adding \( \mathrm{NH}_{4} \mathrm{Cl} \) increases the \( \mathrm{NH}_{4}^{+} \) ion concentration. Since \( \mathrm{NH}_{4} \mathrm{OH} \) is a weak base, the addition of \( \mathrm{NH}_{4}^{+} \) from \( \mathrm{NH}_{4} \mathrm{Cl} \) will influence the equilibrium position of the \( \mathrm{NH}_{4} \mathrm{OH} \leftrightarrow \mathrm{NH}_{4}^{+} + \mathrm{OH}^{-} \) reaction, reducing the \( \mathrm{OH}^{-} \) concentration (common ion effect).
04

Conclusion

The addition of \( \mathrm{NH}_{4} \mathrm{Cl} \) is used to decrease the \( \left[ \mathrm{OH}^{-} \right] \) concentration in the solution due to the common ion effect of \( \mathrm{NH}_{4}^{+} \). Therefore, option (c) is the correct statement.

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Key Concepts

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

Common Ion Effect
The common ion effect is an important concept in chemistry that helps us understand how substances in a solution can influence each other. When a compound like ammonium chloride \( \mathrm{NH}_{4} \mathrm{Cl} \) is dissolved in water, it dissociates into its constituent ions, \( \mathrm{NH}_{4}^{+} \) and \( \mathrm{Cl}^{-} \). When these ions are present, they can affect the equilibrium of other related ions in the solution.
  • The presence of \( \mathrm{NH}_{4}^{+} \) in solution is the key player in the common ion effect in this scenario.
  • It works by shifting the equilibrium according to Le Chatelier's Principle.
  • This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change.
In this case, the addition of more \( \mathrm{NH}_{4}^{+} \) ions from \( \mathrm{NH}_{4} \mathrm{Cl} \) can reduce the formation of \( \mathrm{OH}^{-} \) ions by pushing the equilibrium of the weak base \( \mathrm{NH}_{4} \mathrm{OH} \) to favor the formation of more \( \mathrm{NH}_{4}^{+} \) ions. Therefore, the common ion effect plays a crucial role in controlling the concentration of ions in the solution.
Ammonium Chloride
Ammonium chloride (\( \mathrm{NH}_{4} \mathrm{Cl} \)) is a common compound used in laboratories, particularly in qualitative analysis. It is a white crystalline salt that is highly soluble in water, meaning it breaks apart easily to form ammonium ions (\( \mathrm{NH}_{4}^{+} \)) and chloride ions (\( \mathrm{Cl}^{-} \)). This solubility is key to its role in the common ion effect.
  • The introduction of ammonium chloride into a solution increases the \( \mathrm{NH}_{4}^{+} \) ions concentration.
  • This, in turn, affects reactions involving \( \mathrm{OH}^{-} \) ions, especially when working with weak bases like \( \mathrm{NH}_{4} \mathrm{OH} \).
Ammonium chloride is often added to solutions to create a controlled environment where we can predict and manage reactions better. It essentially helps in the suppression of base ionization due to the presence of common ions, making it a valuable tool for chemists who need precise control over the chemical environment.
Hydroxide Ion Concentration
The hydroxide ion concentration (\( \left[ \mathrm{OH}^{-} \right] \)) in a solution is an important factor in determining its basicity. In many chemical reactions, it's crucial to control this concentration for desired outcomes.
  • A common scenario is in the presence of a weak base like \( \mathrm{NH}_{4} \mathrm{OH} \), where the ability to maintain a low \( \mathrm{OH}^{-} \) concentration can be beneficial.
  • Adding an abundant amount of \( \mathrm{NH}_{4}^{+} \) ions, through compounds like ammonium chloride, effectively lowers the \( \mathrm{OH}^{-} \) concentration.
  • This decrease happens because the system adjusts to maintain equilibrium, favoring the formation of more \( \mathrm{NH}_{4}^{+} \) and less \( \mathrm{OH}^{-} \).
By understanding and manipulating the hydroxide ion concentration, chemists can control the acidity or basicity of the solution they are working with, which is crucial for various chemical processes and analytical methods.

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Most popular questions from this chapter

Which of the following salts will form colourless beads in borax bead test? (a) \(\mathrm{CoCl}_{2}\) (b) \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) (c) \(\mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (d) \(\mathrm{CoC}_{2} \mathrm{O}_{4}\)

\(\mathrm{FeCl}_{3}\) is acidic towards litmus. On treatment with excess of \(\mathrm{NH}_{4} \mathrm{SCN}\) it gives red coloured compound (A) and on treatment with excess of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) in the presence of conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\), it evolves deep red vapours of on passing the vapours of (B) into \(\mathrm{NaOH}\), then adding a solution of acetic acid and lead acetate it gives yellow precipitate of compound of chromium (C). The compound (A) is (a) \(\mathrm{NH}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{4}\right]\) (b) \(\mathrm{Fe}(\mathrm{SCN})_{2}\) (c) \(\mathrm{FeCl}_{3}\) (d) \(\mathrm{Fe}(\mathrm{SCN})_{3}\)

Borax \(\left[\mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7} .10 \mathrm{H}_{2} \mathrm{O}\right]\) when heated on platinum loop it gives a dark transparent glass like bead. The hot bead is dipped in the salt till it reacts with transition metal oxide. It produces characteristic bead of meta borate. $$ \begin{array}{ll} \text { Colour of the bead } & \text { Ion } \\ \text { Blue green or light blue } & \mathrm{Cu}^{2+} \\ \text { Yellow } & \mathrm{Fe}^{2+} \text { or } \mathrm{Fe}^{3+} \\ \text { Green } & \mathrm{Cr}^{3+} \\ \text { Violet } & \mathrm{Mn}^{2+} \\ \text { Dark blue } & \mathrm{Co}^{2+} \\ \text { Brown } & \mathrm{Ni}^{2+} \end{array} $$ Glassy bead is of (a) \(\mathrm{Na}_{3} \mathrm{BO}_{3}\) (b) \(\mathrm{B}_{2} \mathrm{O}_{3}+\mathrm{NaBO}_{2}\) (c) \(\mathrm{SiO}_{2}\) (d) \(\mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7}\)

In the separation of \(\mathrm{Cu}^{2+}\) and \(\mathrm{Cd}^{2+}\) in 2 nd group qualitative analysis of cations, tetrammine copper (II) sulphate and tetrammine cadmium(II) sulphate react with \(\mathrm{KCN}\) to form the corresponding cyano complexes. Which one of the following pairs of the complexes and their relative stability enables the separation of \(\mathrm{Cu}^{2+}\) and \(\mathrm{Cd}^{2+} ?\) (a) \(\mathrm{K}_{2}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]\) less stable and \(\mathrm{K}_{2}\left[\mathrm{Cd}(\mathrm{CN})_{4}\right]\) more stable (b) \(\mathrm{K}_{3}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]\) more stable and \(\mathrm{K}_{2}\left[\mathrm{Cd}(\mathrm{CN})_{4}\right]\) less stable (c) \(\mathrm{K}_{3}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]\) less stable and \(\mathrm{K}_{2}\left[\mathrm{Cd}(\mathrm{CN})_{4}\right]\) more stable (d) \(\mathrm{K}_{2}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]\) more stable and \(\mathrm{K}_{2}\left[\mathrm{Cd}(\mathrm{CN})_{4}\right]\) less stable

Nitrates of all metals are (a) unstable (b) coloured (c) insoluble in water (d) soluble in water

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